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Found 57 results

  1. Hello everyone, I am trying to submit a comment in a comment box and send it to the DB but is not happening. The connection is good as I am logging in and all but no data is sent to the DB when I post the comment. It doesn't show in my comment section either. Form <!--comment section--> <?php if(isset($_SESSION['id'])) { echo "<form method='POST' action='" . setComments($conn) . "'> <input type='hidden' name='uidUsers' value='".$_SESSION['id']."'> <input type='hidden' name='posted' value='" . date('Y-m-d H:i:s') . "'> Comments: <textarea rows = '5' cols = '15' name='body'></textarea><br><br> <button name='commentSubmit' type='submit'>Comment</button> </form>"; }else { echo "Log in to comment!"; } getComments($conn); Function to set and get comments function setComments($conn) { if (isset($_POST['commentSubmit'])){ $user_id = $_POST['uidUsers']; $body = $_POST['body']; $posted = $_POST['posted']; $sql = "INSERT INTO comments (uidUsers, posted, body) VALUES ('$user_id', '$posted', '$body')"; $result = mysqli_query($conn, $sql); } } function getComments($conn) { $sql = "SELECT * FROM comments"; $result = mysqli_query($conn, $sql); while ($row = $result->fetch_assoc()){ $id = $row['uidUsers']; $sql2 ="SELECT * FROM users WHERE uidUsers='$id'"; $result2 = mysqli_query($conn, $sql2); if($row2 = $result2->fetch_assoc()){ echo "<div class='comment-box'><p>"; echo $row2['uidUsers'] . "<br>"; echo $row['posted'] . "<br>"; echo nl2br($row['body']); echo "</p></div>"; } } }
  2. I'm being given an error by mysql, "Warning: mysqli_connect() expects parameter 1 to be string, object given in /storage/...", the relevant lines are $code = $this->GenerateCode($num); $enCoded = password_hash($code, PASSWORD_BCRYPT); $query = "INSERT INTO `vaults` " . "(`Alive`, `Contents`, `Code`, `digits`, `Winning_Player_ID`, `Won_Date`, `debug`)" . "VALUES " . "(CONV('1', 2, 10) + 0, '0', '{$enCoded}', '$num', NULL, NULL, '{$code}')"; Followed by the mysqli_connect($this->db, $query).
  3. Hi, I am making a CMS and I can't get the search page to work right. The CMS is for a new local DJI store and I can't get anything to show up when I search for those who have paid for insurance on their drones. And when I enter a first or last name that multiple people have in common, like Mike or Smith, the info doesn't display correctly, The first customer is displayed correctly, but no one else is. searchcodefirst.php searchcodeinsurance.php searchcodelast.php
  4. Hi everyone, I'm having troubles with query from database. I want to select all from one table and select sum from second table in the same time. Is this possible? Here is my script: $conn = mysqli_connect('localhost', 'root', '', 'test'); if (!mysqli_set_charset($conn, "utf8")) { printf("Error loading character set utf8: %s\n", mysqli_error($conn)); } $sql = "SELECT first_id, first_name FROM first"; $sql = "SELECT sum(total) as SumTotal FROM second"; $result = mysqli_query($conn, $sql); while ($data = mysqli_fetch_array($result)) { echo '<tr>'; echo '<th>'.$data['first_id'].'</th>'; echo '<th>'.$data['first_name'].'</th>'; echo '<th>'.$data['SumTotal'].'</th>'; echo '</tr>'; } mysqli_close($conn); Thanks!
  5. If anyone could help, I'm not getting an error message and I've tried changing the code around a few times and outputting errors. The database does not update and i'm getting no error message output on submit, I assume i'm missing something? <?php session_start(); include_once 'dbconnect.php'; if (!isset($_SESSION['userSession'])) { header("Location: index.php"); } $query = $DBcon->query("SELECT * FROM tbl_users WHERE user_id=".$_SESSION['userSession']); $userRow=$query->fetch_array(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Welcome - <?php echo $userRow['email']; ?></title> <link href="bootstrap/css/bootstrap.min.css" rel="stylesheet" media="screen"> <link href="bootstrap/css/bootstrap-theme.min.css" rel="stylesheet" media="screen"> <link rel="stylesheet" href="style.css" type="text/css" /> </head> <?php include_once 'header.php'; ?> <html> <head></head> <body> <h1>Send Message:</h1> <form action='message.php' method='POST'> <table> <tbody> <tr> <td>To: </td><td><input type='text' name='to' /></td> </tr> <tr> <td>From: </td><td><input type='text' name='from' /></td> </tr> <tr> <td>Message: </td><td><input type='text' name='message' /></td> </tr> <tr> <td></td><td><input type='submit' value='Create Task' name='sendMessage' /></td> </tr> </tbody> </table> </form> </body> </html> <?php if (isSet($_POST['sendMessage'])) { if (isSet($_POST['to']) && $_POST['to'] != '' && isSet($_POST['from']) && $_POST['from'] != '' && isSet($_POST['message']) && $_POST['message'] != '') { $to = $_POST['to']; $from = $userRow['email']; $message = $_POST['message']; $q = "INSERT INTO tbl_messages(id,message,to,from) VALUES('', '$message', '$to', '$from')"; if ($DBcon->query($q)) { $msg = "<div class='alert alert-success'> <span class='glyphicon glyphicon-info-sign'></span> Message Sent ! </div>"; }else { $msg = "<div class='alert alert-danger'> <span class='glyphicon glyphicon-info-sign'></span> Message not Sent! ! </div>"; } } } $DBcon->close();?> <?php php include_once 'footer.php'; ?> </html>
  6. Hi! I'm new at php and I need a help with this code I need to Search an employee from a Database that I already have I have the Search Form and I write the name and the lastname and the details of the employee appear, But I need help. I already have the search, but I tells me when I call the query that I have a fatal error Trying to get property of non-object in in this part of the code $count = $query -> num_rows;and this Fatal error: Call to undefined method EmployeeModel::searchEmployees() in here $employees = $this->model->searchE($_GET['str']); require_once('views/search.php'); where I need to call this function in search.php I'm gonna leave my search.php I need a big help here I'm gonna be really thankfull <form method="get" action="index.php" name="searchform" id="searchform"> <input type="text" name="str" id="str"> <input type="submit" name="submit" id="submit" value="Search"> </form> <?php /*class SearchE {*/ $user = "root"; $password = ""; $host = "localhost"; $dbase = "employees_assign"; $table = "tbl_employees"; $search_term= isset($_GET['str']) ? $_GET['str'] : ''; /*mysqli_connect($host,$user,$password); @mysqli_select_db($dbase) or die("Unable to select database");*/ $connect = mysqli_connect('localhost','root','','employees_assign'); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql = "SELECT * FROM tbl_employees WHERE emp_fname LIKE '%$search_term%' OR emp_lnames LIKE '%$search_term%'"; //$query = mysqli_query($connect,$sql); $query = $connect ->query($sql); //$count= mysqli_num_rows(mysqli_result $query); $count = $query -> num_rows; //$count=$query->num_rows; /*if ($query === false) { error_log($connect->error); die("something failed."); }*/ if ($count == 0) { echo "<fieldset><b>No Results Found for Search Query '$search_term'</b></fieldset>"; } else { print "<table border=1>\n"; while ($row = mysqli_fetch_array($query)){ $emp_fname= $row['emp_fname']; $emp_lname= $row['emp_lname']; print "<tr>\n"; print "</td>\n"; print "\t<td>\n"; print "<font face=arial size=4/><div align=center>$emp_fname</div></font>"; print "</td>\n"; print "\t<td>\n"; echo "<font face=arial size=4/>$emp_lname</font>"; print "</td>\n"; print "</tr>\n"; } print "</table>\n"; } /*}*/ ?>
  7. Hello guys. I got a problem that whenever you register on my page, the registration is successful, no errors, successful redirection to login page, but the registration does not write the information into database and I have no idea why... I'm sure that i'm connecting correctly, to the correct table, with the correct commands, but it kinda does not work... BTW (This registration and login and all worked a few weeks ago, but I got an sudden internal server error, so I had to delete and reupload all files, and I had to change database. I changed the database, created the same table with the same columns, also I overwrote ALL old database information to the new (password, dbname,name) and, so page works fine, but that registration does not...I'm including my code for registration and registration form) Registration process CODE: <?php include_once 'db_connect.php'; include_once 'psl-config.php'; $error_msg = ""; if (isset($_POST['username'], $_POST['email'], $_POST['p'])) { // Sanitize and validate the data passed in $username = filter_input(INPUT_POST, 'username', FILTER_SANITIZE_STRING); $email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_EMAIL); $email = filter_var($email, FILTER_VALIDATE_EMAIL); if (!filter_var($email, FILTER_VALIDATE_EMAIL)) { // Not a valid email $error_msg .= '<p class="error">The email address you entered is not valid</p>'; } $password = filter_input(INPUT_POST, 'p', FILTER_SANITIZE_STRING); // Username validity and password validity have been checked client side. // This should should be adequate as nobody gains any advantage from // breaking these rules. // $prep_stmt = "SELECT id FROM members WHERE email = ? LIMIT 1"; $stmt = $mysqli->prepare($prep_stmt); // check existing email if ($stmt) { $stmt->bind_param('s', $email); $stmt->execute(); $stmt->store_result(); if ($stmt->num_rows == 1) { $error_msg .= '<p class="error">A user with this email address already exists.</p>'; } $stmt->close(); } // check existing username $prep_stmt = "SELECT id FROM members WHERE username = ? LIMIT 1"; $stmt = $mysqli->prepare($prep_stmt); if ($stmt) { $stmt->bind_param('s', $username); $stmt->execute(); $stmt->store_result(); if ($stmt->num_rows == 1) { $error_msg .= '<p class="error">A user with this username already exists.</p>'; } $stmt->close(); } // TODO: // We'll also have to account for the situation where the user doesn't have // rights to do registration, by checking what type of user is attempting to // perform the operation. if (empty($error_msg)) { // Create salted password $passwordHash = password_hash($password, PASSWORD_BCRYPT); // Insert the new user into the database if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password) VALUES (?, ?, ?)")) { $insert_stmt->bind_param('sss', $username, $email, $passwordHash); // Execute the prepared query. if (! $insert_stmt->execute()) { header('Location: ../error.php?err=Registration failure: INSERT'); } } header('Location: ./continue.php'); } } and Registration form : <div class="register-form"> <center><h2>Registration</h2></center> <form action="<?php echo esc_url($_SERVER['PHP_SELF']); ?>" method="post" name="registration_form"> <center><p></p><input type='text' name='username' placeholder="Username" id='username' /><br></center> <center><p></p><input type="text" name="email" id="email" placeholder="Email" /><br></center> <center><p></p><input type="password" name="password" placeholder="Insert Password" id="password"/><br></center> <center><p></p><input type="password" name="confirmpwd" placeholder="Repeat Password" id="confirmpwd" /><br></center> <center><p></p><input type="submit" class="button" value="Register" onclick="return regformhash(this.form, this.form.username, this.form.email, this.form.password, this.form.confirmpwd);" /> </center> </form> </div> Anybody know where is problem?
  8. Im having trouble on my php script. it is working on my computer but when I put it in x10hosting it fails and gives an error 500. I tried tracing the problem and I found out that it happens if I call get_result. Here is the part code: $username = strtolower(filter_input(INPUT_POST, 'username')); $password = filter_input(INPUT_POST, "password"); $remember = filter_input(INPUT_POST, "remember"); $result = array(); include 'Connection.php'; $query = "SELECT Number, Username, Password, Alias, Level FROM user WHERE Username = ?;"; $stmt = $conn->prepare($query); if(!$stmt->prepare($query)) { die( "Failed to prepare statement."); } $stmt->bind_param("s", $username); $stmt->execute(); echo $stmt->error; //error hapens here $selectResult = $stmt->get_result();
  9. Not sure what is going on I tried everything (well, that I could think of) . . . any ideas are welcome (hopefully new ones - getting frustrated :/) if ($mysqli->prepare("INSERT INTO solcontest_entries (title, image,content, user, contest) VALUES ($title, $image, $content, $userid, $contest")) { $stmt2 = $mysqli->prepare("INSERT INTO `solcontest_entries` (title, image, content, user, contest) VALUES (?, ?, ?, ?, ?)"); $stmt2->bind_param('sssss', $title, $image, $content, $userid, $contest); $stmt2->execute(); $stmt2->store_result(); $stmt2->fetch(); $stmt2->close(); } else { die(mysqli_error($mysqli)); } Error I get from die mysqli_error: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' of the site's lead ad, user_61609201, contest_1' at line 1" I have also tried $mysqli->query no change occured. I added the "if else die" statement because it was giving no errors, but not adding it to the database. It gives the error where $content is supposed to be inserted. Various combos and singles I tried for the variable: //$content = cleansafely($_POST['content']); //$content = mysqli_real_escape_string ($mysqli, $_POST['content']); //$content = cleansafely($content); $content = $_POST['content']; If any more information is needed please let me know.
  10. Hi guys, I am trying to display data to a user who submitted it. So, user A submits data to the db and also user B. I want it so that user A cannot see user B's data but only the data they have submitted. Hope that makes sense. Here is the code along with the table of data to be displayed. <div class="panel-body"> <div class="table-responsive"> <table class="table table-striped table-bordered table-hover" id="bookings_table"> <thead> <tr> <th class="text-center">Issue ID</th> <th class="text-center">Driver Name</th> <th class="text-center">Date Submitted</th> <th class="text-center">Fleet Number</th> <th class="text-center">Issue</th> <th class="text-center">Description</th> <th class="text-center">Priority</th> <th class="text-center">Status</th> </tr> </thead> <tbody> <?php $check = isset($_SESSION['username']); if($stmt = $link -> prepare("SELECT issue_id, driver_name, submit_date, fleet_number, issue_name, issue_description, issue_priority, issue_status FROM maintenance_requests WHERE username = ?")) { $stmt -> execute(); $stmt -> bind_result($issue_id, $driver_name, $submit_date, $fleet_number, $issue_name, $issue_description, $issue_priority, $issue_status); while($stmt->fetch()) { ?> <tr class="odd gradeX"> <td class="text-center"><?php echo $issue_id; ?></td> <td class="text-center"><?php echo $driver_name; ?></td> <td class="text-center"><?php echo $submit_date; ?></td> <td class="text-center"><?php echo $fleet_number; ?></td> <td class="text-center"><?php echo $issue_name; ?></td> <td class="text-center"><?php echo $issue_description; ?></td> <td class="text-center"><?php echo $issue_priority; ?></td> <?php if($issue_status == "Pending") { ?> <td class="text-center warning"><?php echo $issue_status;?></td> <?php }else if($issue_status == "Open"){ ?> <td class="text-center danger"><?php echo $issue_status; ?></td> <?php }else if($issue_status == "Repaired"){ ?> <td class="text-center success"><?php echo $issue_status ;} ?></td> </tr> <?php } $stmt -> close(); } mysqli_close($link); ?> If I remove the WHERE clause, it displays data from both users. Placing the WHERE clause shows no data. Can someone have a look and see where I am going wrong and maybe point me in the right direction? Many thanks in advance.
  11. $results = mysql_query("$select", $link_id); while ($query_data = mysql_fetch_row($results)) { $link_id = mysqli_connect("$db_host","$db_user","$db_password","$db_database"); if ($mysqli->connect_errno) { echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; } $link_id = mysql_connect("$db_host","$db_user","$db_password"); if (mysql_select_db("$db_database", $link_id)); else { echo "connection failed."; } I am having to update MySQL to mysqli as my host is upgrading to PHP5.6 I have managed to convert and connect to the database converted to I cannot get the fetch results to work, can anyone help me convert the following code Many Thanks
  12. I am updating all my code from mysql to mysqli. Currently using PHP 5.4 but will update to 5.5 once all this updating is done. Anyway, I have this old function for making data safe for inserting into mysql database. I changed all instances of "mysql" to "mysqli"... function mysqli_prep($value) { $magic_quotes_active = get_magic_quotes_gpc(); $new_enough_php = function_exists("mysqli_real_escape_string") ; //i.e. PHP >= v4.3.0 if($new_enough_php) { //PHP v4.3.0 or higher //undo any magic quote effects so mysqli_real_escape_string can do the work if($magic_quotes_active) { $value = stripslashes($value) ;} $value = mysqli_real_escape_string($connection, $value); } else { //before php v4.3.0 // if magic quotes aren;t already on then add slashes manually if(!magic_quotes_active) { $value = addslashes($value); } // if magic quotes are active, then the slashes already exist } return $value; } When I load that page that calls this function, I get... Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in (mypath) This is my $connection by the way, which works fine on other pages that need it... $connection = mysqli_connect('localhost', 'myusername', 'mypassword', 'mytable'); if (!$connection) { die("database connection failed: " . mysqli_error()); } Any ideas what I'm doing wrong?
  13. I have the following general mySQL query: $query = sprintf( "SELECT a.A, a.B, ..., a.C, FROM a WHERE a.A = %s ORDER by a.A;", implode(" OR a.A = ", $_SESSION['values'])); for some records, B has a value, and for others, B is NULL. That is what I want. I extract the query with the following PHP: for ($i = 0; $i < $nrows; $i++){ $row = mysqli_fetch_assoc($result);//fetches data stored within each row extract($row); echo "<tr>"; foreach($row as $column => $field){ if($field == $...){ ... } elseif($field == $C){ echo"<td> <input type='text' name='C+[]' value='$C'> </td>"; } echo "</tr>" } In the resulting html table, records containing not null B fields are presented accurately, while records with null B fields incur a duplication of field C. This offset occurs at the beginning of the record, pushing the final field in the record outside the table boundaries. I think I've narrowed down the problem to the extract() function, as r_print readouts of every other variable in the script returns the accurate field names and values. But, running print_r on $row after extract() provides an identical printout to other variables in the script. What are some possible ways I can stop the duplication of field C from occurring? Happy to provide more information upon request.
  14. Me again.. I've struggled for the past 2 hours to insert article comments and link them to an existent article on the page. Now, the function that is displaying both comments and articles looks like this: function list_articles() { include('core/db/db_connection.php'); $sql = "SELECT blog.content_id, blog.title, blog.content, blog.posted_by, blog.date, article_comments.comments, article_comments.comment_by FROM blog LEFT OUTER JOIN article_comments ON blog.content_id = article_comments.blog_id WHERE blog.content != '' ORDER BY blog.content_id DESC"; $result = mysqli_query($dbCon, $sql); $previous_blog_id = 0; while ($row = mysqli_fetch_array($result)) { if ($previous_blog_id != $row['content_id']) { echo "<h5 class='posted_by'>Posted by {$row['posted_by']} on {$row['date']}</h5> <h1 class='content_headers'>{$row['title']}</h1> <article>{$row['content']}</article> <hr class='artline'>"; $previous_blog_id = $row['content_id']; } if (!empty($row['comment_by']) && !empty($row['comments'])) { echo "<div class='commented_by'>Posted by: {$row['comment_by']} </div> <div class='comments'>Comments: {$row['comments']}</div> <hr class='artline2'>"; } } } The function I'm running to insert comments into article_comments table function insert_comments($comments, $comment_by, $blog_id) { include('core/db/db_connection.php'); $comment_by = sanitize($comment_by); $comments = sanitize($comments); $sql = "INSERT INTO article_comments (comments, comment_by, blog_id) VALUES ('$comments', '$comment_by', '$blog_id')"; mysqli_query($dbCon, $sql); } This works - it does the insertion, however I have no clue on how I could target the $blog_id variable when the user submits the post... The below is the form I use <?php echo list_articles(); if (!empty($_POST)) { insert_comments($_POST['comments'], $_POST['username'], 11); } ?> <form method='post' action='' class='comments_form'> <input type='text' name='username' placeholder='your name... *' id='name'> <textarea name='comments' id='textarea' placeholder='your comment... *' cols='30' rows='6'></textarea> <input type='submit' name='submit' id='post' value='post'> </form> I bet you noticed that I've manually inserted 11 as a param for the last variable. This links to blog_id 11 (the foreign key) in my article_comments table. It is displaying the comment just fine. Is there any way to target $blog_id without having to insert a number manually? Something like how I am targeting the $comments variable using $_POST['comments'] ? Also, even if I can target that, how do I know which post is the user commenting to? Should I give them the option to choose in a drop-down list ? That seems awkward.. but it's the only solution I can think of.
  15. Hi All, I get the following message when running a multiline query through mysqli. Here is my simplified code : $sql='SELECT CURRENT_USER();'; $sql.='SELECT CURRENT_time();'; $rs = @mysqli_multi_query($link,$sql); if (!$rs) { } else { do { if ($rs = mysqli_store_result($link)) { while ($row = mysqli_fetch_row($rs)) { echo "<br/> ". $row[0]; } mysqli_free_result($rs); } if (mysqli_more_results($link)) { echo "<br/>-----------------<br/>"; } } while (mysqli_next_result($link)); } If I use "while (mysqli_next_result($link) && mysqli_more_results($link));" instead of "while (mysqli_next_result($link))", I get no error message, but the current time (returned by the second query) won't display. Thanks for your help!
  16. I am trying to add a query to my script that updates a value in my database by subtracting "1". When I run the query, I get "Fatal error: Call to a member function free() on boolean in ../path/to/my/script" $sql = "UPDATE table_5 SET chairs = chairs - 1 WHERE chairs > 0 AND chair_model = 'model_33'"; Any idea what I'm doing wrong? Table 5: Chairs | Model Number | 22 | model_33 44 | model_44
  17. Good Day PHP world, I am encountering a problem in php code meant to allow the user to update their profile picture. I am using jquery.min and jquery.js. The code below runs with no errors reported. The file has been successfully uploaded to upload path using this form. upload.php <form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'> <input type="file" name="photoimg" id="photoimg" class="stylesmall"/> </form> ajaximage.php $path = "uploads/"; $valid_formats = array("jpg", "png", "gif", "bmp","jpeg"); if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST") { $name = $_FILES['photoimg']['name']; $size = $_FILES['photoimg']['size']; if(strlen($name)) { list($txt, $ext) = explode(".", $name); if(in_array($ext,$valid_formats)) { if($size<(1024*1024)) // Image size max 1 MB { $actual_image_name = $name.".".$ext; $tmp = $_FILES['photoimg']['tmp_name']; if(move_uploaded_file($tmp, $path.$actual_image_name)) { $query = "UPDATE users SET profile_image='$actual_image_name' WHERE student_id='{$_SESSION['user_id']}'"; $result = mysqli_query($link_id, $query); echo "<img src='uploads/".$actual_image_name."' class='preview'>"; } The problem is the image being uploaded does not display on the Student_home.php <div id="about-img"> <img class="profile-photo" align="middle" src='uploads/".$actual_image_name."' /> </div> But the image uploaded will display when i write directly its filename example <div id="about-img"> <img class="profile-photo" align="middle" src="uploads/107.jpg" /> </div> My problem is i wanted to display the uploaded picture of the specific student on Student_Home.php
  18. Hello my good people I'm doing a content manager system in PHP and mySQLi, following a serie of video tutorials. But in my project I am using a tree menu. In the front office all works smoothly (the tree menu displays all results - menu and submenus - and each button carries the information of the respective page). The problem is in the back office. The menu is there, in the url appears its id page but the CRUD, I can not put it to work. ======= CODE START ======= <!-- Gestor de Conteúdos Start --> <div class="container-fluid"> <!-- Conteúdo Fluido Start --> <div class="row-fluid row-offcanvas row-offcanvas-left"> <?php // INSERT QUERY if(isset($_POST['enviado']) == 1) { $header = mysqli_real_escape_string($dbc, $_POST['header']); $url = mysqli_real_escape_string($dbc, $_POST['url']); $user = $_POST['user']; $idPai = $_POST['idPai']; $menuNomePT = mysqli_real_escape_string($dbc, $_POST['menuNomePT']); $conteudo_pagina_PT = mysqli_real_escape_string($dbc, $_POST['conteudo_pagina_PT']); $menuNomeEN = mysqli_real_escape_string($dbc, $_POST['menuNomeEN']); $conteudo_pagina_EN = mysqli_real_escape_string($dbc, $_POST['conteudo_pagina_EN']); $q = "INSERT INTO menuCAL (header, url, user, idPai, menuNomePT, conteudo_pagina_PT, menuNomeEN, conteudo_pagina_EN) VALUES ('$header', '$_POST[url]', $_POST[user], '$_POST[idPai], '$menuNomePT, '$conteudo_pagina_PT', '$menuNomeEN', '$conteudo_pagina_EN'"; $r = mysqli_query($dbc, $q); ?> <div> <?php if($r) { $message = '<p>A página foi adicionada!</p>'; } else { $message = '<p>A página não foi adicionada, devido ao seguinte erro: '.mysqli_error($dbc); $message .= '<p>' .$q.'</p>'; } ?> </div> <?php } ?> <!-- Menu CAL Start --> <div class="col-sm-3 sidebar-offcanvas"> <?php //call the recursive function to print category listing category_tree(0); //Recursive php function function category_tree($menuPai){ global $dbc; $q = "SELECT * FROM menuCAL WHERE idPai ='".$menuPai."'"; $r = mysqli_query($dbc, $q); while($btnMenu = mysqli_fetch_assoc($r)): $i = 0; if ($i == 0) echo '<ul class="menuCAL">'; echo '<li><a href="?page='.$btnMenu['id'],'">' . $btnMenu['GlyPrincipal'] . $btnMenu['GlySecundario'] . $btnMenu['menuNomePT'], '</a>'; category_tree($btnMenu['id']); echo '</li>'; $i++; if ($i > 0) echo '</ul>'; endwhile; } ?> </div> <!-- Menu CAL End --> <!-- Conteúdo Start --> <div class="span10"> <div class="col-sm-12"> <!-- Título Start --> <h1 class="page-header"> <i class="fa fa-file"></i> Adicionar Página </h1> <ol class="breadcrumb"> <li> <a href="#"><i class="fa fa-pencil"></i> Conteúdos</a> </li> <li class="active"> <i class="fa fa-file"></i> Nova página </li> </ol> <!-- Título End --> </div> <!-- Textos & Formulários Start --> <div class="row"> <div class="col-lg-12"> <p><?php if(isset($message)) { echo $message; } ?></p> <?php // SELECT QUERY if(isset($_GET['id'])) { $q = "SELECT * FROM menuCAL WHERE id = $_GET[id]"; $r = mysqli_query($dbc, $q); $opened = mysqli_fetch_assoc($r); } ?> <!-- Formulário Start --> <form action="adicionar_pagina.php" method="post" role="form"> <!-- Campo header Start --> <div class="form-group"> <label for="header">Header:</label> <input type="text" class="form-control" name="header" id="header" value="<?php echo $opened['header']; ?>" placeholder="Texto descritivo a colocar no topo do website"> </div> <!-- Campo header End --> <!-- Campo Label Start --> <div class="form-group"> <label for="url">URL:</label> <input type="text" class="form-control" name="url" id="url" value="<?php echo $opened['url']; ?>" placeholder="Texto a colocar na URL (SEO)"> </div> <!-- Campo Label End --> <!-- Campo User Start --> <div class="form-group"> <label for="user">Administrador:</label> <select class="form-control" name="user" id="user"> <option value="0">›› Nenhum administrador</option> <?php $q = "SELECT id FROM users ORDER BY first ASC"; $r = mysqli_query ($dbc, $q); while ($user_list = mysqli_fetch_assoc($r)) { $user_data = data_user($dbc, $user_list['id']); ?> <option value="<?php echo $user_data['id'] ?>" <?php if($user_data['id'] == $opened['id']) { echo 'selected';} ?>><?php echo $user_data['fullname']; ?></option> <?php } ?> </select> </div> <!-- Campo User Start --> <!-- Campo ID Menu Pai Start --> <div class="form-group"> <label for="idPai">Adicionar a:</label> <select class="form-control" name="idPai" id="idPai"> <option value="0">›› Seleccione onde adicionar a nova página:</option> <?php $q = "SELECT menuNomePT FROM menuCAL WHERE idPai = 0"; $r = mysqli_query ($dbc, $q); while ($submenus = mysqli_fetch_assoc($r)) { ?> <option value="<?php echo $submenus['idPai']; ?>"><?php echo $submenus['menuNomePT']; ?></option> <?php } ?> </select> </div> <!-- Campo ID Menu Pai Start --> <!-- Campo menuNomePT Start --> <div class="form-group"> <label for="menuNomePT">Título PT:</label> <input type="text" class="form-control" name="menuNomePT" id="menuNomePT" value="<?php echo $opened['menuNomePT']; ?>" placeholder="Insira o título em Português"> </div> <!-- Campo menuNomePT End --> <!-- Campo conteudo_pagina_PT Start --> <div class="form-group"> <label for="conteudo_pagina_EN">Conteúdos PT:</label> <textarea class="form-control" name="conteudo_pagina_PT" rows="12" id="conteudo_pagina_PT" placeholder="Insira os textos em Português"><?php echo $opened['conteudo_pagina_PT']; ?></textarea> </div> <!-- Campo conteudo_pagina_PT End --> <!-- Campo menuNomeEN Start --> <div class="form-group"> <label for="menuNomeEN">Título EN:</label> <input type="text" class="form-control" name="menuNomeEN" id="menuNomeEN" value="<?php echo $opened['menuNomePT']; ?>" placeholder="Insira o título em Inglês"> </div> <!-- Campo menuNomeEN End --> <!-- Campo conteudo_pagina_EN Start --> <div class="form-group"> <label for="conteudo_pagina_EN">Conteúdos EN:</label> <textarea class="form-control" name="conteudo_pagina_EN" rows="12" id="conteudo_pagina_EN" placeholder="Insira os textos em Inglês"><?php echo $opened['conteudo_pagina_PT']; ?></textarea> </div> <!-- Campo conteudo_pagina_EN End --> <button type="submit" class="btn btn-default adic_concluir">Gravar</button> <input type="hidden" name="enviado" value="1"> </div> </form> <!-- Formulário End --> </div> <!-- Debug Panel Start --> <?php if($debug == 1) { include('widgets/debug.php'); } ?> <!-- Debug Panel End --> </div> <!-- Textos & Formulários End --> </div> <!-- Conteúdo End --> </div> <!-- Conteúdo Fluido Start --> </div> <!-- Gestor de Conteúdos End --> ======= CODE END ======= I can not do the INSERT or UPDATE query, using the form so that the "user" and the "idPai" (a category id), can be created or changed in the database. I tried to echo the database result, but nothing append And in the second query (in red in the code), gives me an error (Undefined variable: opened in …). Obviously something is wrong, but i can't see what! :/ The database: Can Someone help me? Please !! Thank U
  19. Hi guys and Gals, This has my head wrecked to be honest. I am trying to upload an image to a directory, which is working. However, I also want to put the file name into MySQL. This will work if the image upload script is removed. With the script enabled, the file uploads but I get "Undefined index: userPic" from the following line: $userPic = mysqli_real_escape_string($mysqli, $_POST['userPic']); Here is the complete code: if(isset($_POST['Submit'])){//if the submit button is clicked $company_name = mysqli_real_escape_string($mysqli, $_POST['company_name']); $company_abn = mysqli_real_escape_string($mysqli, $_POST['company_abn']); $company_email = mysqli_real_escape_string($mysqli, $_POST['company_email']); $address = mysqli_real_escape_string($mysqli, $_POST['address']); $company_phone = mysqli_real_escape_string($mysqli, $_POST['company_phone']); $company_slogan = mysqli_real_escape_string($mysqli, $_POST['company_slogan']); $userPic = mysqli_real_escape_string($mysqli, $_POST['userPic']); // Upload Image if (isset($_FILES["userPic"]["name"])) { $name = $_FILES["userPic"]["name"]; $tmp_name = $_FILES['userPic']['tmp_name']; $error = $_FILES['userPic']['error']; if (!empty($name)) { $location = 'uploads/'; if (move_uploaded_file($tmp_name, $location.$name)){ echo 'Uploaded'; } } else { echo 'please choose a file'; } } $sql="UPDATE company_settings SET company_name='$company_name', company_slogan='$company_slogan', company_abn='$company_abn', company_email='$company_email', address='$address', company_phone='$company_phone', userPic='$userPic'"; $mysqli->query($sql) or die("Cannot update");//update or error } Has anyone got any ideas where I am going wrong (besides not using PDO) and how I can solve it? Thanks in advance.
  20. how i can make a insert using this fuctions I m learning php, as using this functions (mysqli abstract) but after update wont work any more. /** insert data array */ public function insert(array $arr) { if ($arr) { $q = $this->make_insert_query($arr); $return = $this->modifying_query($q); $this->autoreset(); return $return; } else { $this->autoreset(); return false; } } complement /** insert query constructor */ protected function make_insert_query($data) { $this->get_table_info(); $this->set_field_types(); if (!is_array(reset($data))) { $data = array($data); } $keys = array(); $values = array(); $keys_set = false; foreach ($data as $data_key => $data_item) { $values[$data_key] = array(); $fdata = $this->parse_field_names($data); foreach ($fdata as $key => $val) { if (!$keys_set) { if (isset($this->field_type[$key])) { $keys[] = '`' . $val['table'] . '`.`' . $val['field'] . '`'; } else { $keys[] = '`' . $val['field'] . '`'; } } $values[$data_key][] = $this->escape($val['value'], $this->is_noquotes($key), $this->field_type($key), $this->is_null($key), $this->is_bit($key)); } $keys_set = true; $values[$data_key] = '(' . implode(',', $values[$data_key]) . ')'; } $ignore = $this->ignore ? ' IGNORE' : ''; $delayed = $this->delayed ? ' DELAYED' : ''; $query = 'INSERT' . $ignore . $delayed . ' INTO `' . $this->table . '` (' . implode(',', $keys) . ') VALUES ' . implode(',', $values); return $query; } before update this class i used to insert data like this $db = Sdba::table('users'); $data = array('name'=>'adam'); $db->insert($data); this method of insert dont works on new class. if i try like this i got empty columns and empty values. thanks for any help complete class download http://goo.gl/GK3s4E
  21. I need urgent help! For two days I've been trying to find the source code to upload image to database and display, after trying all the codes I kept getting all sorts of errors. I'm using phpmyadmin. Fatal error: Call to undefined function finfo_open() in C:\xampp\htdocs\geology\file_insert.php on line 51 Second help needed is to display or restrict 3 latest posts only. I want to display only 3 latest posts sort by latest date posted. I'm not sure how to loop this through. Another thing is, when the user clicks on the title of a specific article, a new window will appear to display only the data that belong to it (Title, Publisher, Content...) For example, there are 3 articles (rock, mineral, salt) user clicks on Rocks, a new window appear showing all Rocks info. <?php $sql = "select * from article ORDER by article_postdate DESC"; $result = mysqli_query($conn, $sql); while($row = mysqli_fetch_assoc($result)) { ?> <h3><a href="#"><?php echo $row["article_title"]; ?></a></h3> <p class="byline"><span><?php echo $row["article_postdate"]; ?><br> Published By:<?php echo $row["author_id"]; ?></p></a></span></p> <p><?php echo $row["article_details"]; ?></p> <td><a href = "edit.php?article_id=<?php echo $row['article_id']; ?>">More...</a></td> <?php } ?>
  22. Hi, I can I include a date range criteria to query with in the following code? The date field in the table (t_persons) is IncidentDate. $criteria = array('FamilyName', 'FirstName', 'OtherNames', 'NRCNo', 'PassportNo', 'Gender', 'IncidenceCountryID', 'Status', 'OffenceKeyword', 'AgencyID', 'CountryID', 'IncidenceCountryID' ); $likes = ""; $url_criteria = ''; foreach ( $criteria AS $criterion ) { if ( ! empty($_POST[$criterion]) ) { $value = ($_POST[$criterion]); $likes .= " AND `$criterion` LIKE '%$value%'"; $url_criteria .= '&'.$criterion.'='.htmlentities($_POST[$criterion]); } elseif ( ! empty($_GET[$criterion]) ) { $value = mysql_real_escape_string($_GET[$criterion]); $likes .= " AND `$criterion` LIKE '%$value%'"; $url_criteria .= '&'.$criterion.'='.htmlentities($_GET[$criterion]); } //var_dump($likes); } $sql = "SELECT * FROM t_persons WHERE PersonID>0" . $likes . " ORDER BY PersonID DESC"; Kind regards.
  23. Hi I'm using php file to retrieve the amount of records in my database, but all of a sudden I get 0 results. It worked fine a few days ago. I got PHP 5.2.* on my host. Here is the current script that I use. Did I made a mistake somewhere? <?php $con=mysqli_connect("******","*****","*****","****"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql="SELECT * FROM mxit"; if ($result=mysqli_query($con,$sql)) { $rowcount=mysqli_num_rows($result); //printf("Result set has %d rows.\n",$rowcount); // Free result set mysqli_free_result($result); } mysqli_close($con); ?>
  24. I created a recipe site that displays various dishes, and I'm tickled pink that I got it to work (first time doing this!) But I'd like to have pages that displays dishes by its category. The column that houses this is called "category" which has 7 total: Appetizers & Beverages, Soups & Salads, Side Items, Main Dishes, Baked Goods, Desserts, and Cookies & Candy. I have this page connected to a "connection.php" page, but here's the code in question: <!-- Page Title--> <div class="row"> <div class="col-lg-12"> <h1 class="page-header">Appetizers & Beverages</h1> </div> </div> <!-- /Page Title --> <!-- Displayed Data--> <?php $sql = "SELECT id, category, bilde, title FROM oppskrift_table ORDER BY title "; $result = mysqli_query($con, $sql); if(mysqli_num_rows($result) > 0 ){ while($row = mysqli_fetch_assoc($result)){ ?> <div align="center" class="col-md-3"> <a href="#"> <img class="img-responsive img-cat" src="bilder/rBilder/<?=$row['bilde']?>" width="250" alt=""> </a> <h4 style="max-width:250px;"> <a href="#" class="dishes"> <?=$row['title']?> </a> </h4> </div> <?php } } ?> <!-- END Displayed Data--> </div> How can I tweak this so that only the category Appetizers & Beverages shows up on the page?
  25. Hi guys, I've been around here for a few years, but for some reason my other account doesn't seem to 'exist' anymore which was real annoying. I also noticed the captcha here was kind of buggy has anyone else been getting that? I'd enter it in case sensative 9-10 times before it would finally work. Anyways, I've been looking through a lot of research in upgrading my server from Mysql to Mysqli funtions. What I am curious about though is other peoples opinions and thoughts on how to make user input safer. For the time being I've just been using mysql_real_escape_string and htmlspecialchars. I've done quite a bit of research on this and there really isn't much for any guides on how to keep your data clean and safe. I've seen a lot of posts that anymore these two functions are not enough to secure your data. So I'm curious what people in this community are doing (annonomysly) to keep your user input safe. I'm also looking into prepared statements as well with Mysqli. Anyways any responses are much appreciated, would love to chat with you guys about this! Does anyone know if there was some deal with why I can't access my origional account? I entered in all of the only 5 email addresses I use. It said it sent an email to the one, but it never appeared in junk/inbox.
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