How to perform aggregation on firestore collection
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0
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I'm working with firestore for the first time and the dataset in which I have users and games data.
In every game data I've finalScore
There is a foreign key in Games
called userID
for each game that differentiate that this game data belong to.
I want to write a function in index.js
which can take in the userID
and SUM all the finalScore
and return the total score for that user.
So far what I have after searching google is:
exports.myFunctionName = functions.firestore
.document('Users').onWrite((change, context) =>
// ... Your code here
);
This is the first time I'm working on this and want to achieve something for which I didn't find related help. I'm working on an iOS app for which I'm using this as a backend service.
node.js swift google-cloud-platform google-cloud-firestore
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up vote
0
down vote
favorite
I'm working with firestore for the first time and the dataset in which I have users and games data.
In every game data I've finalScore
There is a foreign key in Games
called userID
for each game that differentiate that this game data belong to.
I want to write a function in index.js
which can take in the userID
and SUM all the finalScore
and return the total score for that user.
So far what I have after searching google is:
exports.myFunctionName = functions.firestore
.document('Users').onWrite((change, context) =>
// ... Your code here
);
This is the first time I'm working on this and want to achieve something for which I didn't find related help. I'm working on an iOS app for which I'm using this as a backend service.
node.js swift google-cloud-platform google-cloud-firestore
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm working with firestore for the first time and the dataset in which I have users and games data.
In every game data I've finalScore
There is a foreign key in Games
called userID
for each game that differentiate that this game data belong to.
I want to write a function in index.js
which can take in the userID
and SUM all the finalScore
and return the total score for that user.
So far what I have after searching google is:
exports.myFunctionName = functions.firestore
.document('Users').onWrite((change, context) =>
// ... Your code here
);
This is the first time I'm working on this and want to achieve something for which I didn't find related help. I'm working on an iOS app for which I'm using this as a backend service.
node.js swift google-cloud-platform google-cloud-firestore
I'm working with firestore for the first time and the dataset in which I have users and games data.
In every game data I've finalScore
There is a foreign key in Games
called userID
for each game that differentiate that this game data belong to.
I want to write a function in index.js
which can take in the userID
and SUM all the finalScore
and return the total score for that user.
So far what I have after searching google is:
exports.myFunctionName = functions.firestore
.document('Users').onWrite((change, context) =>
// ... Your code here
);
This is the first time I'm working on this and want to achieve something for which I didn't find related help. I'm working on an iOS app for which I'm using this as a backend service.
node.js swift google-cloud-platform google-cloud-firestore
node.js swift google-cloud-platform google-cloud-firestore
asked Nov 10 at 1:31
Chaudhry Talha
1,44521845
1,44521845
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1 Answer
1
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votes
up vote
1
down vote
Firestore doesn't have any aggregation queries, because they don't scale massively as Firestore requires. If you want to find a total value among documents, you will either have to:
- Query all the documents in the client, and sum them manually, or
- Keep a running total over time, as each new score is known. Then, you can query for that running total in another document when you need it.
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Firestore doesn't have any aggregation queries, because they don't scale massively as Firestore requires. If you want to find a total value among documents, you will either have to:
- Query all the documents in the client, and sum them manually, or
- Keep a running total over time, as each new score is known. Then, you can query for that running total in another document when you need it.
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
add a comment |
up vote
1
down vote
Firestore doesn't have any aggregation queries, because they don't scale massively as Firestore requires. If you want to find a total value among documents, you will either have to:
- Query all the documents in the client, and sum them manually, or
- Keep a running total over time, as each new score is known. Then, you can query for that running total in another document when you need it.
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
add a comment |
up vote
1
down vote
up vote
1
down vote
Firestore doesn't have any aggregation queries, because they don't scale massively as Firestore requires. If you want to find a total value among documents, you will either have to:
- Query all the documents in the client, and sum them manually, or
- Keep a running total over time, as each new score is known. Then, you can query for that running total in another document when you need it.
Firestore doesn't have any aggregation queries, because they don't scale massively as Firestore requires. If you want to find a total value among documents, you will either have to:
- Query all the documents in the client, and sum them manually, or
- Keep a running total over time, as each new score is known. Then, you can query for that running total in another document when you need it.
answered Nov 10 at 1:54
Doug Stevenson
66.2k77997
66.2k77997
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
add a comment |
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
That's helpful. Thank you :)
– Chaudhry Talha
Nov 10 at 4:20
add a comment |
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