In spring data mongodb how to achieve pagination f

Question

In spring data mongodb using mongotemplate or mongorepository, how to achieve pagination for aggregateion

Answer 1

To return a Paged Object with correct value of pageable object , I find this is the best and simple way.

Aggregation aggregation = Aggregation.newAggregation(Aggregation.match(Criteria.where("type").is("project")),
                        Aggregation.group("id").last("id").as("id"), Aggregation.project("id"),
                        Aggregation.skip(pageable.getPageNumber() * pageable.getPageSize()),
                        Aggregation.limit(pageable.getPageSize()));


    PageableExecutionUtils.getPage(mongoTemplate.aggregate(aggregation, Draft.class, Draft.class).getMappedResults(), pageable,() -> mongoTemplate.count(Query.of(query).limit(-1).skip(-1), Draft.class));

Answer 2

In addition to ssouris solution you can use Pageable classes for the results.

public Page<UserListItemView> list(final Pageable pageable) {

    final Aggregation agg = newAggregation(
        skip(pageable.getPageNumber() * pageable.getPageSize()),
        limit(pageable.getPageSize())
    );

    final List<UserListItemView> results = mongoTemplate
        .aggregate(agg, User.class, UserListItemView.class)
        .getMappedResults();

    return new PageImpl<>(results, pageable, results.size())
}

Answer 3

This is an answer to an old post, but I'll provide an answer in case anyone else comes along while searching for something like this.

Building on the previous solution by Fırat KÜÇÜK, giving the results.size() as the value for the "total" field in the PageImpl constructor will not making paging work the way, well, you expect paging to work. It sets the total size to the page size every time, so instead, you need to find out the actual total number of results that your query would return:

public Page<UserListItemView> list(final Pageable pageable) {
    long total = getCount(<your property name>, <your property value>);

    final Aggregation agg = newAggregation(
        skip(pageable.getPageNumber() * pageable.getPageSize()),
        limit(pageable.getPageSize())
    );

    final List<UserListItemView> results = mongoTemplate
        .aggregate(agg, User.class, UserListItemView.class)
        .getMappedResults();

    return new PageImpl<>(results, pageable, total);
}

Now, then, the best way to get the total number of results is another question, and it is one that I am currently trying to figure out. The method that I tried (and it worked) was to almost run the same aggregation twice, (once to get the total count, and again to get the actual results for paging) but using only the MatchOperation followed by a GroupOperation to get the count:

private long getCount(String propertyName, String propertyValue) {
    MatchOperation matchOperation = match(Criteria.where(propertyName).is(propertyValue));
    GroupOperation groupOperation = group(propertyName).count().as("count");
    Aggregation aggregation = newAggregation(matchOperation, groupOperation);
    return mongoTemplate.aggregate(aggregation, Foo.class, NumberOfResults.class).getMappedResults().get(0).getCount();
}

private class NumberOfResults {
    private int count;

    public int getCount() {
        return count;
    }

    public void setCount(int count) {
        this.count = count;
    }
}

It seems kind of inefficient to run nearly the same query twice, but if you are going to page results, the pageable object must know the total number of results if you really want it to behave like paging. If anyone can improve on my method to get the total count of results, that would be awesome!

Edit: This will also provide the count, and it is simpler because you do not need a wrapper object to hold the result, so you can replace the entire previous code block with this one:

private long getCount(String propertyName, String propertyValue) {
    Query countQuery = new Query(Criteria.where(propertyName).is(propertyValue));
    return mongoTemplate.count(countQuery, Foo.class);
}

Answer 4

You can use MongoTemplate

org.spring.framework.data.mongodb.core.aggregation.Aggregation#skip
        and 
org.springframework.data.mongodb.core.aggregation.Aggregation#limit

Aggregation agg = newAggregation(
        project("tags"),
        skip(10),
        limit(10)
);

AggregationResults<TagCount> results = mongoTemplate.aggregate(agg, "tags", TagCount.class);
List<TagCount> tagCount = results.getMappedResults();

Answer 5

As per the answer https://stackoverflow.com/a/39784851/4546949 I wrote code for Java.

Use aggregation group to get count and array of data with other paging information.

    AggregationOperation group = Aggregation.group().count().as("total")
            .addToSet(pageable.getPageNumber()).as("pageNumber")
            .addToSet(pageable.getPageSize()).as("pageSize")
            .addToSet(pageable.getOffset()).as("offset")
            .push("$$ROOT").as("data");

Use Aggregation project to slice as per the paging information.

    AggregationOperation project = Aggregation.project()
            .andInclude("pageSize", "pageNumber", "total", "offset")
            .and(ArrayOperators.Slice.sliceArrayOf("data").offset((int) pageable.getOffset()).itemCount(pageable.getPageSize()))
            .as("data");

Use mongo template to aggregate.

    Aggregation aggr = newAggregation(group, project);
    CustomPage page = mongoTemplate.aggregate(aggregation, Foo.class, CustomPage.class).getUniqueMappedResult();

Create a CustomPage.

    public class CustomPage {
        private long pageSize;
        private long pageNumber;
        private long offset;
        private long total;
        private List<Foo> data;
    }