How to debug in jQuery the easiest way?

Question

For example if I use this code: $('#iddoesnotexist').remove(); I will not receive an error if this id is not in the page, how to make to show an error?

By default this behaviour was chosen because otherwise jQuery would regularly throw NullReference Exceptions, but this is what I am looking for...

• Ignas: I use console.log for debugging :)
• elclanrs: You can use console.log or set breakpoints in devtools, or use firequery, multiple options as you can see
• madth3: I think the title was a bad choice. As I understand it, you look for an option that makes jQuery throws errors on empty matches but many people are just looking at the title and recommending debugging tools...
• user1502679: I took a look in Firebug console log and didn't found the errors for the specific code... (I checked for show strict warnings too.)

Answer 1

You can create your own jQuery extension, I suppose. I would* write it as an assertion:

$.fn.mustExist = function() {
 if(!this.length) {
 throw new Error("Collection was empty.");
 }
 return this;
};

And during development:

$('#iddoesnotexist').mustExist().remove();

_{* Actually, I wouldn't use jQuery; document.getElementById also protects you from this kind of thing ;)}

Answer 2

If you want this behavior to be default for all DOM selection, you can create a wrapper function for the jQuery function.

var my_$ = function(selector) {
 var result = jQuery.apply(this, arguments);
 if (typeof selector === "string" && !result.length)
 throw new Error("No matches found");
 return result; 
};
my_$("#foobar"); // No matches found

Be sure to use your own reference instead of overwriting the jQuery or $ identifiers.

Answer 3

See:

jQuery Debugging and http://jqueryfordesigners.com/debugging-tools/ and http://www.jquery4u.com/utilities/live-jquery-debugging-firebug/ and http://samdutton.wordpress.com/2010/09/16/jquery-debugging-tips/ and http://trainofthoughts.org/blog/2007/03/16/jquery-plugin-debug/

for jQuery debugging articles. And, as mentioned in the comments, the simplest way to debug is using console.log.