jQuery and MySQL

Question

I have taken a jQuery script which would remove divs on a click, but I want to implement deleting records of a MySQL database. In the delete.php:

<?php 

$photo_id = $_POST['id'];

$sql = "DELETE FROM photos
  WHERE id = '" . $photo_id . "'";

$result = mysql_query($sql) or die(mysql_error());

?>

The jQuery script:

$(document).ready(function() {
   $('#load').hide();
});

$(function() {
$(".delete").click(function() {
$('#load').fadeIn();
var commentContainer = $(this).parent();
var id = $(this).attr("id");
var string = 'id='+ id ;

$.ajax({
   type: "POST",
   url: "delete.php",
   data: string,
   cache: false,
   success: function(){
 commentContainer.slideUp('slow', function() {$("#photo-" + id).remove();});
 $('#load').fadeOut();
  }

 });

return false;
 });
});

The div goes away when I click on it, but then after I refresh the page, it appears again...

How do I get it to delete it from the database?

EDIT: Woopsie... forgot to add the db.php to it, so it works now >.<

Where is the connection to the database and prevention of sql injection — Ed Heal
– Ed Heal, Commented Jul 6, 2015 at 21:02

Quotidian · Accepted Answer · 2010-05-21 20:23:28Z

1

There's no way the php could even come close to working. Where is the database? Check out http://www.php.net/manual/en/mysql.examples-basic.php from which you can see there's more to the database than just a query.

<?php
// Connecting, selecting database
$link = mysql_connect('mysql_host', 'mysql_user', 'mysql_password')
    or die('Could not connect: ' . mysql_error());
echo 'Connected successfully';
mysql_select_db('my_database') or die('Could not select database');

// Performing SQL query
$query = 'SELECT * FROM my_table';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

// Printing results in HTML
echo "<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "\t<tr>\n";
    foreach ($line as $col_value) {
        echo "\t\t<td>$col_value</td>\n";
    }
    echo "\t</tr>\n";
}
echo "</table>\n";

// Free resultset
mysql_free_result($result);

// Closing connection
mysql_close($link);
?>

answered May 21, 2010 at 20:23

Quotidian

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2 Comments

MacMac Over a year ago

Are you trying to teach me how to use PHP and MySQL with databases? LOL I already know how to do stuff like that...

Ed Heal Over a year ago

Obviously not. Sql injection

Francisco Soto · Accepted Answer · 2010-05-21 20:24:29Z

You have your data as a GET string, but you are using a POST request, try changing your string variable to an object. Like :

$(document).ready(function() {
   $('#load').hide();
});

$(function() {
$(".delete").click(function() {
$('#load').fadeIn();
var commentContainer = $(this).parent();
var id = $(this).attr("id");
var string = { id : id };

$.ajax({
   type: "POST",
   url: "delete.php",
   data: string,
   cache: false,
   success: function(){
 commentContainer.slideUp('slow', function() {$("#photo-" + id).remove();});
 $('#load').fadeOut();
  }

 });

return false;
 });
});

Plus I am hoping you are preparing your MySQL connection properly in your PHP, you cannot just call mysql_query and hope it will know which database you mean, and how to connect to it by itself :)

Look at @Quotidian answer! :)

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