david,
the syntax needs a single variable there...not a list, or multiple
vars...plus, if you look at http://php.net/list, you'll see it returns
void, and foreach($table as void) doesn't make much sense...
jack
----- Original Message -----
From: David Otton <[EMAIL PROTECTED]>
Date: Tuesday, October 23, 2001 4:14 pm
Subject: Re: [PHP] language question
> On Mon, 22 Oct 2001 16:47:56 +0200, you wrote:
>
> >On Monday 22 October 2001 14:28, David Otton wrote:
> >> why does this work:
> >>
> >> foreach ($table as $row)
> >> list ($a, $b) = $row;
> >>
> >> but this doesn't?
> >>
> >> foreach ($table as list ($a, $b));
> >
> >because the correct syntax is
> > foreach ($table as $key => $val) {
> > ...
> > }
>
> >RTFM :)
>
> Well... no. The manual says "There are two syntaxes; the second is a
> minor but useful extension of the first:
>
> foreach(array_expression as $value) statement
> foreach(array_expression as $key => $value) statement"
>
> Ok, let me rewrite my examples, so they're definitely not about
> dictionaries :
>
> $table is an array of arrays. Each inner array has three values.
>
> Roughly.... $table = ((1,2,3),(4,5,6),(7,8,9))
>
> foreach ($table as $row)
> list ($a, $b, $c) = $row;
>
> foreach ($table as list($a, $b, $c));
>
> I still don't see what it is about list() that precludes it's use in
> this way. Anyone?
>
> >BTW: overwriting $a, $b in each iteration isn't particularly
> useful...
> But it is a minimal example of the construct I don't understand.
>
> djo
>
>
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