在JavaScript中转换数据的最佳做法

Question

I'm currently working with a payload that is string and im parsing is using JSON.parse(payload) . My question is, once I convert the payload I need to access at least 5 properties within the newly created JSON object and it is quite nested. My current implementation is:

 ....then(payload => {
   return ({
       obj1: JSON.parse(payload).field.obj1
       obj2: JSON.parse(payload).field.obj2
       obj3: JSON.parse(payload).field.obj3
       obj4: JSON.parse(payload).field.obj4
       obj5: JSON.parse(payload).field.obj5
     });    
   })

I feel like this is to much repetition and feel this way would work better in terms of readability (even then it is not that clean):

   ....then(payload => {
       let jsonObj = JSON.parse(payload)
       return ({
           obj1: jsonObj.field.obj1
           obj2: jsonObj.field.obj2
           obj3: jsonObj.field.obj3
           obj4: jsonObj.field.obj4
           obj5: jsonObj.field.obj5
         });    
       })

Can anyone suggest the best way to execute this in terms of readability and performance?

NOTE: This is used within a promise.all() so the above will iterate over X promises.

Answer 1

You could use a destructuring assignment and return an object with short hand properties .

let { field: { obj1, obj2, obj3, obj4, obj5 } } = JSON.parse(payload);
return { obj1, obj2, obj3, obj4, obj5 };

Answer 2

Edit

Better yet, just parse the whole thing and return the field object. I think that would work. (I'm a little sleep deprived though!)

....then(payload => {
   return (JSON.parse(payload).field);
})

End Edit

1. Parse the whole response to find the field obj.
2. Stringify the field obj to prepare it to be copied.
3. Return the parsed field obj.

-

....then(payload => {
   return (JSON.parse(JSON.stringify(JSON.parse(payload).field)));
})

Answer 3

If you use Bluebird you can also do (as you already use Promise.all )

const payloadPromises = Promise.map(sourcePromises, JSON.parse);
return Promise
  .all(payloadPromises)
  .then(payloads => payload.map(payload => ({
    obj1: payload.field.obj1,
    obj2: payload.field.obj2,
  }));

But actually just variable (your second approach) is completely fine. Perhaps, you should only use const instead of let in this case.