L PROTECTED]...
> Ok. That makes since.
>
> Thanks
> Steve
> > At 04:20 PM 7/26/2002 -0500, you wrote:
> >var names can only be letters, numbers, and underscores.
> >
> >Jim Grill
> >Support
> >Web-1 Hosting
> >http://www.web-1hosting.net
>
Can anyone tell me what I am doing wrong?
I am essentially trying to do this:
$divid[1] = $div_id1;
$divid[2] = $div_id2;
$divid[3] = $div_id3;
$divid[4] = $div_id4;
$divid[5] = $div_id5;
But I was looking for a tighter way, like the following (which does not work):
for($m=1;$m<=5;$m++){
[EMAIL PROTECTED]>
>To: "PHP" <[EMAIL PROTECTED]>
>Sent: Friday, July 26, 2002 3:53 PM
>Subject: [PHP] filling an array(2)
>
>
> > h. Ok. Can somebody explain this one? Why won't it work correctly?
> >
> > for($m=1;$m<=5;$m++){
> >
for($m=1;$m<=5;$m++){
$div_id[$m]=${'divid'.$m};
}
Regards,
Andrey
- Original Message -
From: "Steve Buehler" <[EMAIL PROTECTED]>
To: "PHP" <[EMAIL PROTECTED]>
Sent: Friday, July 26, 2002 10:48 PM
Subject: [PHP] filling an array
> Ca
}
From Martin:
for($i =0; $i < 5; $i++) {
$offset = $m + 1;
$divid[$m] = $div_id{$offset};
}
At 10:55 PM 7/26/2002 +0300, you wrote:
>- Original Message -
>From: "Steve Buehler" <[EMAIL PROTECTED]>
>To: "PHP" <[EMAIL PROTECTED]>
>Sent: Fr
h. Ok. Can somebody explain this one? Why won't it work correctly?
for($m=1;$m<=5;$m++){
$div_idd[$m]=${'row->sub' . $m . 'd'};
}
Can it not be done with a 3 parter? The columns in the table that $row
gets, are sub1d, sub2d, sub3d, sub4d and sub5d. Or is it the "->" that is
me
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