Mixing
Is saveAll() of MongoRepository inserting data in one bulk?
I want to make the save operation efficient, so I'd like to write a bulk of objects to Mongo once in a while (i.e. when exceeding some capacity)
Would saveAll() do that for me? Should I use BulkOperations instead?
spring mongodb spring-boot
asked Jan 1 at 14:22
yasecoyaseco
43613
add a comment |
I want to make the save operation efficient, so I'd like to write a bulk of objects to Mongo once in a while (i.e. when exceeding some capacity)
Would saveAll() do that for me? Should I use BulkOperations instead?
spring mongodb spring-boot
asked Jan 1 at 14:22
yasecoyaseco
43613
add a comment |
I want to make the save operation efficient, so I'd like to write a bulk of objects to Mongo once in a while (i.e. when exceeding some capacity)
Would saveAll() do that for me? Should I use BulkOperations instead?
spring mongodb spring-boot
asked Jan 1 at 14:22
yasecoyaseco
43613
I want to make the save operation efficient, so I'd like to write a bulk of objects to Mongo once in a while (i.e. when exceeding some capacity)
Would saveAll() do that for me? Should I use BulkOperations instead?
spring mongodb spring-boot
spring mongodb spring-boot
asked Jan 1 at 14:22
yasecoyaseco
43613
asked Jan 1 at 14:22
yasecoyaseco
43613
asked Jan 1 at 14:22
yasecoyaseco
43613
asked Jan 1 at 14:22
yasecoyaseco
43613
asked Jan 1 at 14:22
yasecoyaseco
43613
43613
add a comment |
add a comment |
1 Answer
1
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oldest
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Short answer, yes, but only if all documents are new. If not, it will insert or update one by one.
Take a look at SimpleMongoRepository (MongoRepository's default implementation):
public <S extends T> List<S> saveAll(Iterable<S> entities) {
Assert.notNull(entities, "The given Iterable of entities not be null!");
Streamable<S> source = Streamable.of(entities);
boolean allNew = source.stream().allMatch((it) -> {
return this.entityInformation.isNew(it);
});
if (allNew) {
List<S> result = (List)source.stream().collect(Collectors.toList());
return new ArrayList(this.mongoOperations.insert(result, this.entityInformation.getCollectionName()));
} else {
return (List)source.stream().map(this::save).collect(Collectors.toList());
}
}
Notice that when all documents are new, the repository will use MongoOperations.insert method (MongoTemplate is the implementation), Then, if you look at that method's code you'll realize it does a batch insert:
public <T> Collection<T> insert(Collection<? extends T> batchToSave, String collectionName) {
Assert.notNull(batchToSave, "BatchToSave must not be null!");
Assert.notNull(collectionName, "CollectionName must not be null!");
return this.doInsertBatch(collectionName, batchToSave, this.mongoConverter);
}
Hope that helps!
answered Jan 1 at 14:56
DiegoDiego
28718
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Short answer, yes, but only if all documents are new. If not, it will insert or update one by one.
Take a look at SimpleMongoRepository (MongoRepository's default implementation):
public <S extends T> List<S> saveAll(Iterable<S> entities) {
Assert.notNull(entities, "The given Iterable of entities not be null!");
Streamable<S> source = Streamable.of(entities);
boolean allNew = source.stream().allMatch((it) -> {
return this.entityInformation.isNew(it);
});
if (allNew) {
List<S> result = (List)source.stream().collect(Collectors.toList());
return new ArrayList(this.mongoOperations.insert(result, this.entityInformation.getCollectionName()));
} else {
return (List)source.stream().map(this::save).collect(Collectors.toList());
}
}
Notice that when all documents are new, the repository will use MongoOperations.insert method (MongoTemplate is the implementation), Then, if you look at that method's code you'll realize it does a batch insert:
public <T> Collection<T> insert(Collection<? extends T> batchToSave, String collectionName) {
Assert.notNull(batchToSave, "BatchToSave must not be null!");
Assert.notNull(collectionName, "CollectionName must not be null!");
return this.doInsertBatch(collectionName, batchToSave, this.mongoConverter);
}
Hope that helps!
answered Jan 1 at 14:56
DiegoDiego
28718
add a comment |
Short answer, yes, but only if all documents are new. If not, it will insert or update one by one.
Take a look at SimpleMongoRepository (MongoRepository's default implementation):
public <S extends T> List<S> saveAll(Iterable<S> entities) {
Assert.notNull(entities, "The given Iterable of entities not be null!");
Streamable<S> source = Streamable.of(entities);
boolean allNew = source.stream().allMatch((it) -> {
return this.entityInformation.isNew(it);
});
if (allNew) {
List<S> result = (List)source.stream().collect(Collectors.toList());
return new ArrayList(this.mongoOperations.insert(result, this.entityInformation.getCollectionName()));
} else {
return (List)source.stream().map(this::save).collect(Collectors.toList());
}
}
Notice that when all documents are new, the repository will use MongoOperations.insert method (MongoTemplate is the implementation), Then, if you look at that method's code you'll realize it does a batch insert:
public <T> Collection<T> insert(Collection<? extends T> batchToSave, String collectionName) {
Assert.notNull(batchToSave, "BatchToSave must not be null!");
Assert.notNull(collectionName, "CollectionName must not be null!");
return this.doInsertBatch(collectionName, batchToSave, this.mongoConverter);
}
Hope that helps!
answered Jan 1 at 14:56
DiegoDiego
28718
add a comment |
Short answer, yes, but only if all documents are new. If not, it will insert or update one by one.
Take a look at SimpleMongoRepository (MongoRepository's default implementation):
public <S extends T> List<S> saveAll(Iterable<S> entities) {
Assert.notNull(entities, "The given Iterable of entities not be null!");
Streamable<S> source = Streamable.of(entities);
boolean allNew = source.stream().allMatch((it) -> {
return this.entityInformation.isNew(it);
});
if (allNew) {
List<S> result = (List)source.stream().collect(Collectors.toList());
return new ArrayList(this.mongoOperations.insert(result, this.entityInformation.getCollectionName()));
} else {
return (List)source.stream().map(this::save).collect(Collectors.toList());
}
}
Notice that when all documents are new, the repository will use MongoOperations.insert method (MongoTemplate is the implementation), Then, if you look at that method's code you'll realize it does a batch insert:
public <T> Collection<T> insert(Collection<? extends T> batchToSave, String collectionName) {
Assert.notNull(batchToSave, "BatchToSave must not be null!");
Assert.notNull(collectionName, "CollectionName must not be null!");
return this.doInsertBatch(collectionName, batchToSave, this.mongoConverter);
}
Hope that helps!
answered Jan 1 at 14:56
DiegoDiego
28718
Short answer, yes, but only if all documents are new. If not, it will insert or update one by one.
Take a look at SimpleMongoRepository (MongoRepository's default implementation):
public <S extends T> List<S> saveAll(Iterable<S> entities) {
Assert.notNull(entities, "The given Iterable of entities not be null!");
Streamable<S> source = Streamable.of(entities);
boolean allNew = source.stream().allMatch((it) -> {
return this.entityInformation.isNew(it);
});
if (allNew) {
List<S> result = (List)source.stream().collect(Collectors.toList());
return new ArrayList(this.mongoOperations.insert(result, this.entityInformation.getCollectionName()));
} else {
return (List)source.stream().map(this::save).collect(Collectors.toList());
}
}
Notice that when all documents are new, the repository will use MongoOperations.insert method (MongoTemplate is the implementation), Then, if you look at that method's code you'll realize it does a batch insert:
public <T> Collection<T> insert(Collection<? extends T> batchToSave, String collectionName) {
Assert.notNull(batchToSave, "BatchToSave must not be null!");
Assert.notNull(collectionName, "CollectionName must not be null!");
return this.doInsertBatch(collectionName, batchToSave, this.mongoConverter);
}
Hope that helps!
answered Jan 1 at 14:56
DiegoDiego
28718
edited Jan 1 at 15:29
answered Jan 1 at 14:56
DiegoDiego
28718
answered Jan 1 at 14:56
DiegoDiego
28718
answered Jan 1 at 14:56
DiegoDiego
28718
28718
add a comment |
add a comment |
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