Check key in deep nested object in array

Question

I have array of nested object. I have to check the property of key object and return its value. I have done it using the for loop and checking with children property exist or not. But I think it is not optimal way to do it. what will be most optimal way to do it. Here is the code array of object data. I have to get text for the id 121.

var abc = [
  {
    id: 1,
    text: 'One',
    children: [
      {id: 11, text: 'One One'},
      {id: 12, text: 'One two', 
       children: [ {id: 121, text: 'one two one'} ]}
    ]
  },
  {
    id: 2,
    text: 'two'
  }
];

My approach is very specific to this problem. Here it is

for(var val of abc){

  if(val.id == 121){
    console.log('in first loop',val.text);
    break;
  }

  if(Array.isArray(val.children)){

   for(var childVal of val.children) {
       if(childVal.id == 121){
        console.log('in first child', childVal.text); 
         break;
       }
     if(Array.isArray(childVal.children)){
       for(var nextChild of childVal.children){
         if(nextChild.id == 121){
           console.log('in next child', nextChild.text); 
           break;
         }

       }

       }
   }

  }

}

" I have done it using..." - Please add your approach. And add more details to your "requirement" or at least an example with a given in- and output. — Andreas
– Andreas, Commented Apr 1, 2020 at 7:39

Nenad Vracar · Accepted Answer · 2020-04-01 07:44:19Z

2

You could create recursive function using for...in loop that returns the matched object and then you can get its text property.

var abc = [{"id":1,"text":"One","children":[{"id":11,"text":"One One"},{"id":12,"text":"One two","children":[{"id":121,"text":"one two one"}]}]},{"id":2,"text":"two"}]

function getProp(data, key, value) {
  let result = null;


  for (let i in data) {
    if (typeof data[i] == 'object' && !result) {
      result = getProp(data[i], key, value)
    }

    if (i == key && data[i] == value) {
      result = data
    }
  }

  return result;
}

const result = getProp(abc, 'id', 121)
console.log(result)

answered Apr 1, 2020 at 7:44

Nenad Vracar

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1 Comment

Abhishek Over a year ago

Thanks for this recurring approach. I don't know if we can utilize ES6 features here. Just a thought

Nina Scholz · Accepted Answer · 2020-04-01 09:21:36Z

1

You could take an approach with a short circuit and return a result from a wanted property.

const
    getValue = (object, key, id) => {
        const search = o => {
            if (!o || typeof o !== 'object') return;
            if (o.id === id) return { value: o[key] };
            var value;
            Object.values(o).some(p => value = search(p));
            return value;
        };
        return search(object)?.value;
    };

var array = [{ id: 1, text: 'One', children: [{ id: 11, text: 'One One' }, { id: 12, text: 'One two', children: [{ id: 121, text: 'one two one' }] }] }, { id: 2, text: 'two' }];

console.log(getValue(array, 'text', 121));
console.log(getValue(array, 'text', 3000));

edited Apr 1, 2020 at 9:21

answered Apr 1, 2020 at 9:07

Nina Scholz

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4 Comments

Abhishek Over a year ago

nice !!! I think just need to remove ? from return search(object)?.value;

Nina Scholz Over a year ago

no, without optional chaining operator ?., you would need an assignemnt of the result and check if the result is truthy and then get value from it.

Abhishek Over a year ago

I run the code and it throw the syntax error for the same

Roko C. Buljan Over a year ago

@Abhishek are you using IE?

Roko C. Buljan · Accepted Answer · 2020-04-01 10:45:27Z

0

Given your Nodes have nested Nodes in a children property, using a recursion and Array.prototype.find() to find a Node by ID:

const getNode = (a, id, c = 'children', r) => {
  const rec = a => a.find(o => o.id==id && (r=o) || c in o && rec(o[c]));
  return rec(a) && r;
};


const abc = [{id: 1, text: 'One', children: [{id: 11, text: 'One one'}, {id: 12, text: 'One two', children: [{id: 121, text: 'One two one'}]}]}, {id: 2, text: 'Two' }];
console.log( getNode(abc, 121)?.text ); // One two one

edited Apr 1, 2020 at 10:45

answered Apr 1, 2020 at 10:38

Roko C. Buljan

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