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Posted by ZafT on 11/03/56 11:55
>
> You're not actually capturing the mysql_error() output. Try either:
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> print(mysql_error($link));
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> or this if you want to use the text in your code
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> $qryErr = mysql_error($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
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