异步/等待节点js mongodb中的处理错误

Question

I'm using async function in my node, here I'm calling multiple queries with mongodb

(ie) :

let data1;
let data2;
try{
       data1 = await mongoQuery1
  }catch {
       res.send(error)
  }
try{
      data2 = await mongoQuery2 
  }catch {
       res.send(error)
  }
  res.send({ data1, data2 });

I've confused with this flow. So, basically All I want to send to user is both data1 and data2 but what happens if the query1 fails because, it will be obviously falls in the first catch block and send response to the user, and then continue executing the next try block, and send the response again, which is totally wrong. How can I achieve this, what I want is if any error occured in any catch I want to send the error response back to the user

If all the try block succeeds then only I want to send back the success response to the user, How can I achieve this?

Answer 1

It's not really clear what you are trying to do.

If you want to stop and just send the error then you can return after calling res.send to stop the function continuing.

If you want to send the error with the rest of the data then don't call res.send immediately, store the error in data1 and then send it at the end as normal.

Answer 2

Advice to use Promise.all for your scenario: Both query will be run simultaneously and if there is an error it will immediately throw and got to the catch block.

try {
    const [data1, data2] = await Promise.all([mongoQuery1, mongoQuery2]);
    res.send({ data1, data2 })
} catch {
    res.send(error)
}