jquery function called multiple times

Question

I have 2 js files. In the first one I have this:

var functionName = "video"; 
var cont = 1;

$(function() { 
    window.control = function control() {   
        var tipo1 = functionName + cont + "();";
        var tipo2 = eval(tipo1);
        tipo2;
        cont++;
    });

In the second one:

function video1() {
    control();
}

function video2() {
    control();
}

The first time was fine, but in the second, first execute video1() and then video2(), why?

This isn't going to fix the issue, but please don't use eval(). Try window[tipo1](); instead. Note you may need to change window depending on your logic structure. — Rory McCrossan
– Rory McCrossan, Commented Jul 2, 2015 at 10:56
thanks for the advice, but if I call window[tipo1](); there is an error in console: "window[tipo1] is not a function" — Oscar LT
– Oscar LT, Commented Jul 2, 2015 at 11:15

Curtis · Accepted Answer · 2015-07-02 10:55:13Z

3

Your definition is wrong:

window.control = function control() {

I imagine because of this it's firing control() execution.

Change this to:

window.control = function() {

Also I see no reason for defining this function at DOM ready state. It will just cause confusion and potential reference issues. The definition of a function is only ran at execution point, these should potentially be on DOM ready state depending on their use.

answered Jul 2, 2015 at 10:55

Curtis

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1 Comment

Oscar LT Over a year ago

thanks, but don't use to me. I remove window and I just have function control(). But it is the same, first execute video1 and then video2.

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