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:jblanchard@xxxxxxxxxx] Sent: 07 April 2006 18:12 To: David Clough; php-general@xxxxxxxxxxxxx Subject: RE: 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