Thanks for all these responses, but unless I'm missing something none of them work for what I need. Quotes are irrelevant: with the string "Hello $foo" in $bar echo "$bar" echo $bar both produce Hello $foo and echo '$bar' produces $bar I can't use any of the answers like 'Hello'.$foo because I have to parse the string 'Hello $foo' as it comes from the database: I don't get to construct it. I did hold out more hope for the eval function, but it seems to me that this is for PHP code in a database, not to evaluate variables. If anyone else can provide the silver bullet, I will be very grateful. David. In article <David.Clough-6046F3.17371407042006@xxxxxxxxxxxxx>, David.Clough@xxxxxxxxxxxx (David Clough) wrote: > I've been bashing my head against the wall on this, and would be glad of > help to stop. > > 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 > > Note that $bar is loaded from a database query, so I can't control its > contents: I just have to parse it. > > Any help appreciated. > > David. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
