Spring Data MongoDB-与其他集合的聚合

Pra_A 发表于 Java

Pra_A：

我正在研究Spring Boot v.2.1.3.3.RELEASE和Spring Data MongoDB。由于关键的要求，我假设员工知道多种技术，但我只能像下面这样建模，但是主要的语言是任何人。

因此，我决定将技术集合与在员工集合中与员工和技术相关的某种方式分开。

{
    "_id" : ObjectId("5ec65750fdcd4e960f4b2f24"),
    "technologyCd" : "AB",
    "technologyName" : "My ABC",
    "ltechnologyNativeName" : "XY",
    "status" : "A"
}

因此，我完成了如下所示的关联-

注意：多个技术可以与一名员工关联

一名员工可以与多种技术相关联

员工只能拥有一种主要技术

{
    "_id" : ObjectId("5ec507c72d8c2136245d35ce"),
    "firstName" : "John",
    "lastName" : "Doe",
    "email" : "[email protected]",
    .......
    .......
    .......
    "employeeTechnologyRefs" : [ 
        {
            "technologyCd" : "AB",
            "primaryTechnologySw" : "Y",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "AB",
            "primaryTechnologySw" : "N",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "PR",
            "primaryTechnologySw" : "N",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "PR",
            "primaryTechnologySw" : "N",
            "Active" : "A"
        }
    ],
    "countryPhoneCodes" : [ 
        "+352"
    ],
    ....
    ...
}

我在下面的查询中，如何查询技术文档以获取结果并将其映射并创建最终对象？

Criteria criteria = new Criteria();
criteria.andOperator(
        StringUtils.isNotBlank(firstName) ? Criteria.where("firstName").is(firstName.toUpperCase())
                : Criteria.where(""),
        StringUtils.isNotBlank(lastName) ? Criteria.where("lastName").is(lastName.toUpperCase())
                : Criteria.where(""),
        StringUtils.isNotBlank(email) ? Criteria.where("email").is(email.toUpperCase())
                : Criteria.where(""),
        StringUtils.isNotBlank(technologyCd) ? Criteria.where("employeeTechnologyRefs.technologyCd").is(technologyCd.toUpperCase())
                : Criteria.where(""));

MatchOperation matchStage = Aggregation.match(criteria);

GroupOperation groupOp = Aggregation
        .group("firstName", "lastName", "email","_id")
        .addToSet("employeeTechnologyRefs").as("employeeTechnologyRefs");

ProjectionOperation projectStage = Aggregation.project("employeeTechnologyRefs");

Aggregation aggregation = Aggregation.newAggregation(matchStage, groupOp, projectStage);

AggregationResults<CustomObject> results = mongoTemplate.aggregate(aggregation, mongoTemplate.getCollectionName(Employee.class), CustomObject.class);
System.out.println(results);

结果应如下图所示

{
    "_id" : ObjectId("5ec507c72d8c2136245d35ce"),
    "firstName" : 442,
    "lastName" : "LU",
    "email" : "LUX",
    .......
    .......
    .......
    "employeeTechnologyRefs" : [ 
        {
            "technologyCd" : "AB",
            "technologyName" : "My ABC",
            "ltechnologyNativeName" : "XY",
            "primaryTechnologySw" : "Y",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "AB",
            "technologyCd" : "AB",
            "technologyName" : "My ABC",
            "ltechnologyNativeName" : "XY",
            "primaryTechnologySw" : "Y",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "PR",
            "technologyCd" : "AB",
            "technologyName" : "My ABC",
            "ltechnologyNativeName" : "XY",
            "primaryTechnologySw" : "Y",
            "Active" : "A"
        }, 
        {
            "technologyCd" : "PR",
            "technologyCd" : "AB",
            "technologyName" : "My ABC",
            "ltechnologyNativeName" : "XY",
            "primaryTechnologySw" : "Y",
            "Active" : "A"
        }
    ],
    ....
    
}

迪帕克·辛格（Deepak Singh）：

如果在代码中使用下面的查找操作，则应该能够按预期获得答案，并且代码中不需要进行分组操作。

编辑答案：这就是整个代码的外观。还有一件事，您不需要投影，并且如果您只需要尝试投影特定字段，并且作为查找操作的一部分，请不要使用与集合中相同的字段，否则它将覆盖员工集合中的现有数据。

    Criteria criteria = new Criteria();
    criteria.andOperator(
            StringUtils.isNotBlank(firstName) ? Criteria.where("firstName").is(firstName.toUpperCase())
                    : Criteria.where(""),
            StringUtils.isNotBlank(lastName) ? Criteria.where("lastName").is(lastName.toUpperCase())
                    : Criteria.where(""),
            StringUtils.isNotBlank(email) ? Criteria.where("email").is(email.toUpperCase())
                    : Criteria.where(""),
            StringUtils.isNotBlank(technologyCd) ? Criteria.where("employeeTechnologyRefs.technologyCd").is(technologyCd.toUpperCase())
                    : Criteria.where(""));

    MatchOperation matchStage = Aggregation.match(criteria);

    /*GroupOperation groupOp = Aggregation
            .group("firstName", "lastName", "email","_id")
            .addToSet("employeeTechnologyRefs").as("employeeTechnologyRefs");
            */

    LookupOperation lookupOperation = LookupOperation.newLookup().
                                     from("technology_collection_name").
                                     localField("employeeTechnologyRefs.technologyCd").
                                     foreignField("technologyCd").
                                     as("employeeTechnologyRefsOtherName");

   /* ProjectionOperation projectStage = Aggregation.project("employeeTechnologyRefs");
  */  
// And if you want to project specific field from employee array you can use something like.
ProjectionOperation projectStage = Aggregation.project("employeeTechnologyRefs.fieldName")
    Aggregation aggregation = Aggregation.newAggregation(matchStage, lookupOperation, projectStage);

    AggregationResults<CustomObject> results = mongoTemplate.aggregate(aggregation, mongoTemplate.getCollectionName(Employee.class), CustomObject.class);
    System.out.println(results);

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