有没有更好的方法来处理快递中的错误？

Question

一段时间后，我正在快速构建一个 API。这个应用程序可以帮助用户跟踪他们的吉他套路。以下是我的控制器中的 DELETE 函数示例以及可能引发的异常：

 deleteOneRoutine = async (userId, routineId) => {
  // If the given routineId is found in the DB 
  // AND the userId matches, delete it from DB

  const selectedRoutine = await Routine.findOne({_id: routineId});
  if (!selectedRoutine) {return 400};

  if (selectedRoutine.userId != userId) {return 401};

  const deleted = await selectedRoutine.remove();
  if (!deleted) {return 500};

  return deleted;
}

这是接收这些错误代码的路由：

  routineRouter.delete('/:routineId', async (req, res) => {
   const userId = req.params.userId;
   const routineId = req.params.routineId;
   const deleted = await routineController.deleteOneRoutine(userId, routineId);
   if (deleted === 400) {res.status(400).send('Requested routine could not be found.')}
   else if (deleted === 401) {res.status(401).send('Unauthorized user.')}
   else if (deleted === 500) {res.status(500).send('Server error. Could not delete routine.')}
   else {res.status(200).send(`Successfully deleted routine with ID ${routineId}`)}  
});

我想知道是否有更好的方法来处理这个问题，而不是手动从控制器传回错误代码。任何帮助将不胜感激！

Answer 1

在 Express 中抛出错误并添加错误处理程序会更容易。

router.use((err, req, res, next) => {
  res.status(500).send(err.message);
});

router.get('/path', async (req, res) => {
  throw new Error('Can't delete!');
});

我认为错误中间件必须先出现，但我必须检查。

如果你需要特定的错误类型，你可以继承 Error;如果您希望能够传递代码，则可以在错误上添加一个属性（尽管我建议避免将路由逻辑（例如设置特定的 HTTP 代码）与业务逻辑（例如您的 deleteOneRoutine 函数）混合）

Answer 2

我对返回值的当前类型不太满意——我认为返回一个带有错误属性中的错误或例程 ID 的对象会更有意义。但 IMO 更好的方法是传递 res 以便 deleteOneRoutine 可以自己完成。还要创建一个辅助函数来保持 DRY，并且不要忘记捕获异步函数抛出的可能异常：

deleteOneRoutine = async (userId, routineId, res) => {
  const send = (code, message) => res.status(code).send(message);
  const serverError = () => send(500, 'Server error. Could not delete routine.');
  try {
    const selectedRoutine = await Routine.findOne({_id: routineId});
    if (!selectedRoutine) return send(400, 'Requested routine could not be found.');
    if (selectedRoutine.userId !== userId) return send(401, 'Unauthorized user.');
    const deleted = await selectedRoutine.remove();
    if (!deleted) return serverError();
    send(200, `Successfully deleted routine with ID ${deleted}`);
  } catch(e) {
    serverError();
  }
}

routineRouter.delete('/:routineId', (req, res) => {
  const { userId, routineId } = req.params;
  routineController.deleteOneRoutine(userId, routineId, res);
});