node.js - 从 accessSync 返回 ENOENT 以外的值

Question

当 accessSync 找不到文件/目录而不是 ENOENT 时，我如何返回 false？

单元测试

it.only('should be able to read a file stream only if a file exist', function() {
    let testfile = testpath+'/imageeee.png';
    let ok = FileService.existsAsync(testfile);

    ok = ok.then((result) => {
      console.log('exists: ', result);
      return FileService.createReadStream(testfile);
    });

    ok = ok.then((result) => {
      assert.isNotNull(result.path);
      assert.equal(result.path, testfile);
      assert.isTrue(result.readable, true);
    });
    return ok;
  });

功能

 existsAsync(path) {
    let ok = fs.accessAsync(path, fs.F_OK);
    ok = ok.then(function(error) {
      if (error) {
        return false;
      } else {
        return true;
      }
    });
    return ok;
  },

错误

 Error: ENOENT: no such file or directory, access '/home/j/Work/imageeee.png'

Answer 1

任何引发错误的内容都可以包含在 try...catch block 中以捕获错误并从那里继续。如果这是这里的问题函数:

fs.accessAsync(path, fs.F_OK);

将其包装在 try catch 中，并在错误情况下返回 false:

try {
    fs.accessAsync(path, fs.F_OK);
    // ... other code
    return true;
} catch(e) {
    // log the error ?
    return false;
}