TypeError: Cannot destructure property 'err' of '(intermediate value)' as it is undefined

Question

我有一个带有 Node.js 后端的 React 应用程序。

我在后端有这条路线：

router.put(
  '/testimonial/:testimonialId/:action',
  [authenticate, validateUser],
  async (req, res) => {
    if (req.params.action === 'ACCEPT') {
      const { err, status, data } =
        await notificationController.acceptTestimonial(
          req.params.testimonialId,
          req.user._id
        );
      res.status(status).send({ err, data });
    } else if (req.params.action === 'REJECT') {
      const { err, status, data } =
        await notificationController.rejectTestimonial(
          req.params.testimonialId,
          req.user
        );
      res.status(status).send({ err, data });
    }
  }
);

这些是相应的控制器：(acceptTestimonial/rejectTestimonial)

const acceptTestimonial = async (testimonialId, userId) => {
  try {
    await Users.findOne({ _id: userId }, async function (err, creator) {
      if (err) return { err, status: 404, data: null };
      else {
        const updatedTestimonials = creator.testimonials.map((testimonial) => {
          if (testimonial._id == testimonialId) {
            testimonial.status = 'PUBLISHED';
          }
          return testimonial;
        });

        creator.testimonials = updatedTestimonials;
        await creator.save();
        return { err: null, status: 200, data: creator };
      }
    });
  } catch (err) {
    return { err: err.toString(), status: 500, data: null };
  }
};

const rejectTestimonial = async (testimonialId, userId) => {
  try {
    await Users.findOne({ _id: userId }, async function (err, creator) {
      if (!creator)
        return { err: 'Creator not found!', status: 404, data: null };
      else {
        const updatedTestimonials = creator.testimonials.map((testimonial) => {
          if (testimonial._id === testimonialId) {
            testimonial.status = 'REJECTED';
          }
          return testimonial;
        });
        creator.testimonials = updatedTestimonials;

        await creator.save();
        return { err: null, status: 200, data: creator };
      }
    });
  } catch (err) {
    return { err: err.toString(), status: 500, data: null };
  }
};

当此路由运行并执行操作时，我在控制台中收到此错误

UnhandledPromiseRejectionWarning: TypeError: Cannot destructure property 'err' of '(intermediate value)' as it is undefined.

我无法弄清楚问题出在哪里，解构或传递值有问题吗？

提前致谢。

Answer 1

这是因为，您没有“返回”控制器的 try 块中的任何内容。 return语句实际上是回调函数。

await 用于替换回调 & promise.then 所以你必须在 try 块中 return 一些东西。您可以像这样实现：

const rejectTestimonial = async (testimonialId, userId) => {
    try {
        // await Users.findOne({ _id: userId }, async function (err, creator) {
        //   if (!creator)
        //     return { err: 'Creator not found!', status: 404, data: null };
        //   else {
        //     const updatedTestimonials = creator.testimonials.map((testimonial) => {
        //       if (testimonial._id === testimonialId) {
        //         testimonial.status = 'REJECTED';
        //       }
        //       return testimonial;
        //     });
        //     creator.testimonials = updatedTestimonials;
    
        //     await creator.save();
        //     return { err: null, status: 200, data: creator };
        //   }
        // });

        //  instead of this
        //  do this

        const creator = await Users.findOne({ _id: userId });   // for "_id", you can also simply do: await User.findById(userId);
        
        if (!creator)
            return { err: 'Creator not found!', status: 404, data: null };

        // creator.testimonials.forEach((testimonial) => {
        //     if (testimonial._id === testimonialId)
        //         testimonial.status = 'REJECTED';
        // });

        const updatedTestimonials = creator.testimonials.map((testimonial) => {
            if (testimonial._id === testimonialId)
                testimonial.status = 'REJECTED';

            return testimonial;
        });

        creator.testimonials = updatedTestimonials;
        await creator.save();

        return { err: null, status: 200, data: creator };

    } catch (err) {
        return { err: err.toString(), status: 500, data: null };
    }
}

你的控制器返回 try & catch 中的东西！

Answer 2

我有这个功能

const deleteMsg = async (userId, messagesWith, messageId) => {
try {
    const user = await Chat.findOne({ user: userId });

    const chat = user.chats.find(
        chat => chat.messagesWith.toString() === messagesWith,
    );

    if (!chat) {
        return;
    }

    const messageToDelete = chat.messages.find(
        message => message._id.toString() === messageId,
    );

    if (!messageToDelete) return;

    if (messageToDelete.sender.toString() !== userId) {
        return;
    }

    const indexOf = chat.messages
        .map(m => m._id.toString())
        .indexOf(messageToDelete._id.toString());

    await chat.messages.splice(indexOf, 1);

    await user.save();

    return { success: true };
} catch (error) {
    console.error(error);
    return { success: false };
}

};

并调用它

const { success } = await deleteMsg(userId, messagesWith, messageId);

尽管收到此错误

无法解构“（中间值）”的属性 'success'，因为它未定义。

TypeError: Cannot destructure property 'err' of '(intermediate value)' as it is undefined

TypeError: Cannot destructure property 'err' of '(intermediate value)' as it is undefined

mongodb

node.js

express

mern