Javascript loop an array

Question

I have a javascript array written like this...

var json = [
 {"id":"1", "title":"Test 1", "comment":"This is the first test"},
 {"id":"2", "title":"Test 2", "comment":"This is the second test"}
];

what I am trying to do is get each one of the ids.

I have been trying this

for(x in json[0]){
 alert(x.id); 
}

But no luck, can someone point me in the right direction? Please and thank you :)

• Felix Kling: possible duplicate of for ... in loop with string array outputs indices
• Felix Kling: You probably find it helpful as well to read the for...in documentation.

Answer 1

x in your example is giving you the indexes of you array, not the objects. You could do:

for(x in json) {
 alert(json[x].id); 
}

but to loop through an array you're really better off with a "regular" for loop

for (var i = 0, max = json.length; i < max; i++) {
 alert(json[i].id);
}

Answer 2

Any modern browser will allow you to do it easily:

var ids = json.map(function(i) { return i.id; });
// and now you have an array of ids!

Sadly, "modern" does not include IE 8 and earlier.

You can also do the "mundane" form, which is guaranteed to work in all browsers. I see Adam Rackis has beat me to it though, so I 'll go upvote his answer and you should probably do so as well.

• Adam Rackis: +1 - nice - I really need to start using these ES5 methods more. And Sadly, "modern" does not include IE 8 ftw

Answer 3

This is one possible solution:

var json = [{"id":"1","title":"Test 1","comment":"This is the first test"},{"id":"2","title":"Test 2","comment":"This is the second test"}];
for (var i = 0, len = json.length; i < len; i++) {
 alert(json[i].id);
}

Answer 4

A for(x in y) loop in JavaScript gives you the indexes in that array (e.g., so that x[y] gives you the current element).

The two proper ways to loop through an array in JavaScript are:

for(x = 0; x < y.length; x++) { // (this can only loop through arrays)
 // do something with y[x]
}
for(x in y) { // (this can loop through objects too)
 // do something with y[x]
}