Delete specific contents from a json object using another json object

Question

What I wanna do is tricky to explain...

This is the general idea:

var object1 = [{"number":1111,"test":0,"low":2131},{"number":1234,"test":0,"low":2684},{"number":4214,"test":0,"low":6345}];

var object2 = [{"number":1234}];

Some code here

output: var object1 = [{"number":1111,"test":0,"low":2131},{"number":4214,"test":0,"low":6345}];

I am only ever using the number attribute. The rest do not matter how ever the other attributes are still in there regardless. It is scraped content.

I want to remove a item from object using another object as the reference for what i want removed. Keeping in mind, this is all dynamic so i may have 3 items in one object and 100 in the other object.

I need to check object 1 and see if it contains anything from object 2 and If it does then i wanna delete the object 2 items from object 1 then output the remaining items from object 1. Confusing...

Any help would be appreciated.

https://stackoverflow.com/questions/10024866/remove-object-from-array-using-javascript — Harshana, Mar 22 '19 at 09:53
https://stackoverflow.com/questions/1187518/how-to-get-the-difference-between-two-arrays-in-javascript this help? — paqash, Mar 22 '19 at 09:55
What would happen if the match is partial? Let's say `[{ "number": 1234, "test": 1, "low": 4862 }]`, where *number* matches, but other attributes don't. — 3limin4t0r, Mar 22 '19 at 10:02
I edited my question. The other attributes are not in the second object. Only the number attribute is in the second object. — , Mar 22 '19 at 10:07

Raeesaa · Answer 1 · 2019-03-22T10:14:47.613

0

This is how this can be achieved using differenceWith function of lodash:

let difference = _.differenceWith(object1, object2, (obj1, obj2) => {
   return obj1.number === obj2.number
})

edited Mar 22 '19 at 10:14

answered Mar 22 '19 at 09:54

Raeesaa

3,267
2
22
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ttulka · Answer 2 · 2019-03-22T10:15:38.690

You need a function to compare two object, then you can filter found objects out:

var object1 = [{"number":1111,"test":0,"low":2131},{"number":1234,"test":0,"low":2684},{"number":4214,"test":0,"low":6345}];
var object2 = [{"number":1234,"test":0,"low":2684}];

const objectsEqual = (o1, o2) =>
    Object.keys(o1).length === Object.keys(o2).length 
        && Object.keys(o1).every(p => o1[p] === o2[p]);

var res = object1.filter(o1 => !object2.some(o2 => objectsEqual(o1, o2)));

console.log(res);

If the numer is the unique ID of the object, you can just filter:

var object1 = [{"number":1111,"test":0,"low":2131},{"number":1234,"test":0,"low":2684},{"number":4214,"test":0,"low":6345}];
var object2 = [{"number":1234,"test":0,"low":2684}];

var res = object1.filter(o1 => !object2.some(o2 => o1.number === o2.number));

console.log(res);

Serene Abraham Mathew · Answer 3 · 2019-03-22T10:08:01.270

0

You can achieve it using combination of Array.filter and Array.findIndex function

var object1 = [{"number":1111,"test":0,"low":2131},{"number":1234,"test":0,"low":2684},{"number":4214,"test":0,"low":6345}];

var object2 = [{"number":1234,"test":0,"low":2684}];


var result = object1.filter(function(obj1){

    return object2.findIndex(function(obj2){
        return obj2.number == obj1.number ;
    }) ==-1;

});

console.log(result)

edited Mar 22 '19 at 10:08

answered Mar 22 '19 at 09:55

Serene Abraham Mathew

358
4
16

This doesn't work when `var object2 = [{"number":1234,"test":123,"low":1234}];` – ttulka Mar 22 '19 at 10:05
This was not clear from the original question. As the question changed the requirement - your solution is okay as well. – ttulka Mar 22 '19 at 10:13

R3tep · Accepted Answer · 2019-03-22T10:19:54.943

0

You can reduce your object1 and use find too see if the numbers are into object1

var object1 = [{"appid":1111,"test":0,"low":2131},{"appid":1234,"test":0,"low":2684},{"appid":4214,"test":0,"low":6345}];

var object2 = [{"appid":1234}];

var object3 = object1.reduce((ac, v, i) => {
  return object2.find(function(obj) {
    return obj.appid == v.appid;
  }) ? ac :  [...ac, v];
}, [])

console.log(object3)

edited Mar 22 '19 at 10:19

answered Mar 22 '19 at 09:58

R3tep

12,512
10
48
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`JSON.stringify` doesn't work when object properties are in different order: `JSON.stringify({"number":1234,"test":0,"low":2684}) !== JSON.stringify({"number":1234,"low":2684,"test":0})` – ttulka Mar 22 '19 at 10:03
@3limin4t0r SO demands a minimal example, so we don't know where the objects come from. Maybe from an external resource, serialized as JSON - then it is likely that the order was changed and this solution won't therefore work. It could be fine, I want just to point this fact out. – ttulka Mar 22 '19 at 10:09
For some reason when i implement this code, the output is a empty object... though if i test it in jsfiddle it works using the supplied objects but the ones i get when i am scraping should be the same... im not great are coding so sorry if some of this seems a bit confusing. – Mar 22 '19 at 10:14
1

No errors. Here is a image of the array outputs in chromes console: https://i.imgur.com/NMOMQlR.jpg That is for the original objects not the code from here. So... i assume they are just objects? I could be wrong. This is the code im using to make them: Im parsing the first "object" which is just a string into json turning it into an object. The second one im pushing the contents into a object. – Mar 22 '19 at 10:18
1

@Sean I see the error I updated my post. You need to replace the key number by appid into the find function. – R3tep Mar 22 '19 at 10:18
Thank you very much. This is the new response using your code: https://i.imgur.com/ZGhbXsH.jpg I will post a comment if i have further issues with this specific thing. Iv got a few more tests to run while implementing this into a bigger script. so see how it goes. Cheers – Mar 22 '19 at 10:22

score 0 · Answer 5 · answered Mar 22 '19 at 10:09

let object1 = [{ "number": 1111, "test": 0, "low": 2131 }, { "number": 1234, "test": 0, "low": 2684 }, { "number": 4214, "test": 0, "low": 6345 }];
let object2 = [{ "number": 1234, "test": 0, "low": 2684 }];

// use unique key to findIndex, am using number key here replace this in your case.
object1 = object1.filter(obj => {
    return object2.findIndex(obj1 => {
        return obj.number === obj1.number;
    }) === -1;
});

3limin4t0r · Answer 6 · 2019-03-22T10:42:14.693

Here is another option removing all objects from the object1 array, that match with one or more attributes from the objects in object2.

var object1 = [{"number":1111,"test":0,"low":2131},{"number":1234,"test":0,"low":2684},{"number":4214,"test":0,"low":6345}],
    object2 = [{"number":1234}];

console.log(
  object1.filter(obj1 => {
    return !object2.some(obj2 => {
      for (let attrName in obj1) {
        if (obj1[attrName] === obj2[attrName]) return true;
      }
      return false;
    });
  })
);

// this can be reduced to the following if you don't want to look through
// all attributes, but only a predefined attribute

console.log(
  object1.filter(obj1 => !object2.some(obj2 => obj1.number === obj2.number))
);

Delete specific contents from a json object using another json object

6 Answers6