去重 - MongoDB如何去除组合重复项
高洛峰
高洛峰 2017-04-27 09:02:29
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sql server有下列语句:

// 查找名字性别唯一的学生
select distinct name,sex from student

// 转换成MongoDB的话,我怎么把这种组合列去重,MongoDB好像只支持单个field的去重
db.student.distinct("name"); // 只支持一个field
db.student.distinct("name","sex"); // 错误的,不支持多个field的组合去重

有没有大神解决过类似问题,求指导.................
高洛峰
高洛峰

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reply all(3)
某草草
恩。这个方法试过了。
根据id分组,id指定为组合项的话,因为id不会重复,所以作用相当于把组合项去重了。

我给你们补充下:假如现在需要拿出collection的其他field的话,可以使用$push关键字

      // 根据name和sex分组
      // 把分组后的name,sex,age放到对应的Document下,形成一个数组
      db.student.aggregate(
            [
                  {
                        $group:{
                              _id: {name: "$name", sex: "$sex"},
                              name: {$push: "$name"},
                              sex: {$push: "$sex"},
                              age: {$push: "$age"}
                        }
                  }
            ]
        ). forEach(function(x){
            db.temp.insert(
                  {
                    name: x.name,
                    sex : x.sex,
                    age: x.age
                  }
            );
        });
習慣沉默
collection = db.tb;
result = collection.aggregate( 
            [
                {"$group": { "_id": { market: "$market", code: "$code" } } }
            ]
        );
printjson(result);
洪涛
> db.test.find()
{ "_id" : ObjectId("55de6075024f45b4947d4cc3"), "name" : "zhangsan", "sex" : "FEMALE", "age" : 12 }
{ "_id" : ObjectId("55de6076024f45b4947d4cc4"), "name" : "zhangsan", "sex" : "MALE", "age" : 16 }
{ "_id" : ObjectId("55de6076024f45b4947d4cc5"), "name" : "lisi", "sex" : "FEMALE", "age" : 20 }
{ "_id" : ObjectId("55de6076024f45b4947d4cc6"), "name" : "zhangsan", "sex" : "FEMALE", "age" : 25 }
{ "_id" : ObjectId("55de6076024f45b4947d4cc7"), "name" : "lisi", "sex" : "FEMALE", "age" : 36 }
{ "_id" : ObjectId("55de6077024f45b4947d4cc8"), "name" : "zhangsan", "sex" : "FEMALE", "age" : 28 }
> db.test.aggregate([
...     { $group: { _id: { name: "$name", sex: "$sex" } } }
... ])
{ "_id" : { "name" : "lisi", "sex" : "FEMALE" } }
{ "_id" : { "name" : "zhangsan", "sex" : "MALE" } }
{ "_id" : { "name" : "zhangsan", "sex" : "FEMALE" } }
> db.test.distinct("name");
[ "zhangsan", "lisi" ]
> db.test.distinct("name","sex");
[ "zhangsan", "lisi" ]
> db.test.distinct("name","sex","age");
[ "zhangsan", "lisi" ]

The method above is correct. MongoDB does not support combined deduplication. The example provided on the mongo official website only deduplicates a single field.

Official website example: http://docs.mongodb.org/manual/core/single-purpose-aggregation/

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