Spring Data MongoDB Group by

Question

我在我的项目中使用 Spring Data Mongodb，并参考以下类来查询我对结果分组的查询：

学生班：

@Document(collection = "student")
public class Student {

    @Id
    private String id;

    private String firstName;

    private String lastName;

    //other fields

    //getters & setters

}

学生成绩 (dto)：

public class StudentResults {

    private String firstName;

    private List<String> studentIds; //I need List<Student> here

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public List<String> getStudentIds() {
        return studentIds;
    }

    public void setStudentIds(List<String> studentIds) {
        this.studentIds = studentIds;
    }
}

StudentServiceImpl 类：

public class StudentServiceImpl implements StudentService {
    @Autowired
    private MongoTemplate mongoTemplate;

    public List<StudentResults> findStudentsGroupByFirstName() {
        TypedAggregation<Student> studentAggregation = 
               Aggregation.newAggregation(Student.class,
               Aggregation.group("firstName").
               addToSet("id").as("studentIds"),
               Aggregation.project("studentIds").
               and("firstName").previousOperation());

        AggregationResults<StudentResults> results = mongoTemplate.
             aggregate(studentAggregation, StudentResults.class);

        List<StudentResults> studentResultsList = results.getMappedResults();

        return studentResultsList;
    }
}

使用上面的代码，我可以成功检索List<String> studentIds，但是我需要使用Aggregation.group()检索List<Student> students？你能帮忙吗？

Answer 1

将您的TypedAggregation 部分更改为下面并将students 字段添加到StudentResults

 TypedAggregation<Student> studentAggregation = Aggregation.newAggregation(Student.class,
               Aggregation.group("firstName").
               push("$$ROOT").as("students"));

$$ROOT 将推送整个文档。

更新：

TypedAggregation<Student> studentAggregation = Aggregation.newAggregation(Student.class,
              Aggregation.group("firstName").
                 push(new BasicDBObject
                       ("_id", "$_id").append
                       ("firstName", "$firstName").append
                       ("lastName", "$lastName")).as("students"));