JSON.stringify returns "[object Object]" instead of the contents of the object

Question

Here I'm creating a JavaScript object and converting it to a JSON string, but JSON.stringify returns "[object Object]" in this case, instead of displaying the contents of the object. How can I work around this problem, so that the JSON string actually contains the contents of the object?

var theObject = {name:{firstName:"Mark", lastName:"Bob"}};
alert(JSON.stringify(theObject.toString())); //this alerts "[object Object]"

Alerts don't show objects, only strings, you should be using the console for that. And converting an object to a string does the same, you end up with [object Object], as that is the string representation of an object. — adeneo
– adeneo, Commented May 11, 2013 at 3:43
Ever wondered why [object Object] ? Have a look at this answer : stackoverflow.com/a/25419538/3001704 — kchetan
– kchetan, Commented Nov 15, 2016 at 9:33

Arbel · Accepted Answer · 2023-02-16 06:01:18Z

69

Use JSON.stringify(theObject);

edited Feb 16, 2023 at 6:01

answered May 11, 2013 at 3:40

Arbel

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2 Comments

oxygen Over a year ago

how to get name from json string

Chris22 Over a year ago

@oxygen Per OP example above, Use: console.log(JSON.stringify(theObject.name.firstName.toString())); // output: "Mark"

Kevin Boucher · Accepted Answer · 2013-05-11 03:52:38Z

8

JSON.stringify returns "[object Object]" in this case

That is because you are calling toString() on the object before serializing it:

JSON.stringify(theObject.toString()) /* <-- here */

Remove the toString() call and it should work fine:

alert( JSON.stringify( theObject ) );

answered May 11, 2013 at 3:52

Kevin Boucher

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hjpotter92 · Accepted Answer · 2013-05-11 03:41:11Z

7

theObject.toString()

The .toString() method is culprit. Remove it; and the fiddle shall work: http://jsfiddle.net/XX2sB/1/

answered May 11, 2013 at 3:41

hjpotter92

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Tamil Selvan C · Accepted Answer · 2013-05-11 03:42:26Z

3

Use

var theObject = {name:{firstName:"Mark", lastName:"Bob"}};
alert(JSON.stringify(theObject));

answered May 11, 2013 at 3:42

Tamil Selvan C

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