How to load json array

Question

Im trying to load person data from api : www.swapi.co . However i dont know how can i load movies titles instead of their api adress.

$(function(){

var list = $('.list');
var submit = $('.submit-people');
var people = '1';

submit.on('click',function(){
    people++;
    list.empty();

    $.ajax({

        type: 'GET',
        url: 'https://swapi.co/api/people/' + people + '/',

        success: function(data) {

            list.append('<li>Name:' + data.name + '</li>');
            list.append('<li>Height:' + data.height + '</li>');
            list.append('<li>Mass:' + data.mass + '</li>');
            list.append('<li>Gender:' + data.gender + '</li>');
            list.append('<li>Movies:' + data.films + '</li>');

        }

    })

})

my current code is here : https://jsfiddle.net/wtpy71d7/2/

really thanks for any help or suggestions

For each of the films in data.films, you have to do another query (another $.ajax) to fetch the details (e.g., movie title). This is a common feature of REST APIs. — Charles Wood
– Charles Wood, Commented Jul 25, 2017 at 16:31

Gurkan Yesilyurt · Accepted Answer · 2017-07-25 16:40:11Z

1

This might help you.

success: function(data) {
     ...
     for(var i = 0; i< data.films.length; i++ )
          $.ajax({
             type: "get",
             url: data.films[i],

             success: function (data) {
                list.append('<li>Movies:' + data.title + '</li>');
             }
          });
     }
}

answered Jul 25, 2017 at 16:40

Gurkan Yesilyurt

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Comments

Zach · Accepted Answer · 2017-07-25 16:39:51Z

0

It looks like they are giving you the URL for another API Request. What you could do is make another AJAX call and get each movie details, then parse those results and append them to your list. It could look something like this.

list.append('<li>Movies:' + getMovies(data.films) + '</li>');

Your next ajax call might look something like this:

var getMovies = function(moviesAPI) {

$.ajax({
    type: 'GET',
    url: moviesAPI,
  success: function(data) {
  return data;
  }
})

}

answered Jul 25, 2017 at 16:39

Zach

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