How do you query data by with multiple conditions or indexes? - firebase

Question

I'm trying to design a performant NoSQL schema for my Firebase application, but how would I query multiple checkpoints which have a unique tag?
Consider the following rules:
Tiers can have multiple tags
Tags can have multiple Checkpoints
Tag keys (names) are unique for each tier
Checkpoints can have multiple tags, but only 1 tag per tier
Here's what I have so far:
{
"tiers": {
"1": {
"web": true,
"android": true
},
"2": {
"basics": true,
"angular2": true,
"aurelia": true
},
"3": {
"basics": true,
"components": true
}
},
"tags" : {
"1": {
"web": {
"name": "Web Development",
"children": {
"front-end": true,
"angular2": true,
"aurelia": true
}
},
"android": {
"name": "Android Development",
"children": {
"front-end": true
}
}
},
"2": {
"front-end": {
"name": "Basics",
"parents": {
"web": true,
"android": true
},
"children": {
"angular2": true,
"aurelia": true,
"android-studio": true
}
}
},
"3": {
"angular2": {
"name": "Angular 2.x",
"parents": {...},
"children": {...}
},
"aurelia": {
"name": "Aurelia 1.x"
}
}
},
"checkpoints": {
"<randomKey>" : {
"name": "Angular 2 Quick Start",
"tags": {
"1": "web",
"2": "front-end",
"3": "angular2"
}
}
}
}
Right now I can query for all checkpoints under the Tier 1 web tag with:
ref.orderByChild('tags/1').equalTo("web").once('value', snap => console.log(snap.val()));
but since you can only define one indexOn rule, it's not optimized. At least if I can set the indexOn rule I can at least filter out most of the checkpoints then filter the rest in my code.
What can I do to efficiently query my checkpoints based on multiple tags AND the tiers?
Eventually I'd need to do a query for checkpoints with "tags": {"1": "web" AND "2": "front-end"} which I can't for the life of me figure out how to execute efficiently. I was thinking of doing another table with composite keys (e.g. each tier/tag contains references to ALL child checkpoints), but that would lead to requiring me to add and delete references within each tier.
There must be a better way.

Answer 1

I was vastly over complicating the solution - I ended up:
Removing Tier from the equation all together
Adding each checkpoint within each tag it was related to
Querying all checkpoints related to each selected tag and removing the ones which did not appear in each tag selected
Here's my new schema:
allTags = [
{"abc": true, "def": true, "hij": true},
{"abc": true, "def": true}
];
tags: [
{ "0": [
{"software-development": {name: "Software Development", checkpoints: [ {"abc": true}, {"def": true}, {"hij": true}]}}
]},
{"1": [
{"web": {name: "Web", checkpoints: [ {"abc": true}, {"def": true}]}},
{"android": {name: "Android", checkpoints: [{"hij": true}]}}
]}
];
checkpoints: [
{"abc": { name: "Angular2 Quick Start"}},
{"def": { name: "Building global directives in Angular2"}},
{"hij": { name: "Android Quick Start"}},
];
Creating a new Checkpoint:
public createCheckpoint(tags: any[], checkpoint: any) {
// push checkpoint
let checkpointRef = this.firebase.child('test/checkpoints');
let checkpointKey = checkpointRef.push(checkpoint).key(); // might have to do separate call
// Add checkpoint under each related tag
tags.forEach(tag => {
let tagRef = this.firebase.child('test/tags/' + tag.tier + '/' + tag.key + '/checkpoints/' + checkpointKey);
tagRef.set(true)
});
}
Retrieving all checkpoints based on selected tags:
public getCheckpointss(tags: any[]) {
// tag.tier, tag.key
let checkpointKeysToGet: string[] = [];
tags.forEach(tag => {
let ref = this.firebase.child('test/tags/' + tag.tier + '/' + tag.key + '/checkpoints');
ref.once('value', snap => {
let tagKeys:string[] = this.convertToArray(snap.val());
if (checkpointKeysToGet.length == 0){
checkpointKeysToGet = checkpointKeysToGet.concat(tagKeys);
}
else {
checkpointKeysToGet.forEach(existingTagKey => {
let tagKeysInBothTiers = tagKeys.filter(tagKey => {
return checkpointKeysToGet.includes(tagKey, 0)
});
checkpointKeysToGet = tagKeysInBothTiers;
console.log(checkpointKeysToGet);
});
}
});
});
}
All valid criticisms are welcome as long as you propose a solution to go along with it :)