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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.

Is it possible to link column_id from table A with column_id from table B? For example: If column_id A has a value of 6, column_id B should not be able allow entries if the column_id B is more than the value of column id A. I am running the below function to extract data from these tables: function list_articles() { include('core/db/db_connection.php'); $sql = "SELECT 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.comment_id WHERE blog.content != '' ORDER BY blog.content_id DESC"; $result = mysqli_query($dbCon, $sql); while ($row = mysqli_fetch_array($result)) { echo "<h5 class='posted_by'>Posted by " . $posted_by = $row['posted_by'] . " on " . $row['date'] . "</h5>" . "<h1 class='content_headers'>" . $title = $row['title'] . "</h1>" . "<article>" . $content = $row['content'] . "</article>" . "<hr class='artline'>" . "<div class='commented_by'>" . $row['comment_by'] . "</div>" . "<div class='comments'>" . $row['comments'] . "</div>"; } } Thanks.

hey guys i have a few questions regarding my website that has the use of multiple databases....at the moment my site is ran off one mysqli connection and before executing a query i change the database depending on if its my authentication script, geoip, framework etc. what i'm worried about is performance issues....i could have a new connection for each database or continue to have one connection for the whole site and change database when needed...what is the best practice please? thank you.

Hello, I have inserted a user into my database table through phpMyAdmin using the predefined MD5 function. (I know md5 is not secure and I should use bcrypt istead, but I don't need that type of security, my only purpose is not to store the passwords in plain text) Now my problem is that whenever I try to log the user in, I can never read the hashed password back. This is my code: The function that is testing for the username and password: function login($username, $password) { include('core/db/db_connection.php'); $sql = "SELECT COUNT(user_id) FROM `_users` WHERE username = '$username' AND password = '$password'"; $query = mysqli_query($dbCon, $sql); $user_id = get_user_id($username); $username = sanitize($username); $password = md5($password); // issue return (mysqli_result($query, 0) == 1) ? $user_id : false; // possible issue } The logging processing code: if (empty($_POST) === false) { $username = $_POST['username']; $password = $_POST['password']; if (empty($username) === true || empty($password) === true) { $errors[] = 'Username and/or password fields must not be left blank'; } else if (user_exists($username) === false) { $errors[] = 'Username does not exist! Please register before logging in.'; } else if (user_active($username) === false) { $errors[] = 'You haven\'t activated your account yet'; } else { $login = login($username, $password); if ($login === false) { $errors[] = 'Username/password incorrect'; } else { echo 'ok' . '<br/>'; //set user session //redirect user } } print_r($errors); } How can I read the stored MD5 password to allow my registered users access? Many thanks.

I've written a script to store images in my database. The images have a caption that is also uploaded and stored. This was fairly easy to get working. I have a jquery function setup to add a new file input and caption input every time I click a button. This also works. What is not working is my PHP is not uploading multiple files for some reason. Could someone tell me what I have done wrong? Thanks HTML: <form id="uploadMultiple" method="post" action="/scripts/php/imageUploadTest" enctype="multipart/form-data"> <table class="fileUploadTable" id="uploadArea" cellspacing="0"> <tr> <td> <label for="imageFile">Image:</label> <input type="file" name="imageFile" accept=".jpg,.jpeg,.png,.gif"> </td> <td> <label for="imageCaption">Image Caption:</label> <input type="text" name="imageCaption"> </td> <td width="150px"> </td> </tr> <tr id="uploadSubmission"> <td> <input type="submit" value="Upload More" id="uploadMore"> </td> <td> <input type="submit" value="Submit" name="addImage"> </td> <td width="150px"> </td> </tr> </table> </form> JQuery for adding new elements: $(function() { var scntDiv = $('#uploadArea'); var i = $('#p_scents tr td').size() + 1; $('#uploadMore').live('click', function() { $('<tr><td><label for="imageFile">Image:</label> <input type="file" name="imageFile" accept=".jpg,.jpeg,.png,.gif"></td><td><label for="imageCaption">Image Caption:</label> <input type="text" name="imageCaption"></td><td><a href="#" class="removeUpload" width="150px" style="text-align: center">Remove</a></td></tr>').insertBefore( $('#uploadSubmission') ); i++; return false; }); $('.removeUpload').live('click', function() { if( i > 1 ) { $(this).parents('tr').remove(); i--; } return false; }); }); And finally the PHP: require($_SERVER['DOCUMENT_ROOT'].'/settings/globalVariables.php'); require($_SERVER['DOCUMENT_ROOT'].'/settings/mysqli_connect.php'); $db_name = 'imageUploads'; $tbl_name = 'gallery'; if(!$conn) { die('Could not connect: ' . mysqli_error()); } mysqli_select_db($conn, "$db_name")or die("cannot select DB"); foreach($_FILES['imageFile'] as $file){ $caption = $_POST['imageCaption']; $uploadDir = 'http://www.example.com/images/'.'gallery/'; $fileName = $_FILES['imageFile']['name']; $filePath = $uploadDir . $fileName; if(move_uploaded_file($_FILES["imageFile"]["tmp_name"],$_SERVER['DOCUMENT_ROOT']."/images/gallery/".$_FILES["imageFile"]["name"])) { $query_image = "INSERT INTO $tbl_name(filename,path,caption) VALUES ('$fileName','$uploadDir','$caption')"; if(mysqli_query($conn, $query_image)) { echo "Stored in: " . "gallery/" . $_FILES["imageFile"]["name"]; } else { echo 'File name not stored in database'; } } } I was hoping I had it working properly for multiple images with my `foreach` loop but it only uploads one image even if I have 4 selected. EDIT: I've tried modifying my code to this and it's not working either but looking at tutorials this seems to be more on the right track than my previous code: r equire($_SERVER['DOCUMENT_ROOT'].'/settings/globalVariables.php'); require($_SERVER['DOCUMENT_ROOT'].'/settings/mysqli_connect.php'); $db_name = 'imageUploads'; $tbl_name = 'gallery'; if(!$conn) { die('Could not connect: ' . mysqli_error()); } mysqli_select_db($conn, "$db_name")or die("cannot select DB"); foreach($_FILES['imageFile'] as $key => $name){ $caption = $_POST['imageCaption']; $uploadDir = 'http://www.example.com/images/'.'gallery/'; $fileName = $key.$_FILES['imageFile']['name'][$key]; $file_size = $_FILES['files']['size'][$key]; $file_tmp = $_FILES['files']['tmp_name'][$key]; $file_type= $_FILES['files']['type'][$key]; $filePath = $uploadDir . $fileName; if(move_uploaded_file($file_tmp,$_SERVER['DOCUMENT_ROOT']."/images/gallery/".$fileName)) { $query_image = "INSERT INTO $tbl_name(filename,path,caption) VALUES ('$fileName','$uploadDir','$caption')"; if(mysqli_query($conn, $query_image)) { echo "Stored in: " . "gallery/" . $_FILES["imageFile"]["name"]; } else { echo 'File name not stored in database'; } } } I've also added `[]` in the names of the HTML input elements that needed them.

$results = mysql_query("$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

Hi all, I am trying to get data from MySQL to display in a html table in TCPDF but it is only displaying the ID. $accom = '<h3>Accommodation:</h3> <table cellpadding="1" cellspacing="1" border="1" style="text-align:center;"> <tr> <th><strong>Organisation</strong></th> <th><strong>Contact</strong></th> <th><strong>Phone</strong></th> </tr> <tbody> <tr>'. $id = $_GET['id']; $location = $row['location']; $sql = "SELECT * FROM tours WHERE id = $id"; $result = $con->query($sql); if ($result->num_rows > 0) { while($row = $result->fetch_assoc()) { '<td>'.$location.'</td> <td>David</td> <td>0412345678</td> </tbody>'; } } '</tr> </table>'; Anyone got any ideas?

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!

Hello. Are there any GOOD tutorials about MYSQLI Object orientated coding, and procedural. IE - ONE PAGE that has procedural instructions and DOESNT EVEN BREATHE a sign of object-orientated, and the swapped version. I prefer Procedural. I've found tutorials that have both codes in one page - very confusing.. You use 1 line of procedural, and another line of object orientated & of course its causes errors. I want seperate pages/information. I've been coding for years and now want a new thing added to my site. So I COPY the exact WORKING code i have on a separate script to this new script, and now that same code doesn't work. my current code is here:- $link = mysqli_connect('SITE.com', 'blah', 'Blahpasswrd', 'moreblah'); if ($link->connect_errno) { $GLOBALS["jobs"].="Failed to connect to MySQL: " . $link=>connect_error; save_logs("XXXXXXXXXV06_JOBSCHECK",$GLOBALS["jobs"]); exit(); } $sql = "SELECT * FROM `jobs` WHERE `ID` LIKE '$job%';"; $GLOBALS["jobs"].="\n$sql\n\n";$result = $link->query($sql); $result = $link->query($sql); while($row = mysqli_fetch_assoc($result)) { $text.=print_r($row,true)."\n"; } the error message im getting is:- [10-Jun-2023 23:30:43 Pacific/Auckland] PHP Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, bool given in PATH.php on line 72 [10-Jun-2023 23:35:03 Pacific/Auckland] PHP Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, bool given in PATH.php on line 72 all i want is a list of each line/row of the database that matches.