[Resolved] Inserting into SQL table...?

MastahUK

I am fairly new to php, but I have successfully managed to do this on a few occasions when working on other things. All I did for the bit of code I am working on now, is select the appropriate lines from other scripts and modify it so that the sql query is correct. Here is my script:

<?

$title = $frmtitle;
$category = $frmcategory;
$author = $frmauthor;
$source = $frmsource;

if($title && $category && $author && $source)
{
	include 'contentconfig.php';
	mysql_query("INSERT INTO users ( id , title , category , author , source ) VALUES ( '', '$title', '$category', '$author', '$source')") or die(mysql_error());
}
?>

This gives me this 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 ''author' , 'source' ) VALUES ( '', 'Testing', 'photoshop', 'Tes"

It would be great if someone could help me with this.

DigitalExpl0it

try this

mysql_query("INSERT INTO users SET 
id = '', 
title = '$title', 
category = '$category', 
author = '$author', 
source = '$source'") or die(mysql_error());

MastahUK

Thanks, but I seem to be getting exactly the same error as before.

In case it helps, here is how the table is set up:

id - int(11)
title - tinytext, null
category - tinytext, null
author - tinytext, null
source - longtext

MastahUK

Anyone?? I would have thought this was easy to solve....😕

intrigue

$query = "insert into users values
(ID,'".$title."', '".$category."' , '".$author."' , '".$source."')";

Thats how i would do it and i did test it and it worked for me but i didnt use the same field types as i forgot to check what ya used, if there is still a problem then pm mor or email me and i will try and help
/matt

MastahUK

Thanks intrigue, but I tried that and got the same kind of error.

After doing lots of searching, I finally stumbled across the solution myself. This now works for me:

mysql_query("INSERT INTO data SET id='', title='$title', category='$category', author='$author', source='$source'") or die(mysql_error());

Thread Resolved 🙂