Spring Data Mongo-在嵌入式文档中应用唯一的组合字段

我在做 Spring Boot v2.1.3.RELEASE & Spring Data Mongo .在本例中,我想在电子邮件和;部门名称。结合 电邮及;部门名称 必须是唯一的,并且有任何方法可以将电子邮件发送出去,因为它在每个数组对象中重复?

我在下面试过了,但没用!

@CompoundIndexes({
    @CompoundIndex(name = "email_deptName_idx", def = "{'email' : 1, 'technologyEmployeeRef.technologyCd' : 1}")
})

样本数据

{
    "_id" : ObjectId("5ec507c72d8c2136245d35ce"),
    ....
    ....
    "firstName" : "John",
    "lastName" : "Doe",
    "email" : "john.doe@gmail.com",
    .....
    .....
    .....
    "technologyEmployeeRef" : [ 
        {
            "technologyCd" : "john.doe@gmail.com",
            "technologyName" : "Advisory",
            ....
            .....
            "Status" : "A"
        }, 
        {
           "technologyCd" : "john.doe@gmail.com",
           "technologyName" : "Tax",
           .....
           .....
           "Status" : "A"
       }
    ],
    "phoneCodes" : [ 
        "+352"
    ],
    ....
    ....
}

技术JAVA

@Data
@Builder
@AllArgsConstructor
@NoArgsConstructor
@Document
public class Technology {
    @Indexed(name = "technologyCd", unique = true, sparse = true)
    private String technologyCd;

    @Indexed(name = "technologyName", unique = true, sparse = true)
    private String technologyName;
    private String status;
}

员工技术参考。JAVA

@Data
@Builder
@AllArgsConstructor
@NoArgsConstructor
public class EmployeeTechnologyRef {
    private String technologyCd;
    private String primaryTechnology;
    private String status;
}

受雇者JAVA

@Data
@Builder
@AllArgsConstructor
@NoArgsConstructor
@Document
@CompoundIndexes({
    @CompoundIndex(name="emp_tech_indx", def = "{'employeeTechnologyRefs.primaryTechnology' : 1, 'employeeTechnologyRefs.technologyCd' : 1}" ,unique = true, sparse = true)
})
public class Employee {
    private String firstName;
    private String lastName;
    private String email;
    private List<EmployeeTechnologyRef> employeeTechnologyRefs;
}

我使用了下面的代码,但它没有给我任何重复的错误。我们该怎么做?

Technology java8 = Technology.builder().technologyCd("Java").technologyName("Java8").status("A").build();
Technology spring = Technology.builder().technologyCd("Spring").technologyName("Spring Boot2").status("A").build();
List<Technology> technologies = new ArrayList<>();
technologies.add(java8);
technologies.add(spring);

technologyRepository.saveAll(technologies);

EmployeeTechnologyRef t1 = EmployeeTechnologyRef.builder().technologyCd("Java").primaryTechnology("Y")
        .status("A")
        .build();
EmployeeTechnologyRef t2 = EmployeeTechnologyRef.builder().technologyCd("Spring").primaryTechnology("Y")
        .status("A")
        .build();
List<EmployeeTechnologyRef> employeeTechnologyRefs = new ArrayList<>();
employeeTechnologyRefs.add(t1);
employeeTechnologyRefs.add(t2);
employeeTechnologyRefs.add(t1);

Employee employee = Employee.builder().firstName("John").lastName("Kerr").email("john.kerr@gmail.com")
        .employeeTechnologyRefs(employeeTechnologyRefs).build();
employeeRepository.save(employee);