JS map two object/array

Question

I have 2 objects with format:

obj1 = [
          {
            "code": "in_today",
            "text": "Today"
          },
          {
            "code": "in_week",
            "text": "This week"
          },
          {
            "code": "in_month",
            "text": "This month"
          },
          {
            "code": "normal",
            "text": "Other"
          }
        ]

obj2 stores "code" value which defined in obj1:

obj2 = ["in_today", "in_week", "normal"]

How can I use obj2's values to change obj1 into something likes:

[
    {
        "code": "in_today",
        "text": "Today",
        "selected": true
    },
    {
        "code": "in_week",
        "text": "This week",
        "selected": true
    },
    {
        "code": "in_month",
        "text": "This month",
        "selected": false
    },
    {
        "code": "normal",
        "text": "Other"
        "selected": true
    }
]

What's the best solution for this case? Thanks!

Is the format of the first object fixed or are you creating it yourself? — Martina
– Martina, Commented Apr 9, 2017 at 17:47

Rob M. · Accepted Answer · 2017-04-09 17:44:54Z

1

You can use Array.map to transform your objects based on whether their code is in the obj2 array:

var obj1 = [
  {
    "code": "in_today",
    "text": "Today"
  },
  {
    "code": "in_week",
    "text": "This week"
  },
  {
    "code": "in_month",
    "text": "This month"
  },
  {
    "code": "normal",
    "text": "Other"
  }
]

var obj2 = ["in_today", "in_week", "normal"]

var newObject = obj1.map(function(obj) {
  if (obj2.indexOf(obj.code) > -1) {
    obj.selected = true;
  } else {
    obj.selected = false;
  }
  return obj;
})

console.log(newObject)

Or a bit simpler if ES6 is available:

const newObject = obj1.map((obj) => {
  obj.selected = obj2.includes(obj.code);
  return obj;
})

answered Apr 9, 2017 at 17:44

Rob M.

36.6k6 gold badges56 silver badges51 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

kind user · Accepted Answer · 2017-04-09 17:45:22Z

0

Simple&quick solution using Array#forEach.

var obj1 = [{"code":"in_today","text":"Today"},{"code":"in_week","text":"This week"},{"code":"in_month","text":"This month"},{"code":"normal","text":"Other"}],
    obj2 = ["in_today", "in_week", "normal"];

    obj1.forEach(v => v.selected = obj2.indexOf(v.code) > -1);

    console.log(obj1);

answered Apr 9, 2017 at 17:45

kind user

42k8 gold badges69 silver badges78 bronze badges

Comments

Cassandra Fox · Accepted Answer · 2017-04-09 17:45:55Z

0

You'll want to do something like this:

for (var i = 0; i < obj1.length; ++i) {
  if (obj2.indexOf(obj1[i].code) == -1) {
    obj1[i].selected = false;
  } else {
    obj1[i].selected = true;
  }
}

Essentially, you just loop over obj1, then check if the value of obj1.code is present in obj2, then set selected accordingly.

answered Apr 9, 2017 at 17:45

Cassandra Fox

666 bronze badges

Comments

dev8080 · Accepted Answer · 2017-04-09 17:53:31Z

0

obj1 = [
          {
            "code": "in_today",
            "text": "Today"
          },
          {
            "code": "in_week",
            "text": "This week"
          },
          {
            "code": "in_month",
            "text": "This month"
          },
          {
            "code": "normal",
            "text": "Other"
          }
        ]

obj2 = ["in_today", "in_week", "normal"];
    
    obj1.forEach( function(elem){

     if( obj2.indexOf(elem.code) > -1)
       elem.selected = true;
     else
       elem.selected = false;

    });
    
    console.log( JSON.stringify(obj1) );

edited Apr 9, 2017 at 17:53

answered Apr 9, 2017 at 17:44

dev8080

4,0051 gold badge14 silver badges18 bronze badges

1 Comment

dev8080 Over a year ago

it's a snippet now, runs as expected.

Collectives™ on Stack Overflow

JS map two object/array

4 Answers 4

Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related