Question

您好，我想在函数外部使用回调进行变量外部使用，但是出了点问题，我认为我的回调并不像我想的那样正确：

function latitude(callback){
  var mylat;
  const https = require('https');
     https.get('https://url_of_my_json', (resp) => {
     let data = '';


  // A chunk of data has been recieved.
    resp.on('data', (chunk) => {
    data += chunk;
  });

  // The whole response has been received. Print out the result.
     resp.on('end', () => {
     mylat = JSON.parse(data).results[0].geometry.location.lat;
     callback(mylat);

     });


  }).on("error", (err) => {
   console.log("Error: " + err.message);
 });

}

var mylat = latitude(); // variable i want to use for the rest of code

谢谢

Answer 1

回调的语法不正确。

这里是示例，为了更好地了解回调，请尝试尽可能简单地阅读此示例，只需复制save newfile.js do node newfile即可运行应用程序。

function myNew(next){
    console.log("Im the one who initates callback");
    next("nope", "success");
}


myNew(function(err, res){
    console.log("I got back from callback",err, res);
});

快乐的编码：）

AWS回调变量中的Node.js

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