使用Spring Data Mongodb，是否有可能在不提取和迭代整个集合的情况下获得字段的最大值? [英] Using Spring Data Mongodb, is it possible to get the max value of a field without pulling and iterating over an entire collection?

问题描述

使用mongoTemplate.find()，我指定一个查询，可以通过它调用.limit()或.sort():

Using mongoTemplate.find(), I specify a Query with which I can call .limit() or .sort():

.limit()返回Query对象
.sort()返回一个Sort对象

.limit() returns a Query object
.sort() returns a Sort object

鉴于此，我可以说Query().limit(int).sort()，但这不能执行所需的操作，它只是对有限的结果集进行排序.

Given this, I can say Query().limit(int).sort(), but this does not perform the desired operation, it merely sorts a limited result set.

由于.sort()返回Sort()，我无法调用Query().sort().limit(int)

I cannot call Query().sort().limit(int) either since .sort() returns a Sort()

因此，使用Spring Data，如何执行mongoDB shell中所示的以下操作?也许有一种方法可以传递我尚未找到的原始查询?

So using Spring Data, how do I perform the following as shown in the mongoDB shell? Maybe there's a way to pass a raw query that I haven't found yet?

如果需要的话，我可以扩展Paging接口...只是似乎无济于事.谢谢！

I would be ok with extending the Paging interface if need be...just doesn't seem to help any. Thanks!

> j = { order: 1 }
{ "order" : 1 }
> k = { order: 2 }
{ "order" : 2 }
> l = { order: 3 }
{ "order" : 3 }
> db.test.save(j)
> db.test.save(k)
> db.test.save(l)
> db.test.find()
{ "_id" : ObjectId("4f74d35b6f54e1f1c5850f19"), "order" : 1 }
{ "_id" : ObjectId("4f74d3606f54e1f1c5850f1a"), "order" : 2 }
{ "_id" : ObjectId("4f74d3666f54e1f1c5850f1b"), "order" : 3 }
> db.test.find().sort({ order : -1 }).limit(1)
{ "_id" : ObjectId("4f74d3666f54e1f1c5850f1b"), "order" : 3 }

推荐答案

您可以在sping-data-mongodb中执行此操作.如果索引了排序字段(或@Id字段)，则Mongo将优化排序/限制组合.这会产生非常快的O(logN)或更好的结果.否则，它仍然是O(N)而不是O(N * logN)，因为它将使用top-k算法并避免使用全局排序( Mkyong的示例但我先进行排序，然后将限制设置为一秒.

You can do this in sping-data-mongodb. Mongo will optimize sort/limit combinations IF the sort field is indexed (or the @Id field). This produces very fast O(logN) or better results. Otherwise it is still O(N) as opposed to O(N*logN) because it will use a top-k algorithm and avoid the global sort (mongodb sort doc). This is from Mkyong's example but I do the sort first and set the limit to one second.

Query query = new Query();
query.with(new Sort(Sort.Direction.DESC, "idField"));
query.limit(1);
MyObject maxObject = mongoTemplate.findOne(query, MyObject.class);

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