[英]javascript function with $.get won't return value
I have the following function: 我有以下功能:
function parseLink(link){
var newlink="";
$.get(link,
function(data){
startoffset = data.indexOf("location.replace") + 18;
endoffset = data.indexOf("tiny_fold = 1;") - 8;
newlink= data.substr(startoffset,(endoffset-startoffset));
});
return newlink;
}
I'm using jquery $.get to parse a URL, which works fine if I do it without a function, but the function will return the empty string. 我正在使用jquery $ .get来解析URL,如果我不使用函数就可以正常工作,但是该函数将返回空字符串。 Clearly I'm doing something wrong, but I don't know what;
显然我做错了,但是我不知道是什么。 any help would be highly appreciated.
任何帮助将不胜感激。
The call to $.get is asynchronous. $ .get的调用是异步的。 See the control flow is like this:
看到控制流程是这样的:
parseUrl("http://www.test.com")
$.get(..., function callback() { /* this is called asynchronously */ })
return "";
...
// sometime later the call to $.get will return, manipulate the
// newLink, but the call to parseUrl is long gone before this
callback();
I think what you meant to do is: 我认为您的意思是:
function parseUrl(link, whenDone) {
$.get(link, function () {
var newLink = "";
// Do your stuff ...
// then instead of return we´re calling the continuation *whenDone*
whenDone(newLink);
});
}
// Call it like this:
parseUrl("mylink.com", function (manipulatedLink) { /* ... what I want to do next ... */ });
Welcome to async spaghetti world :) 欢迎来到异步意大利面世界:)
You'll need to pass in a function to be called when $.get
returns. $.get
返回时,您需要传递一个要调用的函数。 Something like: 就像是:
function parseLink(link, callback) {
$.get(link,
function(data) {
startoffset = data.indexOf("location.replace") + 18;
endoffset = data.indexOf("tiny_fold = 1;") - 8;
var newlink= data.substr(startoffset,(endoffset-startoffset));
callback(newlink);
});
}
Then you can call it with: 然后,您可以使用以下命令调用它:
parseLink('foo', function (newlink)
{
//Stuff that happens with return value
}
);
Because the .get()
operates asynchronously, parseLink()
carries on executing and returns the empty newlink
before the AJAX call has returned. 由于
.get()
异步操作,因此parseLink()
继续执行并在AJAX调用返回之前返回空的新newlink
。
You'll need to trigger whatever is working with newlink
from the callback, which may need you to rethink your implementation a little. 您需要从回调中触发对
newlink
进行任何处理,这可能需要您重新考虑一下实现。 What comes next (what should then happen to the populated newlink
)? 接下来是什么(填充的
newlink
发生什么情况)?
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.