How to get object with key and value using Jquery each method..?

Question

I am not able to get a key and value pairs using the each function of jquery. How to get the key and values object using jQuery each function..?

my try:

var result = '{"FirstName":"John","LastName":"Doe","Email":"[email protected]","Phone":"123 dead drive"}';

$.each($.parseJSON(result), function(n, v) {
    console.log({n: v});
});

result i am getting as:

Object {n: "John"} 
Object {n: "Doe"} 
Object {n: "[email protected]"} 
Object {n: "123 dead drive"}

but i am looking for :

    Object {FirstName: "John"} 
    Object {LastName: "Doe"} 
    Object {Email: "[email protected]"} 
    Object {Phone: "123 dead drive"}

what is the correct approach to get this done..?

Thanks in advance.

Arun P Johny · Accepted Answer · 2013-06-19 04:28:49Z

2

You need

$.each($.parseJSON(result), function(n, v) {
    var obj = {};
    obj[n] = v;
    console.log(obj);
});

answered Jun 19, 2013 at 4:28

Arun P Johny

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Comments

Pablo S G Pacheco · Accepted Answer · 2013-06-19 04:49:56Z

1

$.each($.parseJSON(result), function(n, v) {
    console.log('{'+n +': '+ v +'}');    
});

answered Jun 19, 2013 at 4:49

Pablo S G Pacheco

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Comments

som · Accepted Answer · 2013-06-19 04:43:33Z

0

$.each($.parseJSON(result), function(n, v) {
    console.log(n +":"+ v );
});

{} treated as object.You log variable as object.

answered Jun 19, 2013 at 4:43

som

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Comments

Vishal Suthar · Accepted Answer · 2013-06-19 04:45:12Z

0

Here you go:

var resultJSON = '{"FirstName":"John","LastName":"Doe","Email":"[email protected]","Phone":"123 dead drive"}';

var result = $.parseJSON(resultJSON);
$.each(result, function(k, v) {
    alert(k + ': ' + v);
});

Live Demo

answered Jun 19, 2013 at 4:45

Vishal Suthar

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Comments

Colin Brock · Accepted Answer · 2013-06-19 04:47:18Z

-1

will give the desired output

var arr = [];

$.each($.parseJSON(result), function(n, v) {
        arr[n] = v;    
    });

console.log(arr);

edited Jun 19, 2013 at 4:47

Colin Brock

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answered Jun 19, 2013 at 4:39

Rachit Doshi

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3 Comments

tomsseisums Over a year ago

This will not work, because basic array cannot be associative. You need arr = {}; here.

Rachit Doshi Over a year ago

It is giving the desired output

Colin Brock Over a year ago

There is no need for <code> tags, simply indent with 4 spaces.

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