Question

我正在开发一个简单的天气应用程序，该应用程序使用的mysql数据库仅包含城市名称。然后我在数据库中查询所有名称，然后向openweathermap的API发送请求以获取天气信息。

function getCities() {
  con.connect();
  con.query(('SELECT city_name FROM cities'), (err, res) => {
    console.log(res);
    getWeather(cities);
  });
};

async function getWeather(cities) {
  var data = [];
  for (var i = 0; i < cities.length; i++) {
    var url = `http://api.openweathermap.org/data/2.5/weather?q=${cities[i].city_name}&units=metric&appid=271d1234d3f497eed5b1d80a07b3fcd1`;
    await request(url, (err, res, body) => {
      var json = JSON.parse(body);
      var weather = {
        city: json.name,
        temperature: json.main.temp,
        description: json.weather[0].description,
        icon: json.weather[0].icon
      };
      data.push(weather);
    });
  }
  console.log(data);
}

getCities()函数可以正常工作并返回所有城市，但是在getWeather函数中发生的错误尤其是这些错误：

Desktop/WeatherApp/node_modules/request-promise-core/lib/plumbing.js:130
            throw thrownException;
            ^

TypeError: Cannot read property 'temp' of undefined
    at Request.request [as _rp_callbackOrig] (/home/kristijan/Desktop/WeatherApp/app.js:60:41)
    at Request.plumbing.callback (/home/kristijan/Desktop/WeatherApp/node_modules/request-promise-core/lib/plumbing.js:76:39)
    at Request.RP$callback [as _callback] (/home/kristijan/Desktop/WeatherApp/node_modules/request-promise-core/lib/plumbing.js:46:31)
    at Request.self.callback (/home/kristijan/Desktop/WeatherApp/node_modules/request/request.js:185:22)
    at Request.emit (events.js:182:13)
    at Request.<anonymous> (/home/kristijan/Desktop/WeatherApp/node_modules/request/request.js:1161:10)
    at Request.emit (events.js:182:13)
    at IncomingMessage.<anonymous> (/home/kristijan/Desktop/WeatherApp/node_modules/request/request.js:1083:12)
    at Object.onceWrapper (events.js:273:13)
    at IncomingMessage.emit (events.js:187:15)
    at endReadableNT (_stream_readable.js:1094:12)
    at process._tickCallback (internal/process/next_tick.js:63:19)

据我所知，它是在从API获取结果之前跳过了箭头功能的内部？

Answer 1

await仅在等待的函数结果返回promise时才等待异步操作。 request()函数不会返回承诺（它与您传递的回调一起使用），因此await不会等待该结果。您可以使用request-promise库来代替，它确实会返回一个诺言，并且不会向其传递回调。

这是一个例子：

const rp = require('request-promise');

async function getWeather(cities) {
  let data = [];
  for (let i = 0; i < cities.length; i++) {
    let url = `http://api.openweathermap.org/data/2.5/weather?q=${cities[i].city_name}&units=metric&appid=271d1234d3f497eed5b1d80a07b3fcd1`;
    let body = await rp(url);
    let json = JSON.parse(body);
    let weather = {
        city: json.name,
        temperature: json.main.temp,
        description: json.weather[0].description,
        icon: json.weather[0].icon
    };
    data.push(weather);
  }
  console.log(data);
}

请注意，您还可以让请求API自动为您解析JSON：

const rp = require('request-promise');

async function getWeather(cities) {
  let data = [];
  for (let i = 0; i < cities.length; i++) {
    let url = `http://api.openweathermap.org/data/2.5/weather?q=${cities[i].city_name}&units=metric&appid=271d1234d3f497eed5b1d80a07b3fcd1`;
    let json = await rp({uri:url, json: true});
    let weather = {
        city: json.name,
        temperature: json.main.temp,
        description: json.weather[0].description,
        icon: json.weather[0].icon
    };
    data.push(weather);
  }
  console.log(data);
}

NodeJS没有从函数返回数据

1 个答案: