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Found 13 results

  1. here is a code of sign in page I want to add a role access for the student, teacher, and admin I have table name student in the database and a column role see image attached for database include("dbconfig.php"); session_start(); if($_SERVER["REQUEST_METHOD"] == "POST") { // username and password sent from form $name = mysqli_real_escape_string($con,$_POST['name']); $password = mysqli_real_escape_string($con,$_POST['password']); $sql = "SELECT user_id FROM student WHERE name = '$name' and password = '$password'"; $result = mysqli_query($con,$sql); $row = mysqli_fetch_array($result,MYSQLI_ASSOC); // $active = $row['active']; $count = mysqli_num_rows($result); if($count == 1) { //session_register("name"); $_SESSION['login_user'] = $name; header("location: allstudents1.php"); }else { $error = "Your Login Name or Password is invalid"; } } ?>
  2. Hi, I have a problem with importing .csv file. The content of file have some rows with invalid culomn number. Can I skip rows with this error? My code is: <?php $addauto = 1; $delimiter = ';'; $csoport_kod = new mysqli('localhost', 'root', '', '2012'); $csoport_kod->query("DELETE FROM gf1"); $i=1; if (($handle = fopen("ftp://2012.com/gf1.csv", "r")) !== FALSE) { while (($data = fgetcsv($handle, 10000, $delimiter)) !== FALSE) { foreach($data as $i => $content) { $data[$i] = $csoport_kod->real_escape_string($content); } $csoport_kod->query ("INSERT ignore INTO gf1 VALUES('" . implode("','", $data) . "');"); } fclose($handle); } echo 'OK' ?> THX!
  3. I am using this API https://github.com/blocktrail/blocktrail-sdk-php. I am suppose to add this to top of my directory where I want to initialize the API. require 'vendor/autoload.php'; use Blocktrail\SDK\BlocktrailSDK; But the second line gives me an error. use Blocktrail\SDK\BlocktrailSDK; Parse error: syntax error, unexpected 'use' (T_USE) in C:\xampp\htdocs\site\templates\header.php on line 12 Is the "use" a correct syntax for php? Why is it giving me an error?
  4. On my Joomla 2.5 / Virtuemart 3 website I get the following error - Warning: array_push() expects parameter 1 to be array, null given in /var/www/vhosts/pizzacaldo.co.uk/httpdocs/newjoomla/libraries/tcpdf/tcpdf.php on line 17047 Warning: array_push() expects parameter 1 to be array, null given in /var/www/vhosts/pizzacaldo.co.uk/httpdocs/newjoomla/libraries/tcpdf/tcpdf.php on line 17047 Warning: array_push() expects parameter 1 to be array, null given in /var/www/vhosts/pizzacaldo.co.uk/httpdocs/newjoomla/libraries/tcpdf/tcpdf.php on line 17047 Warning: array_push() expects parameter 1 to be array, null given in /var/www/vhosts/pizzacaldo.co.uk/httpdocs/newjoomla/libraries/tcpdf/tcpdf.php on line 17047 Warning: array_push() expects parameter 1 to be array, null given in /var/www/vhosts/pizzacaldo.co.uk/httpdocs/newjoomla/libraries/tcpdf/tcpdf.php on line 17047 I'm told on the Virtuemart forum it might be something to do with the PDF invoice generating (which I or my clients never use - only the admin order confirmation e-mail is used). Can someone please tell me how to fix this? Thanks.
  5. I am using a simple image upload. http://www.w3schools.com/php/php_file_upload.asp It gives me 2 errors like this. Warning: move_uploaded_file(C:/xampp/htdocs/home/upload/images/grandpix 2.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\home\upload\post.php on line 174 Warning: move_uploaded_file(): Unable to move 'C:\xampp\tmp\phpF915.tmp' to 'C:/xampp/htdocs/home/upload/images/grandpix 2.jpg' in C:\xampp\htdocs\home\upload\post.php on line 174 This is my code. Post.php . I see that it's the "$target_dir" issue. How can I fix it? if(isset($_FILES['fileToUpload'])){ if(!empty($_FILES['fileToUpload']['name'])) { $target_dir = $_SERVER['DOCUMENT_ROOT'].'/home/upload/images/'; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION); $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false) { $uploadOk = 1; } else { $errors[] = 'File is not an image.'; $uploadOk = 0; } // Check if file already exists if (file_exists($target_file)) { $errors[] = 'Sorry, file already exists.'; $uploadOk = 0; } // Check file size if ($_FILES["fileToUpload"]["size"] > 500000) { $errors[] = 'Sorry, your file is too large.'; $uploadOk = 0; } // Allow certain file formats if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) { $errors[] = 'Sorry, only JPG, JPEG, PNG & GIF files are allowed.'; $uploadOk = 0; } // Check if $uploadOk is set to 0 by an error if ($uploadOk == 0) { $errors[] = 'Sorry, your file was not uploaded.'; // if everything is ok, try to upload file } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) { echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded."; } else { $errors[] = 'Sorry, there was an error uploading your file.'; } } $insert_image = $db->prepare("INSERT INTO images(user_id, item_id, image_path, date_added) VALUES(:user_id, :item_id, :image_path, :date_added)"); $insert_image->bindParam(':user_id', $userid); $insert_image->bindParam(':item_id', $item_id); $insert_image->bindParam(':image_path', $target_file); $insert_image->bindParam(':date_added', $date_added); if(!$insert_image->execute()) { $errors[] = 'There was a problem uploading the image!'; } else { if(empty($errors)) { $db->commit(); $success = 'Your post has been saved.'; } else { $db->rollBack(); } } } else { $errors[] = 'An image is required!'; } }
  6. I am using this https://github.com/ircmaxell/password_compat I have no problem using it on a local server. But as soon as I test it on a live server, it gives me an error like this. Parse error: syntax error, unexpected '{' in /home/public_html/sub/snippets/password.php on line 19 The error is on the same line as the beginning of the code in that library. Why is this happening?
  7. Respected Users, Few days ago i created a bot site regarding auto comment and response comment etc. but whenever i try to login there using my facebook id and password an error shows up "Login Session Expired". The Problem is with the login script, i am unable to find that errorIts my humble request to you, kindly go through the post and fix me the problem. If u need anything regarding this Error kindly Msg me. Sorry for my poor English. Login PHP Code Posted Below: code removed
  8. This is the code from my previous topic. It was for getting 3 countdown counters to work on the same page. It does that. But I didn't notice this up until now. It gives me this error and it keeps counting up the errors in firebug. TypeError: document.getElementById(...) is null This is the code. Can you tell me what's wrong with it? <script> $( document ).ready(function() { //Create object with the list of due dates //The 'name' will correspond to the field ID to populate the results var dueDates = { 'date1':'<?php echo $global_payment_due_1; ?>', 'date2':'<?php echo $global_payment_due_2; ?>', 'date3':'<?php echo $global_payment_due_3; ?>' }; var timer = setInterval(function() { //Instantiate variables var dueDate, distance, days, hours, minutes, seconds, output; //Set flag to repeat function var repeat = false; // Get todays date and time var now = new Date().getTime(); //Iterate through the due dates for (var dueDateId in dueDates) { //Get the due date for this record dueDate = new Date(dueDates[dueDateId]); // Find the distance between now an the due date distance = dueDate - now; // Time calculations for days, hours, minutes and seconds days = Math.floor(distance / (1000 * 60 * 60 * 24)); hours = Math.floor((distance % (1000 * 60 * 60 * 24)) / (1000 * 60 * 60)); minutes = Math.floor((distance % (1000 * 60 * 60)) / (1000 * 60)); seconds = Math.floor((distance % (1000 * 60)) / 1000); //Determine the output and populate the corresponding field output = "OVERDUE"; if (distance > 0) { output = days + "d " + hours + "h " + minutes + "m " + seconds + "s"; repeat = true; //If any record is not expired, set flag to repeat } document.getElementById(dueDateId).innerHTML = output; //If flag to repeat is false, clear event if(!repeat) { clearInterval(timer); } } }, 1000); }); </script>
  9. I am building a website about Doctor who, and I want to post various quotes from a database, when I wrote the select-query, I got the following error: Quotes Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /customers/e/a/c/doctorwhofans.be/httpd.www/test/Content/Quotes.php on line 106 (which is the one with the while expression). <?php //verbinding met de database require("./connect.php"); //select-query schrijven en uitvoeren $sql = " select * from QuotesTabel"; $result = mysql_query($sql); //alle records weergeven (terwijl er rijen gevonden worden. while ($row = mysql_fetch_assoc($result)) { //toon foto met info require("./ToonFoto.php"); } ?> Can someone help me please?
  10. Hello, I'm coding a log in page and I keep getting this problem with mysqli_result. It's a mysqli server. My error is: Fatal error: Cannot use object of type mysqli_result as array in /home/a7017672/public_html/login.html on line 34 My code is: <?PHP if(isset($_SESSION['loggedin'])) { die("You are already logged in!"); } if(isset($_POST['submit'])) { $email = mysqli_real_escape_string($con,$_POST['username']); $pass = mysqli_real_escape_string($con,$_POST['password']); $mysql = mysqli_query($con,"SELECT * FROM users WHERE email_address = '{$email}' AND password = '{$pass}'"); $mysql2 = mysqli_fetch_array($mysql); if (!$mysql ||mysqli_num_rows($mysql) < 1) { die("Incorrect password!"); } $_SESSION['loggedin'] = "YES"; $_SESSION['email'] = $email; $_SESSION['fname'] = $mysql['first_name']; $_SESSION['lname'] = $mysql['last_name']; $_SESSION['add1'] = $mysql['address_1']; $_SESSION['add2'] = $mysql['address_2']; $_SESSION['county'] = $mysql['county']; $_SESSION['postcode'] = $mysql['postcode']; $_SESSION['tel'] = $mysql['tel_no']; $_SESSION['mobile'] = $mysql['mobile_no']; $_SESSION['team'] = $mysql['team']; $_SESSION['ismanager'] = $mysql['is_manager']; $_SESSION['isadmin'] = $mysql['is_admin']; $_SESSION['sysadmin'] = $mysql['is_sysadmin']; die("You are now logged in!"); } echo "<form method='POST'> <p style='font-size: 14pt;'>Username (Email): <br><input type='text' name='username' maxlength='50' size='30'> <br /> <p style='font-size: 14pt;'>Password:<br><input type='text' name='password' maxlength='50' size='30'><br /> <input type='submit' name='submit' value='Login'> </form>"; ?> Any help is greatly appreciated!
  11. I'm not hoping to much that I can solve this thing, but maybe someone here who knows php well can help me. Screenshot: http://extrazoom.com/image-70122.html?heuln50x50 File: https://mega.nz/#!uYgWmRTL!5ZyabPKnYWjeG2sL_PXyfFaAKaJiv4zceQGrky7fmPk And yes, I replaced "preg_replace" with "preg_replace_callback" and I ended up here: Screenshot:: http://extrazoom.com/image-70124.html?heuln50x50 Thank you and sorry for my bad english, if it's bad .
  12. Hi! I'm getting this PHP syntax error on line 87 of my file. ERROR: PHP Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) The function works fine, if I take the function out of that file and put it into a separate file, it works fine. function get_UserId($username){ include('config.php'); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "SELECT id FROM users WHERE username='$username'"; $result = $conn->query($sql); $row = $result->fetch_assoc(); $id = $row['id']; $conn->close(); return $id; } Thanks
  13. [Linux] PHP Notice: Undefined variable: connection in /var/www/html/popreport/functions.php on line 23 PHP Fatal error: Call to a member function query() on a non-object in /var/www/html/popreport/functions.php on line 23 The fuction output in the following program is called from another file called records records-board.php If you look at the program below you'll see that I did define $connection above line 23 in the file functions.php And for the second error I'm really not getting it because that same foreach loop was working fine with the exact same argument list when it was in the file records-board.php, but now that I've copied most of the code from records-board.php and placed it in functions.php all the sudden my program can't see the variable $connection and has a problem with my foreach loop on line 23. Again, both of those lines worked fine when they were in another file. functions.php <?php //session_start(); // open a DB connectiong $dbn = 'mysql:dbname=popcount;host=127.0.0.1'; $user = 'user'; $password = 'password'; try { $connection = new PDO($dbn, $user, $password); } catch (PDOException $e) { echo "Connection failed: " . $e->getMessage(); } $sql = "SELECT full_name, tdoc_number, race, facility FROM inmate_board WHERE type = 'COURT'"; function output() { foreach($connection->query($sql) as $row) { echo "<tr><td>$row[full_name]</td></tr>"; } } ?> records-board.php <? include 'functions.php'; php output(); ?> Any ideas?
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