有以下两个集合:
用户集合
{
userId:user1,
creationTimeStamp:2019-11-05T08:15:30
status:active
},
{
userId:user2,
creationTimeStamp:2019-10-05T08:15:30
status:active
}文档集合
{
userId:user1,
category:Development
published:true
},
{
userId:user2,
category:Development
published:false
}我想连接这两个集合,并过滤用户,以便在创建时间戳之间,文档属于开发类别,并且不是从活动用户发布的
如何编写mongodb java聚合才能得到这样的结果:
{
userId: user2,
status:active,
category:Development,
published:false
}发布于 2020-04-20 14:31:36
您还没有提到User集合中userId的重复。
所以这个脚本是
[{
$match: {
category: "Development",
published: false
}
}, {
$lookup: {
from: 'user',
localField: 'userId',
foreignField: 'userId',
as: 'joinUser'
}
}, {
$unwind: {
path: "$joinUser",
preserveNullAndEmptyArrays: true
}
}, {
$match: {
"joinUser.status": "active"
}
}, {
$addFields: {
"status": "$joinUser.status"
}
}, {
$project: {
_id: 0,
userId: 1,
category: 1,
published: 1,
status: 1
}
}]和java代码,
包括这些导入
import static org.springframework.data.mongodb.core.aggregation.Aggregation.match;
import static org.springframework.data.mongodb.core.aggregation.Aggregation.lookup;
import static org.springframework.data.mongodb.core.aggregation.Aggregation.unwind;
import static org.springframework.data.mongodb.core.aggregation.Aggregation.project;方法是,
public Object findAllwithVideos() {
Aggregation aggregation=Aggregation.newAggregation(
match(Criteria.where("category").is("Development").and("published").is(false)),
lookup("user","userId","userId","joinUser"),
unwind("joinUser",true),
new AggregationOperation(){
@Override
public Document toDocument(AggregationOperationContext aggregationOperationContext){
return new Document("$addFields",
new Document("status","$joinUser.status")
);
}
},
project("userId","category","published","status")
).withOptions(AggregationOptions.builder().allowDiskUse(Boolean.TRUE).build());
return mongoTemplate.aggregate(aggregation, mongoTemplate.getCollectionName(Document.class), Object.class);
}https://stackoverflow.com/questions/61314016
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