Posted by Joshua Ruppert on 12/17/84 11:55

ZafT wrote:
> >
> > 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

No problem. Just remember, there are no dumb questions just dumb
answers :o)

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