jQuery全局变量值超出功能

Question

var hhhhhhh = ''; 
     function displayLocation(latitude, longitude) { 
      var request = new XMLHttpRequest(); 

      var method = 'GET'; 
      var url = 'http://maps.googleapis.com/maps/api/geocode/json?latlng=' + latitude + ',' + longitude + '&sensor=true'; 
      var async = true; 

      request.open(method, url, async); 
      request.onreadystatechange = function() { 
       if (request.readyState == 4 && request.status == 200) { 
        var data = JSON.parse(request.responseText); 
        var addressComponents = data.results[0].address_components; 
        for (i = 3; i < 4; i++) { 
         hhhhhhh = addressComponents[i].long_name; 
        } 
       } 
      }; 
      request.send(); 
     } 
     console.log(hhhhhhh); 

var hhhhhhh没有从displayLocation函数中获取任何值。在控制台中，在函数内部函数之外，它的值为null。请Help.thank你

Answer 1

您拥有的console.log输出之前调用你的函数

var hhhhhhh = ''; 
 
function displayLocation(latitude, longitude) { 
 
    //your code 
 
    hhhhhhh = "Value"; 
 
} 
 
displayLocation(0, 0); 
 
console.log(hhhhhhh);

Answer 2

你需要调用的函数，然后安慰值：

VAR HHHHHHH = ''; function displayLocation（latitude，longitude）{ var request = new XMLHttpRequest（）;

  var method = 'GET'; 
      var url = 'http://maps.googleapis.com/maps/api/geocode/json?latlng=' + latitude + ',' + longitude + '&sensor=true'; 
      var async = true; 

      request.open(method, url, async); 
      request.onreadystatechange = function() { 
       if (request.readyState == 4 && request.status == 200) { 
        var data = JSON.parse(request.responseText); 
        var addressComponents = data.results[0].address_components; 
        for (i = 3; i < 4; i++) { 
         hhhhhhh = addressComponents[i].long_name; 
        } 
       } 
      }; 
      request.send(); 
     } 

    displayLocation(lat,long) // you need to call the function and then console the value 

     console.log(hhhhhhh);