How can I check if two javascript objects have the same fields?

Question

Given:

I have the following two variables in Javascript:

var x = {
            dummy1: null
            dummy2: null
        };

// Return true
var y = {
            dummy1: 99,
            dummy2: 0
        }

// Return false
var y = "";


// Return false
var y = {
            dummy1: null
        };

// Return false
var y = {
            dummy1: null,
            dummy2: null,
            dummy3: 'z'
        }

// Return false
var y = null;

// Return false
var y = ""

Can anyone suggest to me how I can check if object x has the same field names as y ? Note that I am not checking the values of the parameters.

Please see this answer: http://stackoverflow.com/questions/1068834/object-comparison-in-javascript — hsiegeln, Jun 04 '14 at 08:38
meaning You want to make sure that y have "dummy1" and "dummy2" as x have? While exact values are not important? — przemo_li, Jun 04 '14 at 08:39
@hsiegeln, whereas that could be the answer to the OP's question, it is always good to write the answer here in case the link changes. — Hawk, Jun 04 '14 at 08:40
i know that but i'm saying that it is good practice @hsiegeln — Hawk, Jun 04 '14 at 08:43
@Hawk Actually, if the content of that link serves as an answer to this question, then the good practice is to mark this one as a duplicate and point to that one. But I suspect that the question here is actually somewhat different from that one. — JLRishe, Jun 04 '14 at 08:44
@Hawk no it's not. It's good to discourage duplicate questions and answers. — Ankit Jaiswal, Jun 04 '14 at 08:45
@AnkitJaiswal - I'm checking for fields. Not for fields with the same values. — , Jun 04 '14 at 08:46
Sorry for my delay in updating my question to make it more clear. My laptop battery ran out just when I was about to make the question more clear. What I need to do is to check field names. If there are 5 names in x then I need to check y is an object with fields and then see if those exact same field names appear in variable y. — , Jun 04 '14 at 08:52
@Melina Do you want to include inherited properties in the comparison or just use "own" properties? And what should the result be if either of the inputs is a number, nonempty string or boolean? — JLRishe, Jun 04 '14 at 09:33

Kartik626 · Answer 1 · 2014-06-04T12:33:32.543

4

in javascript, every object contains elements as array. for e.g.

   var bequal = true;

   if(Object.keys(x).length !== Object.keys(y).length) {
        bequal = false;
    }
    else {
   for (var prop in x) {
      if(!y.hasOwnProperty(prop)){
          bequal = false;
          break;
          //both objects are not equal..
      }
   }
  }

edited Jun 04 '14 at 12:33

answered Jun 04 '14 at 08:48

Kartik626

91
6

What if `y` has more keys than `x`? – Bergi Jun 04 '14 at 08:55

JLRishe · Accepted Answer · 2014-06-04T12:52:22.930

4

There are probably better names for these functions, but this should do it:

function hasAllProperties(subItem, superItem) {
    // Prevent error from using Object.keys() on non-object
    var subObj = Object(subItem),
        superObj = Object(superItem);

    if (!(subItem && superItem)) { return false; }

    return Object.keys(subObj).every(function (key) {
        return key in superObj;
    });
}

function allPropertiesShared(x, y) {
    return hasAllProperties(x, y) && 
           hasAllProperties(y, x);
}

edited Jun 04 '14 at 12:52

answered Jun 04 '14 at 08:51

JLRishe

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@JL - Can I ask would this also work and return false if either x or y were null or equal to the empty string? – Jun 04 '14 at 08:55
1

@Melina It will now. :) – JLRishe Jun 04 '14 at 08:56
@JL - Thanks for your very clear and simple solution that uses superItem :-) – Jun 04 '14 at 08:58
1

Just fixed an issue. `super` is a reserved keyword in JavaScript. – JLRishe Jun 04 '14 at 09:00
Uh, shouldn't it be `if (!x || !y) return false;`? You seem to still allow the case where one of them is falsy. – Bergi Jun 04 '14 at 12:51
@Bergi You're correct; I misapplied DeMorgan's law. It should have been `!(x && y)`. – JLRishe Jun 04 '14 at 12:52

thefourtheye · Answer 3 · 2014-06-04T09:15:54.270

2

function hasSameKeys(obj1, obj2) {
    var toString = Object.prototype.toString;
    if (toString.call(obj1).indexOf("Object") === -1
        || toString.call(obj2).indexOf("Object") === -1) {
        return false;
    }
    var keys1 = Object.keys(obj1), keys2 = Object.keys(obj2);

    return obj1.length === obj2.length && 
        keys1.every(function(currentKey) {
            return obj2.hasOwnProperty(currentKey);
    });
}

console.assert(hasSameKeys({a:1, b:2}, {b:3, a:1}) === true);
console.assert(hasSameKeys({a:1, b:2}, "")         === false);
console.assert(hasSameKeys({a:1, b:2}, {a:5})      === false);

The toString check makes sure that the objects being compared are really "Objects", not strings.

edited Jun 04 '14 at 09:15

answered Jun 04 '14 at 09:05

thefourtheye

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I think you meant to use `keys1.length` and `keys2.length`? `obj1.length` is `undefined` in this case. And perhaps it's better to use `in` than `hasOwnProperty`, as that will check inherited keys. – JLRishe Jun 04 '14 at 09:08
2

@JLRishe That is exactly the point, avoiding inherited properties. – thefourtheye Jun 04 '14 at 09:08

score 0 · Answer 4 · answered Jun 04 '14 at 08:46

0

Try this:

Object.keys(y).every(function (key) {
    return key in x;
});

answered Jun 04 '14 at 08:46

jupenur

1,005
6
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What if `x` has more keys than `y`? – thefourtheye Jun 04 '14 at 08:49
Then it's considered different and I need it to return false. – Jun 04 '14 at 08:53
If you also want to make sure `x` *doesn't* contain any keys that are not present in `y`, then you just have to do the same check with the variables reversed. – jupenur Jun 04 '14 at 08:55

Paul · Answer 5 · 2014-06-04T08:55:35.923

0

Use Object.keys()

Here's a simple example to show how this works:

$ nodejs // should also work in Browser
> x = {a:1,b:2,c:3}
{ a: 1, b: 2, c: 3 }
> y = {a:3,b:1,c:4}
{ a: 3, b: 1, c: 4 }
> z={message:"Die Batman!"}   // one of these things is not like the others
{ message: 'Die Batman!' }

> Object.keys(x)
[ 'a', 'b', 'c' ]
> Object.keys(y)
[ 'a', 'b', 'c' ]
> Object.keys(z)
[ 'message' ]

Don't compare Object.keys directly as they are arrays and will not ===

> Object.keys(x)===Object.keys(y)
false

Also, Object.keys() might give a different ordering in the array, resulting in unequal arrays like ['a','b','c'] vs ['b','a','c']. Fix for this is to add Array.sort()

I suggest using JSON.stringify to create string representations of the key lists

> JSON.stringify(Object.keys(x).sort())===JSON.stringify(Object.keys(y).sort())
true

> JSON.stringify(Object.keys(x).sort())===JSON.stringify(Object.keys(z).sort())
false

edited Jun 04 '14 at 08:55

answered Jun 04 '14 at 08:49

Paul

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Are you sure `keys` will return the keys in the same order for both the objects? – thefourtheye Jun 04 '14 at 08:50
Docs says "returns an array of a given object's own enumerable properties, in the same order as that provided by a for...in loop (the difference being that a for-in loop enumerates properties in the prototype chain as well)." so... maybe... I wondered this too. – Paul Jun 04 '14 at 08:51
1

No, I'm pretty sure we can't count on the properties to be in the same order. `for...in` doesn't guarantee any particular order, and the only guarantee of `Object.keys()` is that it will be in the same (implementation specific) order as `for...in`. – JLRishe Jun 04 '14 at 08:54
Well then we can bite the bullet and sort them all. – Paul Jun 04 '14 at 08:56

score 0 · Answer 6 · answered Jun 04 '14 at 08:51

0

You can try something like this

var isEqual = true;
for(var k in x)
{
    if(x[k] != y[k])
        isEqual = false;
}
for(var k in y)
{
    if(x[k] != y[k])
        isEqual = false;
}

answered Jun 04 '14 at 08:51

Ted Xu

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How can I check if two javascript objects have the same fields?

6 Answers6