具有复杂过滤器和复合索引的Firestore [英] Firestore with complex filter and composite indexes
问题描述
我使用Cloud Firestore，这是我的数据结构:
I use Cloud Firestore and this is my data structure :
Firestore-root
|
--- collection (collection)
|
--- uid (document)
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--- name: some_name
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--- some_fields: some_data
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--- origin: [origin_1, origine_2, ...] <- List
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--- category: [category_1, category_2, ...] <- List
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--- meals: [meals_1, meals_2, ...] <- List
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--- some_other_filters: [.....] <- List
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--- all_categories: [origin_1:true, origine_2:true, meals_1:true, ....]
我创建了Map字段all_categories来容纳所有不同的过滤器(来源，类别等)，这样我就可以轻松地执行查询，例如:如果用户选择了foods_4和category_2，category_6和some_other_filter，我可以这样做:
I created the Map field all_categories to hold all different filters (origin, category, ...) so that I can perform queries easily like: if user select meals_4 and category_2, category_6 and some_other_filter I can do:
List<String> filters = getFilters();
query = recipeRepository.getUsersRecipeCollection()
.orderBy(sorted_filter, DESCENDING).limit(5);
for (String item: filters)
query = query.whereEqualTo("all_categories."+item, true);
query.get().addOnCompleteListener(...)
我不能使用 whereArrayContainsAny 函数，因为我需要根据选定的过滤器获取确切的项目.我需要的是"whereArrayContainsExactly .."
I cannot use the whereArrayContainsAny function since I need to get the exact items based on selected filters. What I need is something like "whereArrayContainsExactly.."
因此，例如，如果用户从我的过滤器页面中选择了1、2，..或25个不同的项目(例如:category_1，meats_2，origin_1，...)，这是否意味着我需要创建所有25个组合综合索引的数量(通过我从控制台获得的链接)?还是我为该地图字段创建25个索引?
So, for example, if a user select 1, 2,.. or 25 different items from my filter page (like: category_1, meals_2, origin_1,... ), does it mean I need to create all 25 combinations of the composite indexes (through the link I get from the console) ? or I create 25 indexes of that map fields?
第二次尝试:
我对数据进行非规范化，如下所示:
I denormalised the data as following:
保存所有数据的集合:
过滤器集合:
如果选择饭食_1"和起源_1"，则可以通过 whenAllSuccess 来获取所有文档ID，如下所示:
If I select meals_1 and origin_1, With whenAllSuccess I can get all document ids like the following:
Iterator<String> iterator = filters.iterator();
List<Task> tasks = new ArrayList<>();
while (iterator.hasNext()) {
String filter = iterator.next();
tasks.add(recipeRepository.getFilterCollection("filter_"+filter).get());
}
Tasks.whenAllSuccess(tasks.toArray(new Task[tasks.size()]))
.addOnSuccessListener(new OnSuccessListener<List<Object>>() {
@Override
public void onSuccess(List<Object> objects) {
if (!objects.isEmpty()) {
for (Object querySnapshot: objects) {
QuerySnapshot querySnap = (QuerySnapshot)querySnapshot;
for (DocumentSnapshot id: querySnap.getDocuments()) {
doc_ids.add(id.getId());
}
}
CollectionReference collectionReference = recipeRepository.getUsersRecipeCollection();
List<DocumentReference> listDocRef = new ArrayList<>();
for (String id: doc_ids) {
DocumentReference docRef = collectionReference.document(id);
listDocRef.add(docRef);
}
List<Task<DocumentSnapshot>> tasks = new ArrayList<>();
for (DocumentReference documentReference : listDocRef) {
Task<DocumentSnapshot> documentSnapshotTask = documentReference.get();
tasks.add(documentSnapshotTask);
}
Tasks.whenAllSuccess(tasks).addOnSuccessListener(new OnSuccessListener<List<Object>>() {
@Override
public void onSuccess(List<Object> list) {
Log.e(TAG, ">> list objects: " + list.size() + " data: " + list);
for (Object object : list) {
Recipes recipes = ((DocumentSnapshot) object).toObject(Recipes.class);
if (recipes == null) continue;
Log.e(TAG, ">> recipe name: " + recipes.getName());
}
}
});
}
}
});
通过此操作，我获得了包含进餐_1或起源_1的所有数据(与使用 whereArrayContainsAny 的结果相同)
With this I get all data containing meals_1 OR origin_1 (same result as using whereArrayContainsAny).
我需要获取的餐单和origin_1集合中同时存在的文档ID的列表.
What I need is to get the list of docs ids that exists in both meals_1 and origin_1 collections.
推荐答案
因此，例如，如果用户从我的过滤器页面中选择了1、2，..或25个不同的项目(例如:category_1，meats_2，origin_1，...)，这是否意味着我需要创建所有25个组合综合索引的数量(通过我从控制台获得的链接)?还是我为该地图字段创建25个索引?
So, for example, if a user select 1, 2,.. or 25 different items from my filter page (like: category_1, meals_2, origin_1,... ), does it mean I need to create all 25 combinations of the composite indexes (through the link I get from the console) ? or I create 25 indexes of that map fields?
正如我在for循环中看到的，在每次迭代中，您都在创建一个新的Query对象，这是因为索引是必需的.
As I see in the for loop, at each iteration you are creating a new Query object, and this is because Cloud Firestore queries are immutable. Because you are using in the same query a .orderBy() call along with a .whereEqualTo() call, an index is required.
要回答您的问题，是的，对于您执行的每个非常不同的查询都需要一个索引.
And to answer your question, yes, an index is required for each and very different query that you perform.
另一种解决方案可能是对数据进行非正规化，通过添加每种餐食的单独集合，并且每次添加过滤器时，您都应该为所选集合创建一个新查询.对于NoSQL数据库，这是一种常见的做法.
An alternative solution might be to denormalize the data, by adding separate collections of each type of meal, and every time a filter is added you should create a new query for the selected collection. This is a common practice when it comes to NoSQL databases.
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