Question 70012 in Ajax Question posted in
You can find the official documentation here.

Convert jquery .getJSON to ajax async await for YouTube API – Jquery ajax

I am using YouTube API key to fetch data on search.

const key = "[API Key]";
const url = "https://www.googleapis.com/youtube/v3/search";

$(document).ready(function () {
  const options = {
    part: ["snippet"],
    key: key,
    maxResults: 10,
    q: "developers",
  };

  loadVids();

  function loadVids() {
    $.getJSON(url, options, function (data) {
      const videos = data.items;
      console.log(videos);
    });
  }
});

It is working properly but I want to convert $.getJSON() to async await using vanilla javascript. Please guide me how to do this.

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    I found a solution There is no need of part: ['snippet']. It is working properly without using that.

    const key = "[API Key]";
const url = "https://www.googleapis.com/youtube/v3/search";

const options = {
  part: ["snippet"],
  key: key,
  maxResults: 10,
  q: "developers",
};

async function loadVids() {
  const res = await fetch(
    `${url}?maxResults=${options.maxResults}&q=${options.q}&key=${options.key}`
  );

//can also write the above line in this more sensible way
//const res = await fetch(
//   `https://youtube.googleapis.com/youtube/v3/search?maxResults=${options.maxResults}&q=${options.q}&key=${options.key}`,
// );

  const data = await res.json();
  console.log(data);
}
loadVids();

    More understandable code

    const apiKey = "[API Key]";
const url = "https://www.googleapis.com/youtube/v3/search";

const options = {
  key: apiKey,
  maxResults: 10,
  q: "developers",
};

const { key, q, maxResults } = options;

async function loadVids() {
  const res = await fetch(`${url}?maxResults=${maxResults}&q=${q}&key=${key}`);
  const data = await res.json();
  console.log(data);
}
loadVids();

  2. 0

    You can do it by defining a Promise function like:

    function loadVids(){
  return new Promise((resolve, reject)=>{
    fetch('https://www.googleapis.com/youtube/v3/search', {method: 'GET', body: options})
    .then(response => response.json())
    .then(data => {
       console.log(data));
       resolve();
     })
     .cache(err => reject());     
  });
}

    and then await on it:

    async function f1() {
   var x = await loadVids();
   //other code...
}
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search