Issue
I need to display data from 3 collections and then sort them by date. The problem is that I can’t manage to find the right data structure to set this up.
I have to return an object like this :
data : {
total, // the total number of objects in collections
data // the data from all the three collections
}
Here is the code that I’m trying to setup :
const fillAray = (callb) => {
let total = 0;
let totalData={}
for(let i=0;i<3;i++){
switch(i){
case 0:
historyPart='1st table'
break;
case 1:
historyPart='2nd table'
break;
case 2:
historyPart='3rd table'
break;
}
const collection_ref = admin.firestore().collection(`*****-${historyPart}`);
const user_transactions_ref = collection_ref.doc(uid).collection('transactions');
user_transactions_ref.orderBy('time', 'desc').get().then(
(snapshot) => {
if (!snapshot.empty) {
snapshot.forEach((doc) => {
++total;
});
}
user_transactions_ref.orderBy('time', 'desc').offset(from).limit(size).get().then(
(snapshot) => {
const out = [];
if (!snapshot.empty) {
snapshot.forEach((doc) => {
out.push(doc.data())
});
}
//merging data
Object.assign(totalData,out)
},
(error) => { callback(error, null); }
);
},
(error) => { callback(error, null); }
);
}
sendDataToFront(totalData,total)
// Here I'm supposed to have totalData and total with all datas
}
Somehow the object don’t merge properly … Any ideas ?
Solution
It’s unclear exactly what you’re trying to achieve here, but a generic approach for merging multiple queries would look something like this:
const refA = admin.firestore().collection('foo/A/widgets');
const refB = admin.firestore().collection('foo/B/widgets');
// fetch data in parallel
Promise.all([ refA.get(), refB.get() ])
// merge the results
.then(promiseResults => {
const mergedData = [];
promiseResults.forEach( snapshot => {
snapshot.forEach( doc => mergedData.push(doc.data()) );
});
return mergedData;
})
// sort the results
.then(mergedData => mergedData.sort((a, b) => a.timestamp - b.timestamp))
// use the results
.then(sortedData => {
// do something with data here
console.log(sortedData);
})
// log any errors
.catch(e => console.error(e));
Answered By – Kato
This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0