如何将 try catch function 变成 promise 并使用 promise.all

Question

In Reactjs, I have two try-catch functions listed as below, and I need to wrap them into a promise.all function. Could anyone tell me how please? I know first I have to turn those two functions into promise, but how?

  function one(){
    let isValid = true;
    ...
    if(isValid){
      try{
       const response = await AA.sendMsg({XX})
       return response
       }catch(e){
        console.log(e)
        }
      }
    }
function two(){
    let isValid = true;
    ...
    if(isValid){
      try{
       const response = await BB.sendMsg({YY})
       return response
       }catch(e){
        console.log(e)
        }
      }
    }

Answer 1

sendMsg already returns a Promise (at least, it should, if it doesn't then it makes no sense to use await or introduce Promises).

Wrapping then in extra Promises would be an anti-pattern. Don't do that.

Just pass them to Promise.all:

const [responseAA, responseBB] = await Promise.all([AA.sendMsg(...), BB.sendMsg(...)]);

Answer 2

await is only available if you declare your function as async .

Given that, invoking them will give you a promise without extra work.

so just do:

const result = await Promise.all([one(), two()])