Openweathermap API ASK不起作用

Question

Hello im trying to do simple weather app but i cant get a request from API.

Somone can help me?

$(document).ready(function testCall(){
  $.ajax({
    url: `http://api.openweathermap.org/data/2.5/weather?`,
    dataType: `JSON`,
    id: `7531860`,
    APPID: `8d3beea42a62c24610f3d57a8082a9c0`,
    success: function(data) {
      console.log("lol")
    },
    error: function(){
      console.log("WHy?")
    }

  });
})

Answer 1

Okay, first thing, the error callback function receives arguments you can use to learn a about what occurred. Using your code, slightly modified, I was getting a status code of 0.

error: function(xhr, exception){
  console.log("WHy?")
  console.log(xhr.status);
  console.log(exception);
}

I am not too familiar with using jQuery's ajax, so maybe someone can help you with that if you wish to use jQuery. Below I have provided a way to accomplish what you want using the built in Fetch API. Here is the Fiddle .

function getWeatherData(Id, appId) {
    return fetch(`https://api.openweathermap.org/data/2.5/weather?id=${Id}&APPID=${appId}`)
    .then(res => res.json())
}

const YOUR_ID = null; // Replace Me
const YOUR_APP_ID = null; // Replace Me

getWeatherData(YOUR_ID, YOUR_APP_ID)
.then(weatherData => {
   // Work with your data
   console.log(weatherData);
});

Including your API information was helpful for solving this, but you should now reset it!

EDIT : in rewriting with fetch, I figured out what was wrong. That status code was caused by a typo in the url. You need to request over https, not http. You also need to add the query params into the request, not on the object. See below and the fiddle .

$(document).ready(function testCall(){
  $.ajax({
    url: `https://api.openweathermap.org/data/2.5/weather?id=PUT_YOUR_ID_HERE&APPID=PUT_YOUR_APP_ID_HERE`,
    dataType: `JSON`,
    success: function(data) {
    console.log(data);
    },
    error: function(xhr, exception){
    console.log("WHy?")
    console.log(xhr);
    console.log(xhr.status);
    console.log(exception);
    }
  });
});