I'm using Firebug 1.5.4. When I reference an undefined variable or some such, it breaks right where the problem occurs, and throws me into the debug view where I can see the stack and inspect variables.
However, when I throw my own exception, it just takes me to the console and prints out "uncaught exception: blah". I'd like it to break and let me inspect variables. How can I tell Firebug to do this?
Install Firebug 1.6b1 http://getfirebug.com/releases/firebug/1.6X, Firebug > Console > "the exception" Click the breakpoint selector in the left column. Run your code. Firebug breaks on that line.
Or Firebug > Console > [||] breaks on next error
The respondent was helpful but neglected something very key I was missing; the window.onerror event. Here is the full code:
window.onerror = function(msg) {
debugger;
}
Call Web Developer Debugger (Tools => Web Developer => Debugger or Ctrl + Shift + S), click gear icon and check "Pause on exception":
Or execute debugger; in Web Developer Console!
Official Web Developer Debugger docs: https://developer.mozilla.org/en-US/docs/Tools/Debugger
Put a debugger; statement in your code or use the Script tab of firebug to click on a line number (which inserts a breakpoint).
If you only want to do it when you throw an exception, you could put the debugger statement in a catch block.
来源：https://stackoverflow.com/questions/3754604/how-to-get-firebug-to-break-on-exception