$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)
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]());