I'm trying to refer to a variable using $$, but with a twist. So far, this works:
echo 'temp file: '.$$fld.'<br>';
In this case, $fld equals "img_photo" (although it could be anything).
The above statement could have thus been "echo 'file name:
'.$img_photo.'<br>';" with the same results. So here I actually echo
the value of the variable $img_photo. This is what I want.
But here's my problem:echo 'file name: '.$$fld_name.'<br>';
What I want to do is echo the value of $img_photo_name. But how can I refer to it?
Thanks.
...Rene
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