Or change the quote style to double (")
Just another option
Dan
(If I'm right this time... I really can't afford 88AUD/hr... :) )
-------------------
http://chrome.me.uk
-----Original Message-----
From: Jay Blanchard [mailto:[EMAIL PROTECTED]
Sent: 07 April 2006 18:12
To: David Clough; [email protected]
Subject: RE: [PHP] Parsing variables within string variables
[snip]
I have a variable containing a string that contains the names of
variables, and want to output the variable with the variables it
contains evaluated. E.g.
$foo contains 'cat'
$bar contains 'Hello $foo'
and I want to output $bar as
Hello cat
The problem is that if I use
echo $bar
I just get
Hello $foo
[/snip]
$bar = 'Hello' . $foo;
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
__________ NOD32 1.1475 (20060406) Information __________
This message was checked by NOD32 antivirus system.
http://www.eset.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
