Search Javascript Array of Objects

Question

I'm trying to filter an array of javascript objects by a string. But I want the filter to look at every property and test it to see if the string is valid. AngularJS has a built in filter that does this, but I can't find any solutions for it on SO.

[
    {
        "title":"Mr.",
        "name":"John Smith",
        "firstName":"John",
        "lastName":"Smith"
    },
    {
        "title":"Mr.",
        "name":"Bill Smith",
        "firstName":"Bill",
        "lastName":"SMith"
    }

]

So for example if I entered 'Jo' for the text string it would bring back the object at index 0, which is pretty easy to do if you just want to search by a single property.

Now if I enter 'Mr' it should bring back both items at index 0 and index 1 because we're searching all of the properties.

Hope this makes sense as to what I'm asking.

EDIT: Very sorry was a late night last night and I left off very significants details in the data structure.

{
        "title":"Mr.",
        "name":"John Smith",
        "firstName":"John",
        "lastName":"Smith",
        "contactType":{
            "name":"test"
        },
        "addresses":[
            {"address":"Test Street One"},
            {"address":"Test Street Two"},
        ]
    },
    {
        "title":"Mr.",
        "name":"Bill Smith",
        "firstName":"Bill",
        "lastName":"SMith",
        "contactType":{
            "name":"test"
        },
        "addresses":[
            {"address":"Test Street One"},
            {"address":"Test Street Two"},
        ]
    }

So in this scenario the search would account for any type and any number of nested objects within objects. Sorry for forgetting this part.

"I'm trying to...", show us what you tried.

user1636522
– user1636522

2018-01-27 05:50:50 +00:00
Commented Jan 27, 2018 at 5:50 — user1636522
– user1636522, Commented Jan 27, 2018 at 5:50

Sajeetharan · Accepted Answer · 2018-01-13 07:41:53Z

1

You can loop through the objects of the array and use indexof to find the elements which has the input string,

DEMO

var myarray =[
    {
        "title":"Mr.",
        "name":"John Smith",
        "firstName":"John",
        "lastName":"Smith"
    },
    {
        "title":"Mr.",
        "name":"Bill Smith",
        "firstName":"Bill",
        "lastName":"SMith"
    }

];
var toSearch = "Mr";
var results =[];
for(var i=0; i<myarray.length; i++) {
  for(key in myarray[i]) {
    if(myarray[i][key].indexOf(toSearch)!=-1) {
      results.push(myarray[i]);
    }
  }
  }
  
  console.log(results)

answered Jan 13, 2018 at 7:41

Sajeetharan

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2 Comments

some Over a year ago

Is there a reason you don't use the filter function of arrays?

Sajeetharan Over a year ago

no you can still use filter, just that i explained the logic

31piy · Accepted Answer · 2018-01-13 08:53:34Z

1

You can use the combination of Array#filter and Object.values in order to achieve this:

let data = [{
    "title": "Mr.",
    "name": "John Smith",
    "firstName": "John",
    "lastName": "Smith"
  },
  {
    "title": "Mr.",
    "name": "Bill Smith",
    "firstName": "Bill",
    "lastName": "SMith"
  }

];

let string = 'Jo';
let results = data.filter(item => Object.values(item).some(value => value.includes(string)));

console.log(results);

edited Jan 13, 2018 at 8:53

answered Jan 13, 2018 at 7:49

31piy

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1 Comment

some Over a year ago

You could shorten it to let results = data.filter(item => Object.values(item).some(value => value.includes(string))); since that will be automatically returned.

user1636522 · Accepted Answer · 2018-01-13 07:48:26Z

Loop through all properties for each object :

var items = [{
    "title": "Mr.",
    "name": "John Smith",
    "firstName": "John",
    "lastName": "Smith"
}, {
    "title": "Mr.",
    "name": "Bill Smith",
    "firstName": "Bill",
    "lastName": "SMith"
}];

console.log("Jo :", items.filter(x => contains(x, "Jo")).map(x => x.name));
console.log("Mr :", items.filter(x => contains(x, "Mr")).map(x => x.name));

function contains (x, w) {
  for (let k in x) {
    if (x[k].indexOf(w) !== -1) {
      return true;
    }
  }
  return false;
}

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