node.js request(get) result return

Question

I want to return the result obtained from the request.

The code does not produce the desired result.

Please help me

CODE

const getData = async (gn) => {
  let gameData = undefined;

  gameData = await rq({
    method: "GET",
    url: base_url + "/get_gameData/OXquiz",
  }).then(function (response) {
    return response;
  });

  return gameData;
};

console.log(getData("OXquiz"));

RESULT

Promise { <pending> }

The result I want

{"status":0,"message":[{"ga_win_point":"100","ga_lose_point":"50","ga_time":"10","ga_max_number":"8"}],"url":""}

Does this answer your question? [How do I return the response from an asynchronous call?](https://stackoverflow.com/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call). TL;DR you want `getData("OXquiz").then(console.log)` — Phil, Jul 18 '22 at 05:13
Your entire function can be simplified to `const getData => (gn) => rq({ method: "GET", url: \`${base_url}/get_gameData/${gn}\` })` — Phil, Jul 18 '22 at 05:14

Jawak · Answer 1 · 2022-07-18T05:17:01.433

0

try

getData().then(item => console.log(item))

or

const data = await getData()

edited Jul 18 '22 at 05:17

answered Jul 18 '22 at 05:16

Jawak

1
1

score -1 · Answer 2 · answered Jul 18 '22 at 05:14

-1

Why are you using await and .then at a time. If you want to work asynchronously use the following snippet.

const getData = async (gn) => {
  let gameData = undefined;

  gameData = await rq({
    method: "GET",
    url: base_url + "/get_gameData/OXquiz",
  });

  return gameData;
};

console.log(getData("OXquiz"));

answered Jul 18 '22 at 05:14

Aravind

600
5
12

1

Thank you for your answer. I ran that code and still get "Promise { }" result. ㅠㅠㅠㅠㅠ – ISSAC_TOAST Jul 18 '22 at 05:46

node.js request(get) result return

2 Answers2