VIVA LA NWO VIVA LA NWO - 11 months ago 64
Ajax Question

mysql_real_escape_string() just makes an empty string?

I am using a jQuery AJAX request to a page called

that connects to my database and inserts a row. This is the


// Some config stuff
define(DB_HOST, 'localhost');
define(DB_USER, 'root');
define(DB_PASS, '');
define(DB_NAME, 'quicklike');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('ERROR: ' . mysql_error());
$sel = mysql_select_db(DB_NAME, $link) or die('ERROR: ' . mysql_error());

$likeMsg = mysql_real_escape_string(trim($_POST['likeMsg']));
$timeStamp = time();

die('ERROR: Message is empty');

$sql = "INSERT INTO `likes` (like_message, timestamp)
VALUES ('$likeMsg', $timeStamp)";

$result = mysql_query($sql, $link) or die('ERROR: ' . mysql_error());

echo mysql_insert_id();



The problematic line is
$likeMsg = mysql_real_escape_string(trim($_POST['likeMsg']));
. It seems to just return an empty string, and in my database under the
column all I see is blank entries. If I remove
though, it works fine.

Here's my jQuery code if it helps.

$('#like').bind('keydown', function(e) {
if(e.keyCode == 13) {
var likeMessage = $('#changer p').html();

if(likeMessage) {
cache: false,
url: 'like.php',
type: 'POST',
data: { likeMsg: likeMessage },
success: function(data) {
} else {

All this jQuery code works fine, I've tested it myself independently.

Any help is greatly appreciated, thanks.

Answer Source

Are you 1000% sure that $_POST["likeMsg"] actually contains something?

As for mysql_real_escape_string() returning an empty value, the manual says there is only one situation where that can happen:

Note: A MySQL connection is required before using mysql_real_escape_string() otherwise an error of level E_WARNING is generated, and FALSE is returned. If link_identifier isn't defined, the last MySQL connection is used.

this doesn't seem to be the case here though, as you do have a connection open. Strange.