Question

以下是我要在nodejs中执行的操作。 Rest API将城市名称作为输入。我正在尝试使用输入城市的地理编码API来获取经度和纬度。然后，我正在尝试使用纬度和经度，使用其他API获取最近城市的列表。然后，对于所有这些城市，我正在获取天气报告，对于这些城市，然后，我正在获取是否有水，并将其作为JSON返回。

如您所见，有很多 then ，此练习的目的是避免嵌套回调。

我正在使用async / await，它应该消除了嵌套的then函数。但是我看不到另一种方法。完整的代码段如下。我要修复的丑陋部分是requester.makeRequest（）

以下仅是必要代码的片段，而不是完整的工作代码。任何有关如何解开此问题的帮助将不胜感激。

app.get('/search', function(req, res, next) {
  const requester = {
      lastRequest: new Date(),
      makeRequest: async function(url) {
        const response = await fetch(url);
        const json = await response.json();
        return json;
      }
  };

requester.makeRequest(geocode_url +`?locate=${req.query.q}&json=1`
    + geocode_token)
  .then(function(city){
    var final_result = []
    var lat = city.latt;
    var long = city.longt;
    // request to get list of cities closer to that location,
    //takes latitude and longitude as parameters
    requester.makeRequest(metaweather_url + '?lattlong='
     + lat + ',' + long)
    .then(function(closer_cities) {
      var cities_len = closer_cities.length
      for(i = 0; i < closer_cities.length; i++) {
        woeid = closer_cities[i].woeid
        //request to get weather using woeid parameter
        requester.makeRequest(woeid_url + woeid)
        .then(function(weather) {
          var lattlong = weather.latt_long;
          requester.makeRequest(onwater_url+ lattlong +
          '?access_token=' + water_access_token)
          .then(function(onwater) {
            var temp = Object.assign(weather, onwater)
            final_result.push(temp)
            if (final_result.length == cities_len) {
              res.status(200).json({error: false,
                data: {message: final_result}})
            }
          })
        })
       }
      })
    })
  })

Answer 1

我想你那时还需要一个

requester.makeRequest(geocode_url +`?locate=${req.query.q}&json=1`
    + geocode_token)
  .then(async function(city){
    var final_result = []
    var lat = city.latt;
    var long = city.longt;
    // request to get list of cities closer to that location,
    //takes latitude and longitude as parameters
    closer_cities = await requester.makeRequest(metaweather_url + '?lattlong='+ lat + ',' + long);
    var cities_len = closer_cities.length;
    for(i = 0; i < closer_cities.length; i++) {
      woeid = closer_cities[i].woeid
      //request to get weather using woeid parameter
      weather = await requester.makeRequest(woeid_url + woeid)
      var lattlong = weather.latt_long;
      onwater = await awaitrequester.makeRequest(onwater_url+ lattlong + '?access_token=' + water_access_token)
      var temp = Object.assign(weather, onwater)
      final_result.push(temp)
      if (final_result.length == cities_len) {
        res.status(200).json({error: false, data: {message: final_result}})
      }
    }
  })

编辑：我真的不认为我的回答与您的问题有关

Answer 2

此行：requester.makeRequest ... .then(function(city){

将.then(function(city){替换为var city = await requester.makeRequest，city将具有应许的兑现价值，其余then s则应这样做：

（请注意，await仅在async函数内部使用，可以使用iife）

(async () => {


  var city = await requester.makeRequest(`${geocode_url}?locate=${req.query.q}&json=1${geocode_token}`);

  var final_result = []
  var lat = city.latt;
  var long = city.longt;
  // request to get list of cities closer to that location,
  //takes latitude and longitude as parameters
  var closer_cities = await requester.makeRequest(`${metaweather_url}?lattlong=${lat},${long}`);

  var cities_len = closer_cities.length;

  for (i = 0; i < closer_cities.length; i++) {
    woeid = closer_cities[i].woeid
    //request to get weather using woeid parameter
    var weather = await requester.makeRequest(woeid_url + woeid);

    var lattlong = weather.latt_long;
    var onwater = await requester.makeRequest(`${onwater_url}${lattlong}?access_token=${water_access_token}`);

    var temp = Object.assign(weather, onwater)
    final_result.push(temp)
    if (final_result.length == cities_len) {
      res.status(200).json({
        error: false,
        data: {
          message: final_result
        }
      })
    }
  }

})();

Answer 3

then首先被滥用，因为它会导致回调地狱。 Promise是基于回调的，但它们支持链接，该链接应消除嵌套的回调。

应该是：

  requester.makeRequest(geocode_url +`?locate=${req.query.q}&json=1` + geocode_token)
  .then(function(city){
    var final_result = []
    var lat = city.latt;
    var long = city.longt;

    return requester.makeRequest(metaweather_url + '?lattlong='
     + lat + ',' + long)
  })
  .then(function(closer_cities) {
     ...
  });

如果then中有一个承诺，则应将其返回。这样一来，回调嵌套就不会超过一个级别。

await是then的语法糖，拒绝也应处理：

app.get('/search', function(req, res, next) {
  try {
    ...
    const city = await requester.makeRequest(geocode_url +`?locate=${req.query.q}&json=1`
      + geocode_token);
    var final_result = []
    var lat = city.latt;
    var long = city.longt;

    const closer_cities = await requester.makeRequest(metaweather_url + '?lattlong='
         + lat + ',' + long);
    ...
  } catch (err) {
    next(err)
  }
});

Answer 4

在调用async函数时，您不应该使用.then(...)构造...
只需let result = await myAsynchronousFunction(a, b, c); ...

Node.js：使用异步/等待时如何避免嵌套.then（）

4 个答案: