Javascript - Filter array results into dropdown list

Question

I have an array as shown below where I want to filter the data and insert the filtered list into a dropdown list.

Here is the data:

{"people":[
    {
      "id":"100",
      "name":"name 1",
      "desc":"desc 1",
      "class": "a"
    },
    {
      "id":"192",
      "name":"name 2",
      "desc":"desc 2",
      "class": "b" 
    },
    {
      "id":"324",
      "name":"name 3",
      "desc":"desc 3",
      "class": "b" 
    },
    {
      "id":"324",
      "name":"name 4",
      "desc":"desc 4",
      "class": "a" 
    },
    {
      "id":"324",
      "name":"name 5",
      "desc":"desc 5",
      "class": "a"
    }
]}

What I now need to do is to add the array with "class" of "a" and then insert the id's of these into a dropdown list like:

$('#myselect').append($('<option>').text(value.id).attr('value', value.id));

How can I do this?

Can you mention the expected output array ? do you want to filter the records with class="a" — Pugazh
– Pugazh, Commented Sep 20, 2016 at 13:09
for (var i = 0; i < value.people.length; i++) { if(value.people[i].class == "a") { $('#myselect').append($('<option value="' + value.people[i].id + '">' + value.people[i].id + '</option>')); } } — Dhananjay Sakhare
– Dhananjay Sakhare, Commented Sep 20, 2016 at 13:25

Pugazh · Accepted Answer · 2016-09-20 13:17:47Z

1

Here is an solution using Array.filter()

var obj = {
  "people": [{
    "id": "100",
    "name": "name 1",
    "desc": "desc 1",
    "class": "a"
  }, {
    "id": "192",
    "name": "name 2",
    "desc": "desc 2",
    "class": "b"
  }, {
    "id": "324",
    "name": "name 3",
    "desc": "desc 3",
    "class": "b"
  }, {
    "id": "324",
    "name": "name 4",
    "desc": "desc 4",
    "class": "a"
  }, {
    "id": "324",
    "name": "name 5",
    "desc": "desc 5",
    "class": "a"
  }]
};

obj.people.filter(function(val) {
  if (val.class == "a") {
    $('#myselect').append($('<option>').text(val.id).attr('value', val.id));
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select id="myselect">
</select>

answered Sep 20, 2016 at 13:17

Pugazh

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

You can try something like this:

Logic

Get unique class names.
Loop on this unique values and filter data based on it
Create htmlString based on object and append it to necessary select

If you feel, adding selects in html is an over kill, you can generate them also based on data.

Sample

var data = {
  "people": [{
    "id": "100",
    "name": "name 1",
    "desc": "desc 1",
    "class": "a"
  }, {
    "id": "192",
    "name": "name 2",
    "desc": "desc 2",
    "class": "b"
  }, {
    "id": "324",
    "name": "name 3",
    "desc": "desc 3",
    "class": "b"
  }, {
    "id": "324",
    "name": "name 4",
    "desc": "desc 4",
    "class": "a"
  }, {
    "id": "324",
    "name": "name 5",
    "desc": "desc 5",
    "class": "a"
  }]
}

var uniqueClasses = {};
data.people.forEach(function(p) {
  uniqueClasses[p.class] = 1;
});

Object.keys(uniqueClasses).forEach(function(c) {
  var _html = data.people.filter(function(p) {
    return p.class === c
  }).map(function(p) {
    return "<option value='" + p.id + "'>" + p.name + "</option>"
  }).join("");
  $("#class_" + c).append(_html);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<select id="class_a"></select>
<select id="class_b"></select>

Olivier Boissé · Accepted Answer · 2016-09-20 13:34:16Z

0

You can do like this with lodash library

//Filter with class = 'a'
var peopleWithClassA = _.filter(obj.people, { class: 'a' });

//Append option elem for each filtered people
_.forEach(peopleWithClassA, function(people) {
   $('#myselect').append($('<option>')
                 .text(people.id)
                 .attr('value', people.id));
});

Without lodash :

obj.people
.filter(function(p) {
    return p.class === 'a';
})
.forEach(function(p) {
    $('#myselect').append($('<option>').text(p.id).attr('value', p.id));
});

edited Sep 20, 2016 at 13:34

answered Sep 20, 2016 at 13:17

Olivier Boissé

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