RE: [PHP] inserting a variable into a variable
| From: | scott [gts] | Date: | Wed, 20 Jun 2001 15:37:57 +0000 |
| Subject: | RE: [PHP] inserting a variable into a variable | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-54555@lists.php.net to get a copy of this message | ||
you so totally could use an assoc. array right now:
$cat_adt = "x";
$rm = "y";
$a[ $cat_adt ] = "whatever";
$a[ "$rm"."$cat_adt" ] = "Yeah";
print $a['yx'];
save yourself a lot of trouble and dont bother with
variables-of-variables and trying to get
$rm_$cat_adt == $rm_x
just use a hash... that's what it's there for ;)
> -----Original Message-----
> From: Tom Carter [mailto:subs@roundcorners.com]
> Sent: Wednesday, June 20, 2001 5:24 AM
> To: Hasan Niyaz; php-general@lists.php.net
> Subject: Re: [PHP] inserting a variable into a variable
>
>
> > I have come to a situation where i am having a variable inside another
> variable.
> > for example.
> >
> > $rm_$cat_adt
>
> PHP would read this as trying to prepend the variable $cat_adt to the
> variable $rm_
>
> You seem to be trying to insert the variable $cat into the middle of a
> variable.. possible, but probably better to slightly rethink your naming
> strategy
>
> HTH,
> Tom
>
> >
> > The above is a variable and $cat is again another variable
> > This does not work..
> >
> > Need some help!
> >
> >
> > Thanks,
> > Hasan
> >
>