如今有张表,搜索的历史记录,须要统计每一个字出现次数,若是数据库是关系型的话,很好解决group by count order一会儿就能搞定,如今是Mongo不少事情变得麻烦起来了.....html
首先看mongo中分组的用法:java
db.appUser.aggregate([{"$group":{_id:"$userid",count:{$sum:1}}},{$sort:{count:-1}}])
知道了这个用法,再怎么用Java写出来了呢?mysql
我找啊找,找啊找,找啊找。。。。。。。。。。。spring
终于在spring的官网上找到答案了:sql
http://docs.spring.io/spring-data/data-mongodb/docs/current/reference/html/mongodb
ps:打开网页,页面很长,而后Ctrl+F查找Aggregation数据库
若是是截取再分组:app
> db.user_start_app.aggregate([{"$group":{_id:{"$substr" : ["$citycode",0 ,4]},count:{$sum:1}}},{$sort:{count:-1}}]) { "_id" : "4403", "count" : 9 } { "_id" : "6103", "count" : 6 } { "_id" : "0755", "count" : 3 } { "_id" : "4212", "count" : 1 } >
一 :简单的分组查询--使用Mongo自己提供的AggregationOutput进行分组查询学习
//此查询语句== select eval as eval, count(*) as docsNum from news group by eval having docsNum>=85 order by docsNum desc //具体的mongodb和sql的对照能够参考:http://docs.mongodb.org/manual/reference/sql-aggregation-comparison/ String groupStr = "{$group:{_id:{'eval':'$eval'},docsNum:{$sum:1}}}"; DBObject group = (DBObject) JSON.parse(groupStr); String matchStr = "{$match:{docsNum:{$gte:85}}}"; DBObject match = (DBObject) JSON.parse(matchStr); String sortStr = "{$sort:{_id.docsNum:-1}}"; DBObject sort = (DBObject) JSON.parse(sortStr); AggregationOutput output = mongoTemplate.getCollection("news").aggregate(group, match, sort); System.out.println(output.getCommand());
二:获取和testGroup1方法一样结果的另外一种写法,Spring Data MongoDB隆重登场,语法更加简洁易懂lua
TypedAggregation<News> agg = Aggregation.newAggregation( News.class, project("evaluate") ,group("evaluate").count().as("totalNum") ,match(Criteria.where("totalNum").gte(85)) ,sort(Sort.Direction.DESC, "totalNum") ); AggregationResults<BasicDBObject> result = mongoTemplate.aggregate(agg, BasicDBObject.class); System.out.println(agg.toString());
mongodb官网:
http://docs.mongodb.org/manual/applications/aggregation/
spring-data-mongodb:http://docs.spring.io/spring-data/data-mongodb/docs/current/reference/htmlsingle/#mongo.aggregation