学习节点#9杂耍异步

Question

我正试着通过节点学校的学习节点。

这个问题与前面的问题()相同，因为您需要使用http.get()。但是，这一次将为您提供三个URL作为前三个命令行参数。 您必须收集每个URL提供给您的完整内容，并将其打印到控制台(stdout)。您不需要打印长度，只需将数据打印为字符串；每个URL只需一行。问题是，您必须按照URL作为命令行参数提供给您的相同顺序打印出来。

我不明白为什么我的解决方案不能完全工作，因为在我看来，它看起来是一样的，但功能更强，并且不确定它们的内部测试工作：

1.  ACTUAL:    ""
1.  EXPECTED:  "As busy as a dead horse also lets get some dero. Built like a sleepout no dramas lets get some chook. She'll be right thingo my she'll be right ute. "

2.  ACTUAL:    "She'll be right bizzo no worries she'll be right fair dinkum. We're going aerial pingpong no worries as busy as a gyno. "
2.  EXPECTED:  "She'll be right bizzo no worries she'll be right fair dinkum. We're going aerial pingpong no worries as busy as a gyno. "

3.  ACTUAL:    "He's got a massive pretty spiffy heaps she'll be right brizzie. He hasn't got a fly wire where shazza got us some strewth. She'll be right spit the dummy with it'll be fair go. We're going gobsmacked with as stands out like arvo. He's got a massive bush bash mate she'll be right slacker. "
3.  EXPECTED:  "He's got a massive pretty spiffy heaps she'll be right brizzie. He hasn't got a fly wire where shazza got us some strewth. She'll be right spit the dummy with it'll be fair go. We're going gobsmacked with as stands out like arvo. He's got a massive bush bash mate she'll be right slacker. "

4.  ACTUAL:    ""
4.  EXPECTED:  ""

我的代码：

var http = require('http');
var bl = require('bl');

var result = [];
var urls = process.argv.slice(2);
urls.forEach(function(url, i) {
  http.get(url, function(response) {
    response.pipe(bl(function(err, data) {
      if (err) return console.error(err);
      result[i] = data.toString();
      if (i === urls.length - 1) {
        console.log(result.join('\n'));
      }
    }));
  });
});

正式解决办法：

var http = require('http')
var bl = require('bl')
var results = []
var count = 0

function printResults () {
  for (var i = 0; i < 3; i++)
    console.log(results[i])
}

function httpGet (index) {
  http.get(process.argv[2 + index], function (response) {
    response.pipe(bl(function (err, data) {
      if (err)
        return console.error(err)

      results[index] = data.toString()
      count++

      if (count == 3)
        printResults()
    }))
  })
}

for (var i = 0; i < 3; i++)
  httpGet(i)

基本上，第一个测试从未通过(尽管如果迭代数组中只有一个url (而不是3个)，那么第一个测试就通过了，而不是其他的)。任何洞察力都会很好。我不知道该在哪里问这个问题，也许我只是错过了一些JS的东西，如果这不合适的话，我很抱歉。

Answer 1

您还没有确保下载了所有的urls。

请求不一定会按顺序返回。考虑一下，如果3是第一次回来。您将跳过另外两个urls，只打印出3个。

演示代码对响应的数量进行计数，因此在输出答案之前，可以保证获取所有内容。

Answer 2

我认为您只需要等待直到所有要求的结果结束或任何一个错误。这是我过去的答案：

var http = require('http');
var bl = require('bl');

var urls = process.argv.slice(2)
var count = urls.length;

var results = [];

urls.forEach((url, index) => {
    http.get(url, (res) => {
        res.pipe(bl((err, data) => {
            if (err) throw err;

            results[index] = data.toString();
            count--;

            if (count == 0) {
                results.forEach((result) => {
                    console.log(result)
                });
            }
        }))
    })
})

Answer 3

var http = require('http');

var links = [2, 3, 4];

var buffer = [];

(function render(index) {

http.get(process.argv[links[index]], function (response){

    response.setEncoding('utf8');

    response.on('data', function(chunk){
        if(buffer[index] === undefined) {
            buffer[index] = '';
        }
        buffer[index] += chunk;
    });
    response.on('end', function () {
        var newIndex = index+1;
        if(links[newIndex] !== undefined) {
            render(newIndex);
        } else {
            return renderOutput();
        }
    });
    response.on('error', console.error);

}).on('error', console.error);
})(0); //self-calling function

function renderOutput() {

buffer.forEach(function (elem) {

console.log(elem);

});
}