Question

when it gives a error I want him to do 2 things.

1. echo nl2br($qVraagOp);
2. mysql_error();

so i thougth:

$rVraagOp = mysql_query( $qVraagOp ) or die( echo nl2br($qVraagOp); mysql_error(); );

I can write a function that does these two things and then call that but that a bit redundant.
is there a other way?

Matthy

Answer 1

I would write a function like:

function PrintDebug($DebugData)
{ 
  if($DebugMode == 1)
  { 
      return nl2br($DebugData) . "<br />" . mysql_error();
  }
  else
  {
      return "Ops stuff got messed up!!!";
  }
} 

And I would use it like

$rVraagOp = mysql_query( $qVraagOp ) or die(PrintDebug($qVraagOp));

It’s your job to save the debug status in a DB(think is maintenance mode active or not) and logging to a file would be helpful too. Also I want to point out that I didn’t tested it.