CODEWITHSUNDEEP
jsonobjectnested
peace_love
peace_love

Reputation: 6471

How can I get access to multiple values of nested JSON object?

I try to access to my data json file:

[{"id":1,"name":"Maria","project":[{"id":5,"name":"Animals"},{"id":6,"name":"Cats"}]}

This is my approach:

data[0].name;

But like this I get only the result:

Animals

But I would need the result:

Animals, Cats

Upvotes: 2

Views: 1698

Answers (2)

Feathercrown
Feathercrown

Reputation: 2591

If that's your actual data object, then data[0].name would give you "Maria". If I'm reading this right, though, you want to get all the names from the project array. You can use Array.map to do it fairly easily. Note the use of an ES6 arrow function to quickly and easily take in the object and return its name.

var bigObject = [{"id":1,"name":"Maria","project":[{"id":5,"name":"Animals"},{"id":6,"name":"Cats"}]}];
var smallObject = [{"id":5,"name":"Animals"},{"id":6,"name":"Cats"}];

console.log("Getting the names from the full array/data structure: "+bigObject[0].project.map(obj => obj.name))
console.log("Getting the names from just the project array: "+smallObject.map(obj => obj.name))

EDIT: As per your comment on the other answer, you said you needed to use the solution in this function:

"render": function (data, type, row) {if(Array.isArray(data)){return data.name;}}

To achieve this, it looks like you should use my bottom solution of the first snippet like so:

var data = [{"id":5,"name":"Animals"},{"id":6,"name":"Cats"}];

function render(data, type, row){
  if(Array.isArray(data)){
    return data.map(obj => obj.name);
  }
};

console.log("Render returns \""+render(data)+"\" as an array.");

Upvotes: 1

Ashish
Ashish

Reputation: 4330

You are accessing only the name property of 0th index of project array. To access all object at a time you need to loop over the array. You can use Array.map for this. 

var data = [{"id":1,"name":"Maria","project":[{"id":5,"name":"Animals"},{"id":6,"name":"Cats"}]}]

var out = data[0].project.map(project => project.name).toString()

console.log(out)

Upvotes: 2

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