Question

Say I add to First.js:

$(document).ready(

function () {
    dosomething A
    });

function () {
    dosomething C
    });
});

and to Second.js:

$(document).ready(

function () {
    dosomething B
    });
});

will all 3 functions be executed after DOM is ready?

What will be the case when I register

to First.js:

$(document).ready(
A = function () {
    dosomething A
    });

C = function () {
    dosomething C
    });
});

to Second.js:

$(document).ready(
A = function () {
    dosomething A
    });

});

The later will override the first?

TIA

Answer 1

Your first example is invalid syntax. It will cause the javascript interpreter to throw an exception. You need to pass one and only one function to $(document).ready(fn). You can include multiple function calls inside the one function, but you can only pass one function to .ready().

Your second example is also a syntax error – an extra });. If that is removed, it will work and execute that one function.

Your third example in both first.js and second.js is also a syntax error. You can’t put arbitrary javascript as the parameter to .ready(). It must be one function reference with proper syntax.

Now, what you may have been trying to ask if you actually provided legal syntax in your examples is that all functions you pass to .ready(fn) will be executed when the document is ready. jQuery keeps an array of all functions that have been passed and executes them all when the document becomes ready. The jQuery documentation for .ready() does not specify the calling order if .ready() has been called multiple times with multiple functions, though one could examine the source code and see what the order is likely to be.