Dear Paul, this is exactly the problem: the string including the dollar sign comes from the database.
The problem I have is that the echo statement parses the $bar reference, but not the $foo reference within it. So echo $bar generates Hello $foo which is better than $bar but doesn't get as far as Hello cat What I think I need is to send the results of the first echo to a second echo for parsing. Is there some way of doing that? Thanks for any help... David. In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] (Paul Novitski) wrote: > I wrote: > >You need to EVALUATE the string coming from the database: > > > >Assuming that $sDataField contains the string 'Hello $foo': > > > > $foo = "cat"; > > $sText = eval($sDataField); > > > >RESULT: $sText = "Hello cat" > > > >http://php.net/eval > > > I was assuming that you meant that the string "Hello $foo" -- > including the dollar sign -- came from the database. > > If $foo exists as a native PHP variable, I'd want to see your actual > code to tell why it's not being properly evaluated by the parser. > > Sometimes you need to use curly braces to help the PHP interpreter > differentiate a variable from the surrounding text: > > $bar = "Hello ${foo}amaran"; > > If you escape the dollar sign, the variable won't be evaluated: > > $bar = "Hello \$foo"; > RESULT: "Hello $foo" > > Paul -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
