JavaScript在继续尝试回调和Promise之前不等待处理

Question

我看过很多promise教程和callback教程，我想我只是一无所获。我下面的代码读取目录中的所有文件并处理这些文件。当结束时，我需要它调用writeToFile（）..但是writeToFile在处理完成之前会得到调用。我希望这篇文章能最终结束我的困惑。谢谢。下面的代码：

function go(callbackFunction){
  var fs = require('fs');
  fs.readdir('./filesToProcess', (err, files) => {
    for (var i = 0; i < files.length; i++) {
      //console.log("run the process on : " + files[i]);
      console.log("Processing: " + files[i] + "\n");
      processFile('./filesToProcess/' + files[i]);
    };

  });
  callbackFunction();
}

go(writeToFile);

Answer 1

您必须等待功能完成，让我解释一下承诺

const foo = async () =>{ /* some function that takes time */}


// If I call foo like this "as you did inside the go function"

foo()
console.log("After foo")

//The console.log will print before the foo execute because you didn't wait for it
//The correct way is

await foo()
console.log("After foo")

//or

foo().then(() => console.log("After foo") )

要进一步了解如何使用诺言Here

Answer 2

您可以尝试异步/等待

Async function go(){...
await fs.readdir(...
}

或者如果fs.readdir返回一个诺言，您可以这样做：

function go(callback){
fs.readdir('./filesToProcess', (err, files) => {
    for (var i = 0; i < files.length; i++) {
      //console.log("run the process on : " + files[i]);
      console.log("Processing: " + files[i] + "\n");
      processFile('./filesToProcess/' + files[i]);
    };

  }).then(callback (){})

Answer 3

您要做的就是将callbackFunction移到传递给readdir的回调中，这样，您将确保仅在处理完成后才执行它。

function go(callbackFunction){
  var fs = require('fs');
  fs.readdir('./filesToProcess', (err, files) => {
    for (var i = 0; i < files.length; i++) {
      //console.log("run the process on : " + files[i]);
      console.log("Processing: " + files[i] + "\n");
      processFile('./filesToProcess/' + files[i]);
    };
    **callbackFunction();**    
  });

}

go(writeToFile);