I'm struggling to understand how I can return data from multiple promises to build up an array of data.

Is there anyway I can return the data outside of the promise to push to the data variable?

I have the following:

db_sh.find({
    selector: {sh: req.params.sh_id},
    fields: ['_id', 'sh_id', 'time'],
    sort: ['_id']
}).then(function (result) {
    let data = {};
    console.log('Found: '   result.docs.length);
    if (result.docs.length > 0) {
        for (var i = 0; i < result.docs.length; i  ) {
            let student = result.docs[i];

            Promise
                .all([getMealBooking(student._id), getStudentData(student._id)])
                .then(function(response) {
                    var meal_booking_data = response[0];
                    var student_data = response[1];
                    console.log(meal_booking_data);

                    console.log(student_data);
                })
                .catch(function (err) {
                    return res.send(false);
                });

            data[student.time] = [
                meal_booking_data,
                student_data
            ]
        }
    }
    /** Sort Data Oldest First*/
    data = Object.keys(data).sort().reduce((a, c) => (a[c] = data[c], a), {});
    console.log(data);
    res.send(data);
});

I have two promises (getMealBooking() & getStudentData()): and I am using Promise.all() to return me the results of both of these promises. I have tried to return the data but I cannot get the results to build up the data array.

Any help to be able to build up a list of all my data would be great.

CodePudding user response：

You need two Promise.alls - one to iterate over each student, and a nested one to fetch the getMealBooking and getStudentData for each student.

Put everything into an async function (that catches and sends false if needed) to make the control flow easier to understand.

const { docs } = await db_sh.find({
    selector: { sh: req.params.sh_id },
    fields: ['_id', 'sh_id', 'time'],
    sort: ['_id']
});
if (docs.length === 0) {
    // no data; stop here
    res.send({});
    return;
};
const data = {};
await Promise.all(
    docs.map(student => (
        Promise.all([getMealBooking(student._id), getStudentData(student._id)])
            .then(([mealBookingData, studentData]) => {
                data[student.time] = [mealBookingData, studentData];
            })
    ))
);
const sortedData = Object.keys(data).sort().reduce((a, c) => (a[c] = data[c], a), {});
res.send(sortedData);

CodePudding user response：

Another Promise.all() is needed for the loop that contains the Promise.all() you've already figured out. It's better to factor a little so you can see what's happening.

function getStudentMealAndData(student) {
  return Promise
    .all([getMealBooking(student._id), getStudentData(student._id)])
    .then(function(response) {
      var meal_booking_data = response[0];
      var student_data = response[1];
      console.log(meal_booking_data);
      console.log(student_data);
      return { student, meal_booking_data, student_data };
    })
    .catch(function (err) {
      return res.send(false);
    });
}

This simplifies the then block a bit...

}).then(function (result) {
    console.log('Found: '   result.docs.length);
    let promises = []
    for (var i = 0; i < result.docs.length; i  ) {
      let student = result.docs[i];
      promises.push(getStudentMealAndData(student));
    }
    return Promise.all(promises);
}).then(results => {
  // results are an array of [{ student, meal_booking_data, student_data }, ...]
  let data = results.reduce((acc, s) => {
    acc[s.student.time] = [ s.meal_booking_data, s.student_data ];
    return acc;
  }, {});
  data = Object.keys(data).sort().reduce((a, c) => (a[c] = data[c], a), {});
    console.log(data);
    res.send(data);
});

CodePudding user response：

let arr = [];
const datas = await Promise.all([
    getMealBooking(),
    getStudentData()
]);
arr.push(datas[0]); //adds getMealBooking() results
arr.push(datas[1]); // adds getStudentData() results