Question

I’m currently trying to make a page via php which allows the user to update data in my database. I’m experiencing two problems: first when I run my code I get the “Error: Query was empty”, however updates were made to the database and this leads me to my second problem. Fields that were left empty (a user doesn’t have to enter data into all the fields if they only have one or two things to update) become blank after the updates are made. This is because my current script updates all elements, but is there any way I can have it where if the user leaves an input field blank, nothing gets changed when the database is updated?

Here is my code:

if (isset($_POST['submit'])) {
    $id = $_POST['id'];
    $lastname = $_POST['lastname'];
    $firstname = $_POST['firstname'];
    $color = $_POST['color'];
    $number = $_POST['number'];

    // need id to be filled and need at least one other content type for changes to be made
    if (empty($id) || empty($lastname) and empty($firstname) and empty($major) and empty($gpa)) {
        echo "<font color='red'>Invalid Submission. Make sure you have an ID and at least one other field filled.  </font><br/>";   
    } else {    
        // if all the fields are filled (not empty)
        // insert data to database  
        mysql_query ("UPDATE students SET lastname = '$lastname', firstname = '$firstname', favoritecolor = '$color', favoritenumber = '$number' WHERE id = '$id'");

        if (!mysql_query($sql,$con)) {
            die('Error: ' . mysql_error());
        }

        // display success message
        echo "<font color='blue'>Data updated successfully.</font>";
        // Close connection to the database
        mysql_close($con);
    }
} 

Answer 1

To answer your question, you need to catch the query’s result and check for errors on that.

$query = mysql_query(/*query*/);
if (!$query)
    //error handling

Be sure to read up on SQL injections, as per my comment.