CODEWITHSUNDEEP
javascriptphpsql
Chad Cardiff
Chad Cardiff

Reputation: 479

Run PHP SQL query command when link clicked on

I want there to be a SQL query to take place when a link is clicked on, so far this is what I have:

mainpage.php:

function deleteImage1() {
    $.post("delete_image.php", { num:1, id:<?php echo $id; ?> }, function(data,status){
  alert("Data: " + data + "\nStatus: " + status);
  });
    return false;
}

<a href="#" onclick="deleteImage1()">Delete Image</a>

delete_image.php:

<?php

// $connect stuff here 

    $num = $_GET['num'];
    $id = $_GET['id'];

    if ($num === '1') {
    $image_num == '';
    } else {
    $image_num == $num;
    }

    $sqlCommand = mysql_query("UPDATE alpacas1 SET image$image_num='' WHERE id=$id");

if (!$sqlCommand) {
    die('Invalid query: ' . mysql_error());
} else {
 echo "Updated successfully!";
}
?>

Now, when I click on the "Delete Image" link, it try's to run the function, and it does, but returns as this in the popup:

Data: Invalid query: You have an error in your SQL syntax; check the manual that corresponds with your MySQL server version for the right syntax to use near " at line 1.
Status: success

But when I reload the page, (or look in the database), the image field has not been changed to '' (or null).

Also, I have tested the SQL query so it is not that. I have done so by visiting delete_image.php in my browser with the id & num variables being "posted" (ex. delete_image.php?num=1&id=20) and by visiting it IN BROWSER, it SUCCESSFULLY deletes the image and gives me the message "Updated successfully!".

I'm guessing it's something with my Javascript as I am a beginner! Thanks for your help!

Upvotes: 0

Views: 2238

Answers (1)

Jason
Jason

Reputation: 1962

I should've noticed this earlier...

You're using the jQuery .post(), yet you're accessing the variables as if they are sent via GET. You should either use $_POST or $_REQUEST to pull in your data:

$num = $_POST['num'];
$id = $_POST['id'];

Without the proper variables, your query reads:

UPDATE alpacas1 SET image='' WHERE id="

Thus the MySQL error is returned.

Upvotes: 2

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