我想在兩個實體之間有一個JoinTable,這兩個實體是從同一個所有者實體派生的。因此,使用@ManyToMany關係我有@JoinTable(如在下面DDL)加入這個實體時:JPA:與@JoinTable的多對多關係每個關係都有相同的列
CREATE TABLE IF NOT EXISTS `local_services`.`service_provided_on` (
`provider_id` INT UNSIGNED NOT NULL,
`service_id` INT UNSIGNED NOT NULL,
`service_point_no` INT UNSIGNED NOT NULL,
`work_station_no` INT UNSIGNED NOT NULL,
PRIMARY KEY (`provider_id`, `service_id`, `service_point_no`, `work_station_no`),
INDEX `fk_provider_service_has_work_station_work_station1_idx` (`service_point_no` ASC, `work_station_no` ASC, `provider_id` ASC),
INDEX `fk_provider_service_has_work_station_provider_service1_idx` (`provider_id` ASC, `service_id` ASC),
CONSTRAINT `fk_service_provided_on_provider_service`
FOREIGN KEY (`provider_id` , `service_id`)
REFERENCES `local_services`.`provider_service` (`provider_id` , `service_id`)
ON DELETE CASCADE
ON UPDATE CASCADE,
CONSTRAINT `fk_service_provided_work_station`
FOREIGN KEY (`service_point_no` , `work_station_no` , `provider_id`)
REFERENCES `local_services`.`work_station` (`service_point_no` , `work_station_no` , `provider_id`)
ON DELETE CASCADE
ON UPDATE CASCADE)
ENGINE = InnoDB
正如你可以看到有2個外鍵和他們每個人使用相同的PROVIDER_ID列。我想定義使用這個@JoinTable,給定提供者提供的服務屬於它的WorkPlace(WorkStation)。顯而易見的是,服務商提供的ID爲ex。 5只能在屬於id爲5的提供者的工作場所提供。所以最好的做法是在每個ForeignKeys之間共享這個@JoinColumn。並嘗試例如插入工作場所/服務與不匹配的提供商ID以引發一些異常!
我嘗試做這樣的事情:
@ManyToMany(fetch = FetchType.LAZY)
@JoinTable(name = "service_provided_on",
joinColumns = {
@JoinColumn(name = "provider_id", referencedColumnName = "provider_id", nullable = false, columnDefinition = "BIGINT UNSIGNED"),
@JoinColumn(name = "service_id", referencedColumnName = "service_id", nullable = false, columnDefinition = "INT UNSIGNED")
},
inverseJoinColumns = {
@JoinColumn(name = "provider_id", referencedColumnName = "provider_id", insertable = false, updatable = false),
@JoinColumn(name = "service_point_no", referencedColumnName = "service_point_no", nullable = false, columnDefinition = "INT UNSIGNED"),
@JoinColumn(name = "work_station_no", referencedColumnName = "work_station_no", nullable = false, columnDefinition = "INT UNSIGNED")
}
)
但它顯然是行不通的,提高的例外是這樣的:
Caused by: org.hibernate.MappingException: Repeated column in mapping for collection: pl.salonea.entities.WorkStation.providedServices column: provider_id"}}
我考慮重新命名爲一個外鍵此PROVIDER_ID前。 work_station_provider_id,但然後我會被允許插入不匹配的provider_ids,也許我可以定義一些CONSTRAINT來阻止這種行爲(如何在JPA中定義這個)?它可以工作,但我會有相同的提供冗餘列provider_id