Къде ми е грешката?

[ih999]

Значи имам някаква грешка но не мога да я намеря не иска да търси ..Ето index.html:

<html>
<head>
<title>Ajax , etc</title>
<script type="text/javascript" src="js/jquery-1.5.2.min.js"></script>
<script>
$(document).ready(function() {
$("#q").autocomplete("ftw.php" , {minChars: 3,formatItem: liFormat,maxItemsToShow: 10});
});

function liFormat(row,i,num) {
var result = '<li>' + row[0] + '</li>';
return result;
}
</script>
</head>
<body>

<input type="text" id="q" />



</body>

</html>

ftw.php

<?php
//header("Content-type: text/plain; charset=utf-8");
$q = $_GET['q'];

mysql_connect('localhost' , 'root' , '') or die (mysql_error());
mysql_select_db('catalog') or die (mysql_error());
mysql_query("SET NAMES cp1251");

$query = mysql_query("SELECT title FROM films WHERE title LIKE = '%$q%");

while($res = mysql_fetch_assoc($query)) {
echo "$row[title]";
}
?>

Някой може ли да ми помогне?

 

[heminei]

$query = mysql_query("SELECT title FROM films WHERE title LIKE = '%$q%'")or die(mysql_error());

Пробвай така?

 

[ih999]

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= '%%'' at line 1

тфа ко ше рече ? :Д

 

[ih999]

Оправих го но не иска да търси..

ftw.php

<?php
//header("Content-type: text/plain; charset=utf-8");
$q = $_GET['q'];

mysql_connect('localhost' , 'root' , '') or die (mysql_error());
mysql_select_db('catalog') or die (mysql_error());
mysql_query("SET NAMES cp1251");

//$query = mysql_query("SELECT title FROM films WHERE title LIKE = '%$q%");
$query = mysql_query("SELECT title FROM films WHERE title LIKE '%$q%'")or die(mysql_error());
while($res = mysql_fetch_assoc($query)) {
echo "$row[title]";
}
?>

 

[tedo96]

<?php
//header("Content-type: text/plain; charset=utf-8");
$q = $_GET['q'];

mysql_connect('localhost' , 'root' , '') or die (mysql_error());
mysql_select_db('catalog') or die (mysql_error());
mysql_query("SET NAMES cp1251");

//$query = mysql_query("SELECT title FROM films WHERE title LIKE = '%$q%");
$query = mysql_query("SELECT title FROM films WHERE title LIKE '%$q%'")or die(mysql_error());
while($row = mysql_fetch_assoc($query)) {
echo $row[title];
}
?>