I realized that may not make any sense. Following up the eample relates to your original query like this..
> MySQL field `name` = hi my name is $name > In my script I have: > $name = "Shawn"; $data = 'hi my name is $name'; $code = 'echo "'.$data.'";'; $name = 'Shawn'; eval($code); // prints "hi may name is Shawn". Hope that makes it more clear. - Kevin ----- Original Message ----- From: "Kevin Stone" <[EMAIL PROTECTED]> To: "PHP-GENERAL" <[EMAIL PROTECTED]> Sent: Monday, July 14, 2003 1:23 PM Subject: Re: [PHP] Re: Eval var from query > 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 > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
