春季数据mongodb group by

Question

我在项目中使用Spring数据Mongodb，并在查询结果时参考以下类：

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，我能够List<String> studentIds成功检索到，但是我需要List<Student> students使用Aggregation.group()？进行检索。你能帮我吗？

Answer 1

将您的TypedAggregation零件更改为下面，并添加students字段到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"));