为什么这不会收到返回值？ [重复]

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这个问题在这里已有答案:

How do I return the response from an asynchronous call? 31 answers

如何从异步调用返回响应? 31个答案

I have this in my HTML code

我的HTML代码中有这个

<script type="text/javascript">
    var URL = replace(); 
    alert(URL);
</script>

If I put this in header, alert shows “http://example.com/result"

如果我把它放在标题中,则会显示“http://example.com/result”

function replace() { 
　url = 'http://example.com'; 
　$.get(url, function(data){ 
　　var url_gen = data.responseText; 
　}); 
　return “http://example.com/result"; 
}

But if I use return like this, it won't do anything. Why?

但如果我像这样使用返回,它将不会做任何事情。为什么?

function replace() { 
　url = 'http://example.com'; 
　$.get(url, function(data){ 
　　var url_gen = data.responseText; 
　}); 
　return url_gen; 
}

Just to make sure, If I insert alert(url_gen) in front of return url_gen; in 2nd code, alert shows 'http://example.com' so url_gen has an actual value in it!

只是为了确保,如果我在返回url_gen之前插入alert(url_gen);在第二个代码中,警报显示'http://example.com',因此url_gen中有一个实际值!

Note: Here, assuming that url_gen is the string which is web-scraped from a web page.

注意:这里,假设url_gen是从网页上进行网页抓取的字符串。

2 个解决方案

#1

Make url_gen global like below:

使url_gen全局如下:

function replace() { 
url = 'http://example.com'; 
var url_gen ='';
$.get(url, function(data){ 
url_gen = data.responseText; 
}); 
return url_gen; 
}

#2

Because 1) url_gen only defined in $.get(), can't be accessed outside; 2) $.get() is async.

因为1)url_gen只在$ .get()中定义,不能在外面访问; 2)$ .get()是异步的。

Workaround: return directly from $.get():

解决方法:直接从$ .get()返回:

function replace() { 
　url = 'http://example.com'; 
　$.get(url, function(data){ 
　  var url_gen = data.responseText; 
    return url_gen;
　}); 
　 
}

#1