RE: [PHP] inserting a variable into a variable

2001-06-20 Thread Peter Houchin - SunRentals Australia 

why don't you do something like 

$var = $foo  $baa;

?

-Original Message-
From: Hasan Niyaz [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 20, 2001 12:07 PM
To: [EMAIL PROTECTED]
Subject: [PHP] inserting a variable into a variable


Hello all,

I have come to a situation where  i am having a variable inside another variable.
for example.

$rm_$cat_adt

The above is a variable and $cat is again another variable
This does not work..

Need some help!


Thanks,
Hasan

Re: [PHP] inserting a variable into a variable

2001-06-20 Thread mailing_list 

 I have come to a situation where  i am having a variable inside another
 variable.
 for example.
 
 $rm_$cat_adt

This works for me:

?
  $cat=def;
  ${abc_.$cat._ghi}=TEST;
  echo 'abc_.$cat._ghi = '.${abc_.$cat._ghi}.br\n;
  echo 'abc_def_ghi = '.$abc_def_ghi.br\n;
?

michi

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RE: [PHP] inserting a variable into a variable

2001-06-20 Thread Ray Hilton 

What on earth are you trying to do? :)

I'm not sure what your getting at, do you mean defining a variable based
on another variables name?  Perhaps try using an array, and specifying
$rm($cat)

Sorry, im not too sure what you mean

Ray Hilton
-
[EMAIL PROTECTED]
http://rayh.co.uk

-Original Message-
From: Hasan Niyaz [mailto:[EMAIL PROTECTED]]
Sent: 20 June 2001 04:07
To: [EMAIL PROTECTED]
Subject: [PHP] inserting a variable into a variable


Hello all,

I have come to a situation where  i am having a variable inside another
variable. for example.

$rm_$cat_adt

The above is a variable and $cat is again another variable
This does not work..

Need some help!


Thanks,
Hasan

Re: [PHP] inserting a variable into a variable

2001-06-20 Thread Tom Carter 

 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

Re: [PHP] inserting a variable into a variable

2001-06-20 Thread Hasan Niyaz 

Michi,

Yours did the trick...
thank you and everyone else who contributed.

Hasan

- Original Message - 
From: [EMAIL PROTECTED]
To: Hasan Niyaz [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Wednesday, June 20, 2001 11:46 AM
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
 
 This works for me:
 
 ?
   $cat=def;
   ${abc_.$cat._ghi}=TEST;
   echo 'abc_.$cat._ghi = '.${abc_.$cat._ghi}.br\n;
   echo 'abc_def_ghi = '.$abc_def_ghi.br\n;
 ?
 
 michi
 
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 GMX - Die Kommunikationsplattform im Internet.
 http://www.gmx.net
 
 --
 GMX Tipp:
 
 Machen Sie Ihr Hobby zu Geld bei unserem Partner 11!
 http://profiseller.de/info/index.php3?ac=OM.PS.PS003K00596T0409a
 
 



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RE: [PHP] inserting a variable into a variable

2001-06-20 Thread scott [gts] 

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:[EMAIL PROTECTED]]
 Sent: Wednesday, June 20, 2001 5:24 AM
 To: Hasan Niyaz; [EMAIL PROTECTED]
 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
 
 

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Re: [PHP] inserting a variable into a variable

2001-06-20 Thread Jonathan Sharp 

Or, you could do...

$cat = 'butch';
${rm_{$cat}_adt} = 'is cool'; // This would set the variable $rm_butch_adt
equal to 'is cool'

-js
- Original Message -
From: scott [gts] [EMAIL PROTECTED]
To: php [EMAIL PROTECTED]
Sent: Wednesday, June 20, 2001 8:37 AM
Subject: RE: [PHP] inserting a variable into a variable


 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:[EMAIL PROTECTED]]
  Sent: Wednesday, June 20, 2001 5:24 AM
  To: Hasan Niyaz; [EMAIL PROTECTED]
  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
  
 

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 To unsubscribe, e-mail: [EMAIL PROTECTED]
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 To contact the list administrators, e-mail: [EMAIL PROTECTED]



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