Hi,
Yeah, I'm really want to do that, since I'm working with the elements of the
original array ( skipped that part in sample code ).
I've tried your suggestion, but it gives the same result, so on just one input
is just gives back '1'.
What troubles me, that foreach gives an inconsistent working. Why is 'foreach'
checking element count at all and working differently with different element
counts? That's not normal is my opinion. 'foreach' shouldn't do this:
if ( count($elements) == 1 ) then loop 1;
else loop normally;
and that's what is does now, since when it's more than one element it's working
like a while loop, with checking the condition before ( and after ) every run.
( if 'foreach' would check that the current run is the last one before
executing the current loop, the results would be the same with each case )
Cheers,
Tamas
-----Original Message-----
From: Robert Cummings [mailto:[email protected]]
Sent: Tuesday, July 05, 2011 4:06 PM
To: Dajka Tamas
Cc: [email protected]
Subject: Re: [PHP] Foreach question
On 11-07-05 09:40 AM, Dajka Tamas wrote:
> Hi all,
>
>
>
> I've bumped into an interesting thing with foreach. I really don't know, if
> this is normal working, or why it is, so I got curious.
>
>
>
> The script:
>
>
>
> foreach ( $cats as&$c ) {
>
> echo $c['id'];
>
> if ( $c['id']< 5 ) {
>
> $c['id']++;
>
> $cats[] = $c;
>
> }
>
> }
That's a bizarre loop... you're feeding references to elements of the
array back into the array over which the loop is iterating. If you
REALLY want to do what you are doing, then do the following:
<?php
foreach( array_keys( $cats ) as $key )
{
$c = &$cats[$key];
echo $c['id'];
if( $c['id'] < 5 )
{
$c['id']++;
$cats[] = $c;
}
}
?>
Cheers,
Rob.
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