Debug output by Checkbox

Hello,

i dont want to output the debug by default. I want the user to choose if the dialog should be displayed or not by a form (checkbox). If the box is checked , the debug should be 1, otherwise 0.

$validator->debug=1;

How can i build this in php?

Thanks!

This is an example script. If you want to make it an option, just add a field to the example script form and let the user pick if he wants to have debug enabled or not.

Just let me know if you have difficulty to do that.