| Peter Beckman 2007-02-13, 3:58 am |
| On Tue, 13 Feb 2007, bedul wrote:
> sry i don't get what u mean??
>
This is how you assign a variable by reference. $rate should be a
reference to $mydata[$prefix], not a copy. If I change the value of
$rate, the value of $mydata[$prefix] is also changed, and vice versa.
[color=darkred]
>
> the reason mydata2 empty was because it don't have value in it!!
>
> full source plz
> why u don't try this
>
> $txt.="<ol>";
> foreach($mydata as $nm=>$val){
> $txt.="\n<li> $nm = $val";
> $txt2="<br>\$mydata[$nm] = $val";
> }
> $txt.="</ol>";
>
> print $txt;
Because I'm trying to point out a problem with PHP, where setting a
reference when the other side is undefined or not set, PHP creates a
reference to a previously non-existent array item, just by setting a
reference. I don't think that should happen.
Your code doesn't set anything by reference.
>
> we should cross check again.
I don't know what you mean.
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Peter Beckman Internet Guy
beckman@purplecow.com http://www.purplecow.com/
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