CODEWITHSUNDEEP
javascriptjquery
Sriram
Sriram

Reputation: 2999

What does [object Object] mean? (JavaScript)

One of my alerts is giving the following result: 

[object Object]

What does this mean exactly? (This was an alert of some jQuery object.)

Upvotes: 86

Views: 238807

Answers (8)

netrolite
netrolite

Reputation: 2703

In my case I got [object Objects] when I did the following:

const person2 = {
    name: "Jo",
    age: 27,
    address: {
        city: "Some city",
        state: "Some state"
    }
}

const requestedPerson = person2

const { 
    name,
    age, 
    address, 
    favFood = "Not defined"
} = requestedPerson

console.log(`Address: ${address}`);

And it was the same as using:

console.log("Address: " + address)

I got it to work by passing it as the second argument:

console.log("Address:", address)

Upvotes: 1

Ahmedakhtar11
Ahmedakhtar11

Reputation: 1458

In my case I was getting [Object, Object] because I was doing

console.log("particular_object" + particular_object)

Instead of

console.log("particular_object")
console.log(particular_object)

I guess adding another string in the same console.log of an object prevents the object from loading..

But most cases you just have to do:

JSON.stringify(particular_object))

UPDATE: Cool comment/clarification by @stevejboyer below!

If you are getting this error from passing text plus an object in the same console.log method by using a plus sign(+); You can actually use a comma to avoid any errors and still console text along with your object.

By doing this instead:

console.log("particular_object", particular_object)

Upvotes: 2

ncubica
ncubica

Reputation: 8485

As @Matt already explained the reason for [object object], I would like to elaborate on how to inspect the object's value. There are three options that come to my mind:

  • JSON.stringify(JSONobject)
  • console.log(JSONobject)
  • or iterate over the object

Basic example.

var jsonObj={
    property1 : "one",
    property2 : "two",
    property3 : "three",
    property4 : "fourth",
};

var strBuilder = [];
for(key in jsonObj) {
  if (jsonObj.hasOwnProperty(key)) {
    strBuilder.push("Key is " + key + ", value is " + jsonObj[key] + "\n");
  }
}

alert(strBuilder.join(""));
// or console.log(strBuilder.join(""))

https://jsfiddle.net/b1u6hfns/

Upvotes: 23

jack1536
jack1536

Reputation: 181

Another option is to use JSON.stringify(obj)

For example:

exampleObj = {'a':1,'b':2,'c':3};
alert(JSON.stringify(exampleObj))

https://www.w3schools.com/js/js_json_stringify.asp

Upvotes: 3

John Michelin
John Michelin

Reputation: 41

If you are popping it in the DOM then try wrapping it in 

<pre>
    <code>{JSON.stringify(REPLACE_WITH_OBJECT, null, 4)}</code>
</pre>

makes a little easier to visually parse.

Upvotes: 3

Guillaume250
Guillaume250

Reputation: 522

Alerts aren't the best for displaying objects. Try console.log? If you still see Object Object in the console, use JSON.parse like this > var obj = JSON.parse(yourObject); console.log(obj)

Upvotes: -1

Matt
Matt

Reputation: 75317

It means you are alerting an instance of an object. When alerting the object, toString() is called on the object, and the default implementation returns [object Object].

var objA = {};
var objB = new Object;
var objC = {};

objC.toString = function () { return "objC" };

alert(objA); // [object Object]
alert(objB); // [object Object]
alert(objC); // objC

If you want to inspect the object, you should either console.log it, JSON.stringify() it, or enumerate over it's properties and inspect them individually using for in.

Upvotes: 79

jetsie
jetsie

Reputation: 135

The alert() function can't output an object in a read-friendly manner. Try using console.log(object) instead, and fire up your browser's console to debug.

Upvotes: 10

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