On Tue, Jul 28, 2009 at 2:48 PM, Matt Neimeyer<m...@neimeyer.org> wrote:
> Background: I'm converting a webapp from Visual FoxPro as a backend to
> MySQL... However one part of our app (a system status checker) is
> common code between the versions.
>
> I've got the following function... In English (in case it's not
> apparent), if the version of the app is 2.0 or higher, then use MySQL
> functions, otherwise use the ODBTP function to connect to VFP.
>
> function my_fetch_array($result)
> {
> global $Version;
> if(version_compare($Version,"2.0",">="))
> { $Ret = mysql_fetch_array($result); if(!$Ret) { } else {
> return $Ret; } }
> else { $Ret = odbtp_fetch_array($result); if(!$Ret) { } else {
> return $Ret; } }
> }
>
> This feels like a hack but works perfectly. Data is returned and all
> is right with the world. Until I added in extra "error reporting".
> When I change the if(!$Ret) portion as such...
>
> if(!$Ret) { die("myError".mysql_error()); } else { return $Ret; }
>
> It ALWAYS dies... and I see "myError" on the screen... If I change it
> like such...
>
> if(!$Ret) { } else { echo "notError"; return $Ret; }
>
> I always see the "notError" on the screen and $Ret gets returned.
>
> WHY does adding the die() inside the { } change the way the if is evaluated?
>
> By the way I've tested this on 4.4.x on OSX and Windows, and on 5.2.5
> on Windows...
>
> Thanks
>
> Matt
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
What's the output of var_dump($Ret) right before that if statement?
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
