JPA連接兩個輔助表

Question

我有這五個表：

REF_CT          REF_CT_L        REF_ST        REF_ST_L
======          ========        ======        ========
CT_ID           CT_ID           ST_ID         ST_ID
ST_ID           LANGUAGE_ID     NAME          LANGUAGE_ID
NAME            NAME            REF_CODE      NAME
REF_CODE        DESCRIPTION                   DESCRIPTION


REF_LANGUAGE
============
LANGUAGE_ID
LOCALE_CODE
NAME

通過使用JPA，我創建了以下三個實體類：

1. 類common.contact.City

    @Entity @Table(name = "REF_CT") @SecondaryTable(name = "REF_CT_L") public class City implements Reference { @Id @Column(name = "CT_ID") @TableGenerator(name = "TABLE_GEN", table = "SYS_SEQ", pkColumnName = "SEQ_NAME", valueColumnName = "SEQ_VALUE", pkColumnValue = "CITY_SEQ") @GeneratedValue(strategy = GenerationType.TABLE, generator = "TABLE_GEN") private Integer id; @Column(name = "NAME") private String name; @Column(table = "REF_CT_L", name = "NAME") private String value; @Column(name = "REF_CODE") private String refCode; @ManyToOne(targetEntity = Language.class) @JoinColumn(table = "REF_CT_L", name = "LANGUAGE_ID") private Language language; @ManyToOne(targetEntity = State.class, optional = true) @JoinColumn(name = "ST_ID", referencedColumnName = "ST_ID") private State state; 

2. 類common.contact.State

    @Entity @Table(name = "REF_ST") @SecondaryTable(name = "REF_ST_L") public class State implements Reference { @Id @Column(name = "ST_ID") @TableGenerator(name = "TABLE_GEN", table = "SYS_SEQ", pkColumnName = "SEQ_NAME", valueColumnName = "SEQ_VALUE", pkColumnValue = "STATE_SEQ") @GeneratedValue(strategy = GenerationType.TABLE, generator = "TABLE_GEN") private Integer id; @Column(name = "NAME") private String name; @Column(table = "REF_ST_L", name = "NAME") private String value; @Column(name = "REF_CODE") private String refCode; @OneToMany(targetEntity = City.class, mappedBy = "state") private Set<City> cities; @ManyToOne(targetEntity = Language.class) @JoinColumn(table = "REF_ST_L", name = "LANGUAGE_ID", unique = true, referencedColumnName = "LANGUAGE_ID") private Language language; 

3. 類common.system.Language

    @Entity @Table(name = "SYS_LANGUAGE") public class Language implements BaseEntity { @Id @Column(name = "LANGUAGE_ID") @TableGenerator(name = "TABLE_GEN", table = "SYS_SEQ", pkColumnName = "SEQ_NAME", valueColumnName = "SEQ_VALUE", pkColumnValue = "LANGUAGE_SEQ") @GeneratedValue(strategy = GenerationType.TABLE, generator = "TABLE_GEN") private Integer id; @Column(name = "LOCALE_CODE") private String locale; @Column(name = "NAME") private String name; 

但是當我嘗試運行此查詢時

select city from City city where city.id = 1 

我收到此錯誤：

org.hibernate.HibernateException: More than one row with the given identifier was found: 10, for class: common.contact.State

僅供參考，添加了以下數據：

REF_CT

CT_ID    ST_ID    NAME    REF_CODE
=====    =====    ====    ========
  1        10    CHERAS     CHE

REF_CT_L

CT_ID    LANGUAGE_ID    NAME       DESCRIPTION
=====    ===========    ====       ===========
  1           1      BANDAR CHERAS  CHERAS BARU
  1           2      CHERAS TOWN    NEW CHERAS  

REF_ST

ST_ID    NAME     REF_CODE
=====    ====     ========
  10    SELANGOR    SEL

REF_ST_L

ST_ID    LANGUAGE_ID     NAME      DESCRIPTION
=====    ===========     ====      ===========
  10          1        SELANGOR    SELANGOR BARU
  10          2        SERIANGOR   NEW SERIANGOR

REF_LANGUAGE

LANGUAGE_ID    LOCALE_CODE    NAME
===========    ===========    ====
     1           ms_En       Malay
     2           en_US       English

生成的SQL是：

SELECT state0_.st_id AS st1_50_0_,
       state0_.NAME AS name50_0_, state0_.ref_code AS ref4_50_0_,
       state0_1_.language_id AS language2_51_0_, state0_1_.NAME AS name51_0_
  FROM ref_st state0_ LEFT OUTER JOIN ref_st_l state0_1_
       ON state0_.st_id = state0_1_.st_id
 WHERE state0_.st_id = 10;

謝謝你的幫助！

Answer 1

您的代碼產生了LEFT JOIN，因為您將City-> State關系設置為可選。 將其更改為不可為空，您將獲得一個內部聯接，該聯接應解析返回的多行。

common.contact.City->

@ManyToOne(targetEntity = State.class, optional = false)
@JoinColumn(name = "ST_ID", referencedColumnName = "ST_ID", nullable = false)
private State state;

Answer 2

“ REF_ST”與“ REF_ST_L”具有一對多關系。 這意味着“ REF_ST_L”不能映射為輔助表（這意味着一對一關系）。

對於“ REF_CT”->“ REF_CT_L”，相同的邏輯有效。