Spring Data MongoDB-以及与其他集合的聚合

Question

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，但主要语言将是任何人。

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方式分开。

{
    "_id" : ObjectId("5ec65750fdcd4e960f4b2f24"),
    "technologyCd" : "AB",
    "technologyName" : "My ABC",
    "ltechnologyNativeName" : "XY",
    "status" : "A"
}
所以，我完成了如下所示的关联-

注意：多个技术可以与一个雇员相关联

一个员工可以与多种技术关联

员工只能拥有一种主要技术

{
    "_id" : ObjectId("5ec507c72d8c2136245d35ce"),
    "firstName" : 442,
    "lastName" : "LU",
    "email" : "LUX",
    .......
    .......
    .......
    "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"
    ],
    ....
    ...
}

我在下面的查询中使用了，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"
        }
    ],
    "countryPhoneCodes" : [ 
        "+352"
    ],
    ....

}

我正在使用Spring Boot 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，但我必须像下面那样建模，但是主要语言是...

Answer 1

如果您在代码中使用以下查找操作，则应该能够按预期方式获得答案，并且您无需在代码中进行分组操作。