Question

我是nodejs的新手我正在尝试追加功能。当我在提示他之后运行下面简单地接受用户输入的代码时，关于fav歌手并放入一个带有歌手名字的文件。并附加fav歌曲，似乎如果用户进入退出，它应该附加到文件但是，它不是??? 同时如果添加了appendFileSync（fn的同步版本）出口的更改？

代码：

var readline = require ('readline');
var fs = require('fs');
var rl = readline.createInterface(process.stdin, process.stdout);
var singer = {
  name:'',
  songs: []
};

rl.question("What is your fav singer? ", function(answer) {
  singer.name = answer;

  //creating a file for this singer
  fs.writeFileSync(singer.name+".md", `${singer.name}\n====================\n\n`); 

  rl.setPrompt(`What's your fav songs for ${singer.name} ?`);
  rl.prompt();

  rl.on('line', function(song) {
    singer.songs.push(song.trim());
    fs.appendFile(singer.name+".md", `* ${song.trim()} \n`);
    //***** WHY IF ITS EXIT ITS NEVER APPEND IT TO THE FILE
    console.log(`* ${song.trim()} \n`);

    if (song.toLowerCase().trim() === 'exit') {
      fs.appendFile(singer.name+".md", "Bye"); 
      //***** WHY ITS NEVER APPEND IT TO THE FILE
      rl.close();       
    } else {
      rl.setPrompt(`what else would ${singer.name} say? ('exit'to leave)`);
      rl.prompt();
    }
  });
});

rl.on ('close', function() {
  console.log ("%s is a real person that says %j", singer.name, singer.songs);
  process.exit();
});

Answer 1

因为fs.appendFile是异步的。当你调用它时，io操作只排队。不幸的是，你的脚本会在真正的io操作发生之前退出。

因此。你必须使用fs.appendFileSync或fs.appendFile进行回调（第3个参数），并且在调用回调时，进行所有进一步的活动。

NodeJs：fs.appendFile函数似乎错过了给出的最后一个String

1 个答案: