CODEWITHSUNDEEP
javascriptjqueryjavascript-objects
Sangram Anand
Sangram Anand

Reputation: 10844

Javascript - getting a particular object from list based on property value

I have a javascript json result list - self.data.allOrganizationsList

self.data.allOrganizationsList contains list of organization whose properties are 

customentityaccess, id, isLabsAccessAllowed, memberCount, organizationname, organizationwebsite

i want to access an organization object which has id = 402881702121fec10121520080930001

I am trying to get this from the below statement, but since each item in the list is an object, i am unable to get the desired result.

self.data.allOrganizationsList.id['402881702121fec10121520080930001']`
//(or)
self.data.allOrganizationsList["id"]['402881702121fec10121520080930001']

Other alternative is to use the $.each(self.data.allOrganizationList, function(index, item)).

But can i achieve the above result without iterating the - self.data.allOrganizationList

From the $.grept()

var myvar = $.grep(response.permissionedOrganizations, function(v) {
            return v.id === '40288170227c142201227eb56c5c000a';
        })[0];
console.log(myvar);

displays undefined in console. Am i missing something.

Upvotes: 0

Views: 2500

Answers (4)

Aki Suihkonen
Aki Suihkonen

Reputation: 20027

If you need to search more than one object based on the value, you can simply iterate the list once: 

for (var i in list) {
    var id = list[i].id;
    my_table[id] = list[i];  // for unique ids
}

For non unique ids make my_table as an associative array of arrays:

if (!my_table[id]) my_table[id] = [list[i]];
else my_table[id].push(list[i]);

In the latter case my_table['12399123'] can return an array of results.

Upvotes: 0

Gadde
Gadde

Reputation: 1471

jQuery.grep( array, function(elementOfArray, indexInArray) [, invert] )

Finds the elements of an array which satisfy a filter function. The original array is not affected. check

Upvotes: 2

pimvdb
pimvdb

Reputation: 154818

With your structure, you want $.grep to filter out the correct element:

var arr = [
  {id: 1, value: 2},
  {id: 3, value: 4}
];

$.grep(arr, function(v) { return v.id === 3; })[0];  // second element

Note that this does iterating behind the scenes - there's not a way to get the correct element by using [key] notation as that's not how your structure is defined.

Upvotes: 4

Evan Trimboli
Evan Trimboli

Reputation: 30082

If it's an array, then it is not keyed by the id, so you need to loop over each item and check the id property, as you described above.

For you to be able to check by id it would have to be something like:

var allOrganizationsList =    
    {
        "id1": {
            "memberCount": 1
        },
        "id2": {
            "memberCount": 2
        }
    };

allOrganizationsList["id1"].memberCount;

Upvotes: 0

Related Questions