mysql – 如何使用JPA引用的表的主键更新一个表中的外键？
作者：互联网
我有两张表如下,
用户
+--------+---------------+------------+--------+-----------+------------+--------------+----------------+----------------+-------------+
| USERID | EMAIL | FIRST_NAME | HONORS | LAST_NAME | LOGIN_TYPE | PHONE_NUMBER | PROFILE_PIC | RECENT_CONV_ID | LOCATION_ID |
+--------+---------------+------------+--------+-----------+------------+--------------+----------------+----------------+-------------+
| 1 | asf@gmail.com | ghj | 0 | ert | 0 | 9879878 | http://vvv.com | NULL | NULL |
——– ————— ———— ——– ——- —- ———— ————– —————- —- ———— ————-
USER_LOCATION
+------------+-------+---------+----------+------------+-----------+-------+
| LOCATIONID | CITY | COUNTRY | LATITUDE | LOCAL_ADDR | LONGITUDE | STATE |
+------------+-------+---------+----------+------------+-----------+-------+
| 1 | xyz | mm | 10 | asfdasf | 10 | qqq |
+------------+-------+---------+----------+------------+-----------+-------+
下面是两个表的CREATE TABLE查询,
CREATE TABLE `USER` (
`USERID` bigint(20) NOT NULL AUTO_INCREMENT,
`EMAIL` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL,
`FIRST_NAME` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL,
`HONORS` bigint(20) NOT NULL,
`LAST_NAME` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL,
`LOGIN_TYPE` int(11) NOT NULL,
`PHONE_NUMBER` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL,
`PROFILE_PIC` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL,
`RECENT_CONV_ID` bigint(20) DEFAULT NULL,
`LOCATION_ID` bigint(20) DEFAULT NULL,
PRIMARY KEY (`USERID`),
KEY `USER_N50` (`RECENT_CONV_ID`),
KEY `USER_N49` (`LOCATION_ID`),
CONSTRAINT `USER_FK1` FOREIGN KEY (`RECENT_CONV_ID`) REFERENCES `RECENT_CONVERSATION` (`ID`),
CONSTRAINT `USER_FK2` FOREIGN KEY (`LOCATION_ID`) REFERENCES `USER_LOCATION` (`LOCATIONID`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1 |
CREATE TABLE `USER_LOCATION` (
`LOCATIONID` bigint(20) NOT NULL AUTO_INCREMENT,
`CITY` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin DEFAULT NULL,
`COUNTRY` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin DEFAULT NULL,
`LATITUDE` double DEFAULT NULL,
`LOCAL_ADDR` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin DEFAULT NULL,
`LONGITUDE` double DEFAULT NULL,
`STATE` varchar(255) CHARACTER SET latin1 COLLATE latin1_bin DEFAULT NULL,
PRIMARY KEY (`LOCATIONID`),
UNIQUE KEY `USER_LOCATION_U1` (`LOCAL_ADDR`,`CITY`,`STATE`,`COUNTRY`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1
现在,我想使用USER_LOCATION的LOCATIONID更新USER中的LOCATION_ID.我如何使用JPA实现它？
我的Java课程：
@Entity(name="USER")
public class User {
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private long userId;
@Column(name="PHONE_NUMBER", nullable=false)
private String phoneNumber;
@Column(name="FIRST_NAME", nullable=false)
private String firstName;
@Column(name="LAST_NAME", nullable=false)
private String lastName;
@Column(name="EMAIL", nullable=false)
private String email;
@Column(name="PROFILE_PIC", nullable=false)
private String profilepic;
@Column(name="LOGIN_TYPE", nullable=false)
private int loginType;
@Column(name="HONORS", nullable=false)
private long honors;
@ManyToOne(cascade={CascadeType.PERSIST})
@JoinColumn(name="LOCATION_ID")
private UserLocation userLocation;
@OneToMany(cascade=CascadeType.PERSIST)
@JoinColumn(name="RECENT_CONV_ID")
private RecentConversation recentConversation;
}
@Entity(name="USER_LOCATION")
@Table(name="USER_LOCATION", uniqueConstraints=@UniqueConstraint(columnNames={"LOCAL_ADDR", "CITY", "STATE", "COUNTRY"}))
@NamedQuery(name="addUserLocation", query="SELECT l FROM USER_LOCATION l " +
"WHERE l.local_addr = :lo_addr AND " +
"l.city = :city AND " +
"l.state = :state AND " +
"l.country = :country")
public class UserLocation {
@Id
@GeneratedValue
private long locationId;
@Column(name="LATITUDE")
private Double latitude;
@Column(name="LONGITUDE")
private Double longitude;
@Column(name="LOCAL_ADDR")
private String local_addr;
@Column(name="CITY")
private String city;
@Column(name="STATE")
private String state;
@Column(name="COUNTRY")
private String country;
@OneToMany(mappedBy="userLocation")
private Collection<User> users = new HashSet<User>();
}
请注意,我正在尝试实现的业务规则是,USER_LOCATION中应该没有基于UNIQUE KEY USER_LOCATION_U1的重复条目.此外,如果同一位置有多个用户,则USER中的LOCATION_ID应更新为该USER_LOCATION.非常感谢.
更新：
我的测试用例,
public class UserTest extends TestCase{
EntityManager em;
public void testUsersFromLocation() {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("TalkExchange");
em = emf.createEntityManager();
User user = createNewUser();
em.getTransaction().begin();
// em.persist(user.getUserLocation());
em.merge(user);
em.flush();
em.detach(user.getUserLocation());
em.contains(user.getUserLocation());
em.contains(user);
em.getTransaction().commit();
getUsersAtLocation();
}
public User createNewUser() {
User user = new User();
user.setEmail("asf@gmail.com");
user.setFirstName("fgfg");
user.setLastName("uiu");
user.setLoginType(0);
user.setPhoneNumber("7777");
user.setProfilepic("http://vvv.com");
user.setUserId(234);
UserLocation userLocation = createUserLocation();
user.setUserLocation(userLocation);
// UserLocation userLocation = getExistingUserLocation();
// user.setUserLocation(userLocation);
userLocation.getUsers().add(user);
return user;
}
public User createNewUser() {
User user = new User();
user.setEmail("asf@gmail.com");
user.setFirstName("fgfg");
user.setLastName("uiu");
user.setLoginType(0);
user.setPhoneNumber("7777");
user.setProfilepic("http://vvv.com");
user.setUserId(234);
UserLocation userLocation = createUserLocation();
user.setUserLocation(userLocation);
// UserLocation userLocation = getExistingUserLocation();
// user.setUserLocation(userLocation);
userLocation.getUsers().add(user);
return user;
}
public UserLocation createUserLocation() {
UserLocation userLocation = new UserLocation();
userLocation.setCity("wrwer");
userLocation.setCountry("MM");
userLocation.setLatitude(new Double(10));
userLocation.setLongitude(new Double(10));
userLocation.setLocal_addr("dfdfd");
userLocation.setState("kjlkj");
// create a query to find out whether the above UserLocation exists in the database.
// if(exists)
// use the existing location
// else
// use add the new location
addLocationRule(userLocation);
return userLocation;
}
}
解决方法:
您应该从此关系中删除CascadeType.PERSIST
@ManyToOne(cascade={CascadeType.PERSIST})
@JoinColumn(name="LOCATION_ID")
private UserLocation userLocation;
级联持久意味着无论何时保存新的User实例,它都会尝试保存新的UserLocation.这显然不是你想要的ManyToOne关系.
您应该在创建User对象之前创建UserLocation实例,然后在用户实例中重用相同的用户位置.
标签：mysql,jpa,datanucleus 来源： https://codeday.me/bug/20190709/1409843.html