如何自动递增主键一次多个创建操作？
如何自动递增主键一次多个创建操作？
提问于 2015-12-10 16:20:52
这里Customer_id是join列。我在oneToMany关系中有两个表customer和address。因为两者都有主键是自动递增的。
以下是几个要点：
在创建新的relationship.
- Primary OneToMany中，对应表中的单向
键是自动创建的
在Customer.java中
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name="customer_id", unique=true, nullable=false)
private long customerId;
@OneToMany(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
@JoinTable(name="address", joinColumns={@JoinColumn(name="customer_id", referencedColumnName="customer_id")}, inverseJoinColumns={@JoinColumn(name="account_id", referencedColumnName="account_id")})
private List<Address> addresses;在Address.java中
@Column(name="customer_id", unique=true, nullable=false)
private int customerId;表：
CREATE TABLE `customer` (
`customer_id` smallint(5) unsigned NOT NULL AUTO_INCREMENT,
`user_name` varchar(45) NOT NULL,
`password` varchar(45) NOT NULL,
`encrypt_key` varchar(200) NOT NULL,
`first_name` varchar(45) NOT NULL,
`last_name` varchar(45) NOT NULL,
`email` varchar(50) DEFAULT NULL,
`active` tinyint(1) NOT NULL DEFAULT '1',
`create_date` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`last_update` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`self_description` varchar(2000) NOT NULL,
PRIMARY KEY (`customer_id`),
KEY `idx_last_name` (`last_name`)
) ENGINE=InnoDB AUTO_INCREMENT=101 DEFAULT CHARSET=utf8;
CREATE TABLE `address` (
`account_id` smallint(5) unsigned NOT NULL AUTO_INCREMENT,
`customer_id` smallint(5) unsigned NOT NULL,
`address_type` varchar(15) NOT NULL, -- Office, Branch-1, Branch-2,
`door_num` varchar(50) NOT NULL,
`landmark` varchar(150) DEFAULT NULL,
`street` varchar(50) DEFAULT NULL,
`area_name` varchar(25) NOT NULL,
`district` varchar(25) NOT NULL,
`city` varchar(25) NOT NULL,
`postal_code` varchar(10) DEFAULT NULL,
`phone1` varchar(20) NOT NULL,
`phone2` varchar(20),
`create_date` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`last_update` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`account_id`),
KEY `idx_fk_city` (`city`),
CONSTRAINT `fk_customer_id` FOREIGN KEY (`customer_id`) REFERENCES `customer` (`customer_id`) ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=201 DEFAULT CHARSET=utf8;虽然我正在创建一个新的客户，但我是这样的错误。
SqlExceptionHelper:144 - SQL错误: 1452，SQLState: 23000 2015年-12-10 13:39:32错误SqlExceptionHelper:146 -无法添加或更新子行:外键约束失败(
mobily.address，CONSTRAINTfk_customer_id外键(customer_id) update CASCADE时引用customer(customer_id) )
因为：Address对象中的custmer_id以'0‘的形式出现。但那是
保存前的Customer对象。是这样的吗：
customer Customer (id=90)
active 1
addresses ArrayList<E> (id=95)
elementData Object[10] (id=117)
[0] Address (id=119)
accountId 0
addressType "main office" (id=121)
areaName "area3" (id=122)
city "city3" (id=123)
customerId 0
district "district3" (id=124)
doorNum "89" (id=125)
landmark "landmark3" (id=126)
phone1 "646432365465" (id=127)
phone2 "4534542355675" (id=128)
postalCode "453245" (id=129)
street "street3" (id=130)
modCount 1
size 1
customerId 0
email "bbb@gmail.com" (id=104)
encryptKey "wwwwwfsad" (id=109)
firstName "xxx" (id=110)
lastName "yyy" (id=111)
password "user3" (id=112)
selfDescription "user3" (id=113)
theme Theme (id=114)
userName "user3" (id=116) 回答 1
Stack Overflow用户
发布于 2015-12-10 16:27:10
从SQL表中删除自动增量。并以这种方式配置您的模型：
@Id
@Column(name = "primary_key_id")
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "chat_seq_gen")
@SequenceGenerator(name = "chat_seq_gen", sequenceName = "chat_seq")
private Long chatid;然后创建一个名为chat_seq的序列(对于这个类)，应该就是这样。这就是你要找的东西吗？如果不是，让我知道，我会删除我的答案。
编辑
另外，在您的Address.java中，您应该这样指代客户：
@ManyToOne
@JoinColumn(name = "customerid", nullable = false)
private Customer customerAddresses;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/34196831
复制相关文章
相似问题