Rob*_*ter 5 spring hibernate pascalcasing spring-data-jpa
我用Spring Data JPA和Hibernate创建了一个Spring Boot 1.3.5 POC(这个版本的Spring Boot中的4.3.11.Final).我的后端数据库是Microsoft SQL Server,我们对数据库对象的标准命名约定是pascal case(例如MySchema.MyTable.MyColumn).我使用javax.persistence.Table和javax.persistence.Column注释来设置名称,并将spring.jpa.hibernate.naming-strategy = org.hibernate.cfg.EJB3NamingStrategy添加到我的application.properties文件中.
一切都很完美.我甚至更新到Spring Boot 1.3.6没有任何问题.
现在我转到使用Hibernate 5.0.9.Final的Spring Boot 1.4.0.RELEASE,并且不推荐使用spring.jpa.hibernate.naming-strategy属性,而选择spring.jpa.hibernate.naming.strategy.我更改了该属性名称,但保留了EJB3NamingStrategy值.我还更改了其他已弃用的元素:
现在,生成的SQL使用默认的camel情况来强调命名约定,而不是我使用EJB3NamingStrategy的pascal情况.
//application.properties
spring.data.jpa.repositories.enabled=true
spring.data.solr.repositories.enabled=false
spring.data.mongodb.repositories.enabled=false
spring.datasource.driver-class-name=com.microsoft.sqlserver.jdbc.SQLServerDriver
spring.jpa.hibernate.naming.strategy=org.hibernate.cfg.EJB3NamingStrategy
#spring.jpa.hibernate.naming-strategy=org.hibernate.cfg.EJB3NamingStrategy
//hibernate.properties
hibernate.dialect=org.hibernate.dialect.SQLServer2012Dialect
hibernate.format_sql=true
//Principal.java
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.validation.constraints.Size;
import org.hibernate.envers.AuditTable;
import org.hibernate.envers.Audited;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
@Entity
@Table(name="Principal", schema="Security")
@Audited
@AuditTable(value = "Principal", schema = "Audit")
public class Principal {
private static final Logger LOG = LoggerFactory.getLogger(Principal.class);
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "Id",
nullable = false)
private Long id;
@Column(name = "Username",
nullable = false,
unique = true)
@Size(min = 1, max = 64)
private String name;
@Column(name = "FirstName",
nullable = false)
@Size(min = 1, max = 64)
private String firstName;
@Column(name = "LastName",
nullable = false)
@Size(min = 1, max = 128)
private String lastName;
@Column(name = "IsEnabled",
nullable = false)
private boolean enabled;
//getters/setters omitted for brevity
}
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原始控制台输出:
Hibernate:
select
principal0_.Id as Id1_8_,
principal0_.IsEnabled as IsEnable2_8_,
principal0_.FirstName as FirstNam3_8_,
principal0_.LastName as LastName4_8_,
principal0_.Username as Username5_8_
from
Security.Principal principal0_
where
principal0_.Username=?
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新的控制台输出:
Hibernate:
select
principal0_.id as id1_7_,
principal0_.is_enabled as is_enabl2_7_,
principal0_.first_name as first_na3_7_,
principal0_.last_name as last_nam4_7_,
principal0_.username as username5_7_
from
security.principal principal0_
where
principal0_.username=?
2016-08-05 09:19:22.751 WARN 5032 --- [ XNIO-2 task-8] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 207, SQLState: S0001
2016-08-05 09:19:22.751 ERROR 5032 --- [ XNIO-2 task-8] o.h.engine.jdbc.spi.SqlExceptionHelper : Invalid column name 'is_enabled'.
2016-08-05 09:19:22.768 ERROR 5032 --- [ XNIO-2 task-8] io.undertow.request : UT005023: Exception handling request to /springbootsecurity/login
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我进行了广泛的搜索,发现了对ImplicitNamingStrategy和PhysicalNamingStrategy的引用; 但插入那些似乎不起作用,可能不是正确的方法.我也看到了创建自己的NamingStrategy的参考.那是我必须走的路吗?
是否有一个不同的Hibernate 5设置,它将使用我在@Table和@Column注释中提供的确切名称?我定义注释的方式有问题吗?
我想说我最后发布了一个愚蠢的问题,但我谈到了创建自定义命名策略的每一个方向.然而,我的答案就是简单地使用Hibernate的PhysicalNamingStrategyStandardImpl.
添加到application.properties:
spring.jpa.hibernate.naming.implicit-strategy=org.hibernate.boot.model.naming.ImplicitNamingStrategyJpaCompliantImpl
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
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从我天真的分析,我假设这是有效的,因为我使用@Table和@Column注释.PhysicalNamingStrategyStandardImpl似乎只是将这些注释中的名称用作数据库对象名称.
所以我的Hibernate生成的查询再次格式化为:
Hibernate:
select
principal0_.Id as Id1_7_,
principal0_.IsEnabled as IsEnable2_7_,
principal0_.FirstName as FirstNam3_7_,
principal0_.LastName as LastName4_7_,
principal0_.Username as Username5_7_
from
Security.Principal principal0_
where
principal0_.Username=?
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阅读@AmanTuladhar的链接和该帖子中的这个链接是它最终点击给我的地方.谢谢!
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