异步并等待并承诺捕获错误-Java 学习之路

我有一个关于在异步和等待中捕获用户错误的问题 .

假设我有一条仅为单个用户提取的路由 .

routes.js

routes.get('/getuserbyid/:id', (req, res) => {

    const id = req.params.id;

    accountController.getById(id)
        .then((result) => {
            res.json({
                confirmation: 'success',
                result: result
            });
        })
        .catch((error) => {
            res.json({
                confirmation: 'failure',
                error: error
            });
        });
});

我有一个控制器来获取请求 . accountController.js

export const getById = async (id) => {

        try {
            const user = await users.findOne({ where: {
                    id: id
                }});

            if (user === null) {
                return 'User does not exist';
            }
            return user;
        } catch (error) {
            return error;
        }
}

所以无论发生什么，我都会获得空或单个记录 . 它仍然是成功的 . 在promises中，我可以拒绝null，因此它将显示在路径中的catch块中 . 现在用异步和等待 . 如何在错误块中实现相同的null？

export const getById = (id) => {

    return new Promise((resolve, reject) => {

        users.findOne({ where: {
            id: id
        }})
            .then((result) => {

                if (result === null) {
                    reject('User does not exit');
                }
                resolve(result);
            })
            .catch((error) => {
                reject(error);
            });
    });

}

3 回答

首先，避免使用Promise反模式 . 你有一个返回Promise的函数，不需要在_140726中包装它 .

然后你的功能可能如下所示：

export const getById = (id) => {
  return users.findOne({ where: { id } })
    .then((user) => {
      if (user === null)
        throw 'User does not exit';

      return user;
    });
}

和async / await版本是

export const getById = async (id) => {
  const user = await users.findOne({ where: { id } });

  if(user === null)
    throw 'User does not exist';

  return user;
}

回复于 2024-04-29T15:31:10+08:00

async 函数始终返回 Promise .

在 async 函数中使用 throw 构造会拒绝返回的 Promise .

考虑

function getValue() {
  return Promise.reject(null);
}

getValue().catch(e => {
  console.log('An raised by `getValue` was caught by a using the `.catch` method');
  console.log(e);
});


(async function main() {
  try {
    await getValue();
  } catch (e) {
    console.log('An raised by `getValue` was caught by a catch block');
    console.log(e);
  }
}());

async function getValue() {
  throw null;
}

getValue().catch(e => {
  console.log('An raised by `getValue` was caught by a using the `.catch` method');
  console.log(e);
}); 

(async function main() {
  try {
    await getValue();
  } catch (e) {
    console.log('An raised by `getValue` was caught by a catch block');
    console.log(e);
  }
}());

异步方法中的 try 块处理同步和异步错误 .

考虑

function throws() {
  throw Error('synchronous failure');
}


async function rejects() {
  throw Error('asynchronous failure');
}

(async function main() {
  try {
    throws();
  } catch (e) {
    console.log('asynchronously handled', e.message);

  }

  try {
    await rejects();
  } catch (e) {
    console.log('asynchronously handled', e.message);
  }
}());

要记住的一个关键点是，如果您忘记了拒绝使用它们的承诺，则不会捕获异步错误 . 这是忘记 .then 内部的 return 或 .catch 回调的类比

回复于 2024-04-29T15:31:10+08:00

如果你的值变空，你可以抛出异常 .

export const getById = async (id) => {
        const user = await users.findOne({ where: {
                id: id
            }});

        if (!user) {
            throw Error('User does not exist..');
        }
        return user;        
}

回复于 2024-04-29T15:31:10+08:00

异步并等待并承诺捕获错误

3 回答

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