Node js。从 fs.watch() 返回的观察程序'error'侦听事件未触发

Question

我将"错误"事件侦听器添加到从fs.watch()返回watcher。

但是当我观察不存在的文件时error事件处理程序不会触发。

const filename = 'test/target.txt';
const filenameNotExist = 'test/not-exist-target.txt';
console.log('Now watching the target.txt for changes...');
const watcher = fs.watch(filenameNotExist || filename);
watcher.on('change', (eventType, fname) => {
  switch (eventType) {
    case 'rename':
      console.log(`file ${fname} renamed`);
      break;
    case 'change':
      console.log(`file ${fname} changed`);
      break;
    default:
      break;
  }
});
// error event handler does not trigger.
watcher.on('error', err => {
  console.log('filename is not correct'); 
  if (err) throw err;
});

标准输出给我这个输出：

Now watching the target.txt for changes...
fs.js:1384
    throw error;
    ^
Error: watch test/not-exist-target.txt ENOENT
    at _errnoException (util.js:1041:11)
    at FSWatcher.start (fs.js:1382:19)
    at Object.fs.watch (fs.js:1408:11)
    at Object.<anonymous> (/Users/elsa/workspace/Training.nodejs/examples/api/fs/watcher.js:10:20)
    at Module._compile (module.js:573:30)
    at Object.Module._extensions..js (module.js:584:10)
    at Module.load (module.js:507:32)
    at tryModuleLoad (module.js:470:12)
    at Function.Module._load (module.js:462:3)
    at Function.Module.runMain (module.js:609:10)

没有filename is not correct日志。哪种情况将触发事件处理程序error观察程序？

Answer 1

fs.FSWatcher 文档说它在成功的fs.watch()调用时返回。检查返回的对象，也许它是一个错误(ENOENT(，而不是您期望的 FSWatcher。

(实际上，我相信会抛出一个 ENOENT，所以你的watcher.on(...)调用永远不会被执行。您可以使用 try-catch 来确保。