On Thu, 2009-07-16 at 15:41 -0400, Miller, Terion wrote:
> Why is this an invalid argument?
>
> foreach(($row['inType']) as $inType){
>
> echo $inType,'<br>';}
>
> I am trying to output results from a data base that may have multiple
> results for the same name....
>
> So trying to use an array and foreach that is the right track ...right?
>
>
I imagine $row is the array, and ['inType'] is an element of the array.
This is not how you use a foreach. Can you show where you are getting
$row from?
Thanks
Ash
www.ashleysheridan.co.uk
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