如今有張表,搜索的歷史記錄,須要統計每一個字出現次數,若是數據庫是關係型的話,很好解決group by count order一會兒就能搞定,如今是Mongo不少事情變得麻煩起來了.....html
首先看mongo中分組的用法:java
db.appUser.aggregate([{"$group":{_id:"$userid",count:{$sum:1}}},{$sort:{count:-1}}])
知道了這個用法,再怎麼用Java寫出來了呢?mysql
我找啊找,找啊找,找啊找。。。。。。。。。。。spring
終於在spring的官網上找到答案了:sql
http://docs.spring.io/spring-data/data-mongodb/docs/current/reference/html/mongodb
ps:打開網頁,頁面很長,而後Ctrl+F查找Aggregation數據庫
若是是截取再分組:app
> db.user_start_app.aggregate([{"$group":{_id:{"$substr" : ["$citycode",0 ,4]},count:{$sum:1}}},{$sort:{count:-1}}]) { "_id" : "4403", "count" : 9 } { "_id" : "6103", "count" : 6 } { "_id" : "0755", "count" : 3 } { "_id" : "4212", "count" : 1 } >
一 :簡單的分組查詢--使用Mongo自己提供的AggregationOutput進行分組查詢學習
//此查詢語句== select eval as eval, count(*) as docsNum from news group by eval having docsNum>=85 order by docsNum desc //具體的mongodb和sql的對照能夠參考:http://docs.mongodb.org/manual/reference/sql-aggregation-comparison/ String groupStr = "{$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:獲取和testGroup1方法一樣結果的另外一種寫法,Spring Data MongoDB隆重登場,語法更加簡潔易懂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官網:
http://docs.mongodb.org/manual/applications/aggregation/
spring-data-mongodb:http://docs.spring.io/spring-data/data-mongodb/docs/current/reference/htmlsingle/#mongo.aggregation