ajax 调用后，如何更改包装器 JavaScript 函数中的全局变量? [英] How do I change a global variable in the wrapper JavaScript function after an ajax call?

问题描述

function ajax_test(str1){ 
  var url = "None" 
  jq.ajax({
    type:'post', 
    cache: false, 
    url: 'http://....' + str1, 
    success: function(data, status, xhr){ 
      url=data; 
    }, 
    error: function (xhr, status, e) {  
    }, 
    async: true, 
    dataType: 'json' 
  }); 
  return url 
} 

如何设置全局变量url为返回的成功ajax数据?

How can I set the global variable url to be the returned success ajax data?

推荐答案

在 Javascript 中，函数不可能返回异步结果.该函数通常会在发出 AJAX 请求之前返回.

In Javascript, it is impossible for a function to return an asynchronous result. The function will usually return before the AJAX request is even made.

您总是可以强制您的请求与 async: false 同步，但这通常不是一个好主意，因为它会导致浏览器在等待结果时锁定.

You can always force your request to be syncronous with async: false, but that's usually not a good idea because it will cause the browser to lock up while it waits for the results.

解决这个问题的标准方法是使用回调函数.

The standard way to get around this is by using a callback function.

function ajax_test(str1, callback){  
   jq.ajax({ 
     //... your options
     success: function(data, status, xhr){  
       callback(data);
     }
   });  
}  

然后你可以这样称呼它:

and then you can call it like this:

ajax_test("str", function(url) {
  //do something with url
});

这篇关于ajax 调用后，如何更改包装器 JavaScript 函数中的全局变量?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！