Node.js的：如何将变量传递给异步回调？ [英] Node.JS: How to pass variables to asynchronous callbacks?

问题描述

我敢肯定，我的问题是基于一个缺乏node.js的非同步编程的理解但在这里不用。

I'm sure my problem is based on a lack of understanding of asynch programming in node.js but here goes.

例如：我有我想要抓取的链接列表。当每个非同步请求返回我想知道它是哪个URL。但是，$ P $的psumably因为竞争情况，每个请求将返回设置为在列表中的最后一个值的URL。

For example: I have a list of links I want to crawl. When each asynch request returns I want to know which URL it is for. But, presumably because of race conditions, each request returns with the URL set to the last value in the list.

var links = ['http://google.com', 'http://yahoo.com'];
for (link in links) {
    var url = links[link];
    require('request')(url, function() {
        console.log(url);
    });
}

期望的输出：

http://google.com
http://yahoo.com

实际输出：

http://yahoo.com
http://yahoo.com

所以我的问题可以是：

So my question is either:

1. 我如何通过URL（按价值计算）的回调函数？ OR


2. 什么是让他们按顺序运行链的HTTP请求的正确方法？ OR


3. 别的东西我失踪？

PS：对于1，我不希望它检查回调的参数，但回调知道有关变量的自上而下的一般方式加以解决。

PS: For 1. I don't want a solution which examines the callback's parameters but a general way of a callback knowing about variables 'from above'.

推荐答案

您网​​址变量的作用域为为循环，如JavaScript只支持全球和功能范围界定。所以，你需要为你的请求创建一个功能范围来电捕捉网​​址由循环的每个迭代值使用即时功能：

Your url variable is not scoped to the for loop as JavaScript only supports global and function scoping. So you need to create a function scope for your request call to capture the url value in each iteration of the loop by using an immediate function:

var links = ['http://google.com', 'http://yahoo.com'];
for (link in links) {
    (function(url) {
        require('request')(url, function() {
            console.log(url);
        });
    })(links[link]);
}

BTW，嵌入要求中环的中间是不是好的做法。它应该可能被重新写为：

BTW, embedding a require in the middle of loop isn't good practice. It should probably be re-written as:

var request = require('request');
var links = ['http://google.com', 'http://yahoo.com'];
for (link in links) {
    (function(url) {
        request(url, function() {
            console.log(url);
        });
    })(links[link]);
}

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