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Nepareiza sintakse?


Sephy
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Ar INSERT strada tadas vertibas, bet ar UPDATE SET nestrada, kur kļūda?

 

$insert = mysql_query("UPDATE news SET newsname = '".$_POST['title']."', SET newstext='".$_POST['add']."', SET address='".$_POST['address']."' WHERE newsname='$derp'") or die(mysql_error());

 

un error

 

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 'SET newstext='Ierakstiet raxta tekstu!t' at line 1

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if (isset($_POST['submit'])) {
$derp=$_POST['galerija'];
$derp= mysql_real_escape_string($derp);
//parbauda vai visilauki aizpilditi!
if (!$_POST['title'] | !$_POST['add'] | !$_POST['address'] ) {
echo('Aizpildiet visus laukus!');
}
$_POST['title'] = mysql_real_escape_string($_POST['title']);
$_POST['add'] = mysql_real_escape_string($_POST['add']);
$_POST['address'] = mysql_real_escape_string($_POST['address']);
$insert = mysql_query("UPDATE news SET newsname = '".$_POST['title']."', newstext='".$_POST['add']."', address='".$_POST['address']."' WHERE '$derp'=newsname") or die(mysql_error());
$add_member = mysql_query($insert);

Edited by Sephy
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Uztaisi

 

$data=array_map('mysql_real_escape_string',$_POST);

 

un $data tev saturēs visus POST datus eskeipotā veidā- varēsi lietot $data['title'];Protams labāk būtu uztaisīt query funkciju, kura automātiski eskeipo parametrus.

Edited by codez
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