Heim > Fragen und Antworten > Hauptteil
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.................
某草草2017-04-27 09:04:29
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: "$age"}
}
}
]
). forEach(function(x){
db.temp.insert(
{
name: x.name,
sex : x.sex,
age: x.age
}
);
});
習慣沉默2017-04-27 09:04:29
collection = db.tb;
result = collection.aggregate(
[
{"$group": { "_id": { market: "$market", code: "$code" } } }
]
);
printjson(result);
PHP中文网2017-04-27 09:04:29
> 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" ]
楼上的方法是正确的,mongoDB确实不支持组合去重,mongo官网提供的示例也只是对单个字段去重。
官网示例: http://docs.mongodb.org/manual/core/single-purpose-aggregation/