Forum Moderators: coopster

Message Too Old, No Replies

sql error

         

kkonline

7:25 am on Mar 22, 2008 (gmt 0)

10+ Year Member



hi
$sql="INSERT INTO $table (title,trusted,link,content,mood,mime,path,name,ip,date)VALUES('$title',1,'$link','$content','$mood','$mimeinfo','$pathinfo','$imgnameinfo','$ip','$date')";

what is the error in the above query?

bkeep

8:54 am on Mar 22, 2008 (gmt 0)

10+ Year Member



replace the ' with " in your values since you are using variables

so something like this

$sql="INSERT INTO $table (title,trusted)VALUES("$title",1)";

Best Regards,
Brandon

eelixduppy

11:24 am on Mar 22, 2008 (gmt 0)



What is the error message you are receiving when you try to run this query? If you do not know, then you should try adding some error reporting code to find out. It would look like the following:

$result = mysql_query($query) or die([url=http://www.php.net/mysql-error]mysql_error[/url]());