|
Posted by ZafT on 12/17/56 11:55
>
> You're not actually capturing the mysql_error() output. Try either:
>
> print(mysql_error($link));
>
> or this if you want to use the text in your code
>
> $qryErr = mysql_error($link);
>
> Josh
>
Josh,
Thanks - I do apparently have an error in my syntax. I'll get on that and
try to fix it before bugging the group again. Your help is quite
appreciated.
Shane
Navigation:
[Reply to this message]
|