Got it! eval( '$newdata = "'.$data.'";');
Thanks! Shawn "Shawn McKenzie" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > Thanks Kevin! That works great. It outputs: hi my name is Shawn > > Now if I want to assign $data to another var, let's say $newdata and have it > eval the $name var inside of that. How would that work? > > Meaning I want to $newdata = hi my name is Shawn > > Thanks! > Shawn > > "Kevin Stone" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > The string you send to eval() must be valid PHP code. So try this.. > > > > eval( 'echo "'.$data.'";'); > > > > - Kevin > > > > > > ----- Original Message ----- > > From: "Shawn McKenzie" <[EMAIL PROTECTED]> > > To: <[EMAIL PROTECTED]> > > Sent: Monday, July 14, 2003 1:15 PM > > Subject: [PHP] Re: Eval var from query > > > > > > > eval($data) > > > > > > returns Parse error: parse error, unexpected T_STRING in > > > C:\apps\apache2\htdocs\test\query.php(23) : eval()'d code on line 1 > > > > > > Thanks! > > > Shawn > > > > > > "Shawn McKenzie" <[EMAIL PROTECTED]> wrote in message > > > news:[EMAIL PROTECTED] > > > > How can I evaluate a var that is from a text field of a database? > > > Example: > > > > > > > > MySQL field `name` = hi my name is $name > > > > > > > > In my script I have: > > > > > > > > $name = "Shawn"; > > > > > > > > After fetching a query result as an associative array I have the > > contents > > > of > > > > the `name` field in $data > > > > > > > > If I echo $data I get: hi my name is $name > > > > > > > > I would like to get: hi my name is Shawn > > > > > > > > TIA, > > > > Shawn > > > > > > > > > > > > > > > > > > > > -- > > > PHP General Mailing List (http://www.php.net/) > > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > > > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
