Question

我正在尝试将函数的返回值存储到变量中，但它似乎不起作用。我已经尝试了一切我可能想到的......也许我只是在做一些非常愚蠢的事情：D 我正在使用express，request，body-parser和mongoose在NodeJS中运行代码。

var requestNews = function(){
    request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){
    if (!error && response.statusCode == 200){
        var parsedData = JSON.parse(body);
        var title       =  parsedData.articles[randomNum].title;
        var description =  parsedData.articles[randomNum].description;
        var url         =  parsedData.articles[randomNum].url;
        var urlToImage  =  parsedData.articles[randomNum].urlToImage;
        var publishedAt =  parsedData.articles[randomNum].publishedAt;
        var articleObj  = {
            title: title, 
            description: description, 
            url: url, 
            urlToImage: urlToImage, 
            publishedAt: publishedAt

        };
        articleObjStr = JSON.stringify(articleObj);
        return articleObjStr;
    }
});
};


app.get("/index", function(req, res){
    var randomNew = requestNews();
    console.log(randomNew); // LOGS UNDEFINED
    res.render("index", {randomNew: randomNew});
});

任何想法？

感谢您的帮助！

Answer 1

请求是异步操作，您无法分配这样的值。您可以使用callback或promise。

以下是一个代码段代码，介绍如何使用回调实现此目的

var requestNews = function(callback){
  request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){
    if (!error && response.statusCode == 200){
      var parsedData = JSON.parse(body);
      var title       =  parsedData.articles[randomNum].title;
      var description =  parsedData.articles[randomNum].description;
      var url         =  parsedData.articles[randomNum].url;
      var urlToImage  =  parsedData.articles[randomNum].urlToImage;
      var publishedAt =  parsedData.articles[randomNum].publishedAt;
      var articleObj  = {
        title: title, 
        description: description, 
        url: url, 
        urlToImage: urlToImage, 
        publishedAt: publishedAt

      };
      articleObjStr = JSON.stringify(articleObj);
      callback(null, articleObjStr);
    } else {
      callback(error)
    }
  });
};


app.get("/index", function(req, res, next){
  requestNews(function(err, data) {
    if (err) return next(err);
    console.log(data);
    res.render("index", {randomNew: data});
  });

});

Answer 2

使用回调：

var requestNews = function(callback){
    request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){
    if (!error && response.statusCode == 200){
        var parsedData = JSON.parse(body);
        var title       =  parsedData.articles[randomNum].title;
        var description =  parsedData.articles[randomNum].description;
        var url         =  parsedData.articles[randomNum].url;
        var urlToImage  =  parsedData.articles[randomNum].urlToImage;
        var publishedAt =  parsedData.articles[randomNum].publishedAt;
        var articleObj  = {
            title: title, 
            description: description, 
            url: url, 
            urlToImage: urlToImage, 
            publishedAt: publishedAt

        };
        articleObjStr = JSON.stringify(articleObj);
        return callback(null, articleObjStr);
    } else {
      return callback(error);
    }
});
};


app.get("/index", function(req, res){
    requestNews(function(err, data){
      if(err) console.log(err);
      else {
        var randomNew = data;
        console.log(randomNew); // LOGS UNDEFINED
        res.render("index", {randomNew: randomNew});
      }
    });

});

Answer 3

您正在尝试从异步方法获取返回值，这是不可能的。为了完成您的操作，您可以执行以下操作：

app.get("/index", function(req, res){
     requestNews(req, res);
});

var requestNews = function(req, res){
request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){
if (!error && response.statusCode == 200){
    var parsedData = JSON.parse(body);
    var title       =  parsedData.articles[randomNum].title;
    var description =  parsedData.articles[randomNum].description;
    var url         =  parsedData.articles[randomNum].url;
    var urlToImage  =  parsedData.articles[randomNum].urlToImage;
    var publishedAt =  parsedData.articles[randomNum].publishedAt;
    var articleObj  = {
        title: title, 
        description: description, 
        url: url, 
        urlToImage: urlToImage, 
        publishedAt: publishedAt

    };
    articleObjStr = JSON.stringify(articleObj);
   // return articleObjStr; do not return value from here it can not be retrived
   res.render("index", {randomNew: articleObjStr});
   }
  });
};

Answer 4

requestNews函数没有返回任何内容。

var requestNews = function(){

      // Missing return Here.
      return request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){

      // All you code here.    
      return articleObjStr;
    }
  });
};

问题是您正在重新传递给传递给请求方法的函数中的值，如果这是一个回调，它也将无法工作。

如果是回叫，请以这种方式进行：

var requestNews = function(onNewsRetrived){ // Use a callback here
          request("https://newsapi.org/v1/articles?source=google-news&sortBy=top&apiKey=2cc11b1813c942*************", function(error, response, body){

      onNewsRetrived(articleObjStr); // Trigger your call back get you got the info.
    }
  });
};

app.get("/index", function(req, res){
   requestNews(funtion(randomNew) {

    console.log(randomNew); // Should Work
    res.render("index", {randomNew: randomNew});

   });       
});

Answer 5

不会触发回调功能。这就是未定义值在日志中打印的原因。

Javascript / NodeJS：将函数的返回值存储到变量中

5 个答案: