.each not working on an Array. But .isArray returns true?

Question

I have created an object which contains a few items, including one which contains multiple objects each of which contains an array. Here is how it is structured.

$.myVar = {
    cp : "",
    ps : {
        m1 : ["001", "002", "003"],
        m2 : ["002", "004"]
    }
};

My scripts keep crashing saying that $.myVar.ps["m1"] has no method each.

When I got into Chrome's console to investigate, I run the following and get the displayed output.

$.myVar.ps["m1"]
["001", "002", "003"]
$.myVar.ps["m1"].each( function (i, p) {alert(i)})
TypeError: Object 001,002,003 has no method 'each'

Also if I run the following, it proves that m1 is an array.

$.isArray($.myVar.ps["m1"])
true

So it seems to agree with m1 is an array, but it refuses to treat it as such. Any idea what I'm doing wrong?

Domenic · Accepted Answer · 2012-05-22 16:36:57Z

19

each is not a native Array method; it is a method of jQuery objects, i.e. those created by the $ function. You can either do

$($.myVar.ps.m1).each(function (i, el) { /* ... */ });

(not recommended, because it creates an unnecessary jQuery object when it wraps the array in $(...)) or you can just use $.each:

$.each($.myVar.ps.m1, function (i, el) { /* ... */ });

The most recommended route, if you are using a modern browser (IE >=9), or using es5-shim, is to use the standard Array.prototype.forEach method:

$.myVar.ps.m1.forEach(function (el, i) { /* ... */ });

Note the different argument order (IMO better since you can then leave out the index if you don't need it).

edited May 22, 2012 at 16:36

answered Nov 24, 2010 at 19:06

Domenic

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kennytm · Accepted Answer · 2010-11-24 19:06:15Z

7

.each is only defined for jQuery objects. For pure Javascript arrays, use the "static method" $.each.

$.each($.myVar.ps["m1"], function(i,p) { alert(i); });

answered Nov 24, 2010 at 19:06

kennytm

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Daniel · Accepted Answer · 2010-11-24 19:05:27Z

-2

each is not a method of an array in javascript. try:

$($.myVar.ps["m1"]).each

answered Nov 24, 2010 at 19:05

Daniel

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1 Comment

ThiefMaster Over a year ago

Don't construct invalid jQuery objects; use $.each() instead.

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