MjM8082 Posted August 21, 2011 Share Posted August 21, 2011 Having update statement trouble... The error I'm getting is... "Could not query the database - : " Not sure what I'm missing, I had this working before also, but now it's not. Here is my code for the update page... <html> <head> <title>Update User</title> </head> <body> <form method="post" action="update_user2.php"> <?php $dbc = mysqli_connect('localhost', 'se266_user', 'pwd', 'se266') or die(mysql_error()); //delete users echo '<b>Delete or Update User</b>.<br />'; if (isset($_POST['remove'])) { foreach($_POST['delete'] as $delete_id) { $query = "DELETE FROM users WHERE course_id = $delete_id"; mysqli_query($dbc, $query) or die ('can\'t delete user'); } echo 'user has been deleted.<br />'; } if (isset($_POST['update'])) { $course_id = $_POST['course_id']; $course_name = $_POST['course_name']; $student_id = $_POST['student_id']; $query = "UPDATE users SET course_name ='". $course_name ."' WHERE course_id = $course_id"; $updres = mysqli_query($query); if(!$updres) { die(" Could not query the database - : <br/>". mysqli_error() ); } else { echo 'course has been updated.<br />'; } } //display users info with checkbox to delete $query = "SELECT * FROM users"; $result = mysqli_query($dbc, $query); while($row = mysqli_fetch_array($result)) { echo '<input type="checkbox" value="' .$row['course_id'] . '" name="delete[]" />'; echo ' ' .$row['course_name'] .' '. $row['student_id']; echo '<br />'; } mysqli_close($dbc); ?> <form method="POST" action="update_user2.php"> <label for="course_id">Course ID:</label> <input type="text" id="course_id" name="course_id" /> <br /> <label for="course_name">Course Name:</label> <input type="text" id="course_name" name="course_name" /> <br /> <label for="course_name">Student ID:</label> <input type="text" id="student_id" name="student_id" /> <br /> <input type="submit" name="remove" value="Remove" /> <input type="submit" name="update" value="Update" /> <br><br> </form> </body> </html> Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted August 21, 2011 Share Posted August 21, 2011 If you were developing with error reporting on, you'd be getting error regarding the parameters you're passing (or in this case, not passing) to mysqli_query() and mysqli_error(). Quote Link to comment Share on other sites More sharing options...
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