Question

我正在尝试使用Spring Jpa从postgres获取记录。 Postgres中的所有表和列都是大写的，但是spring jpa以小写的形式生成查询，尽管我在实体类中提到了大写的表名。我已经在这个类似的问题上经历了很多关于stackoverflow的帖子，并且每个人都建议将命名策略添加到我所做的yaml文件中，但仍然以小写形式生成表名。我已经使用了以下策略（每次都使用，但对我来说没有任何效果）。我错过了别的什么吗？

spring.jpa.properties.hibernate.naming.physical-strategy: org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
spring.jpa.properties.hibernate.naming_strategy: org.hibernate.cfg.EJB3NamingStrategy
spring.jpa.properties.hibernate.naming.implicit-strategy: org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl

Caused by: **org.postgresql.util.PSQLException: ERROR: relation "**table_name**" does not exist**
  Position: 258
        at org.postgresql.core.v3.QueryExecutorImpl.receiveErrorResponse(QueryExecutorImpl.java:2455)
        at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:2155)
        at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:288)
        at org.postgresql.jdbc.PgStatement.executeInternal(PgStatement.java:430)
        at org.postgresql.jdbc.PgStatement.execute(PgStatement.java:356)
        at org.postgresql.jdbc.PgPreparedStatement.executeWithFlags(PgPreparedStatement.java:168)
        at org.postgresql.jdbc.PgPreparedStatement.executeQuery(PgPreparedStatement.java:116)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:498)
        at org.apache.tomcat.jdbc.pool.StatementFacade$StatementProxy.invoke(StatementFacade.java:114)
        at com.sun.proxy.$Proxy201.executeQuery(Unknown Source)
        at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:60)
        ... 179 common frames omitted

这是我添加到yaml文件中的属性：

spring.jpa.properties.database-platform: org.hibernate.dialect.PostgreSQLDialect
spring.jpa.properties.hibernate.id.new_generator_mappings: true
spring.jpa.properties.hibernate.cache.use_second_level_cache: true
spring.jpa.properties.hibernate.cache.use_query_cache: false
spring.jpa.properties.hibernate.generate_statistics: true
spring.jpa.properties.hibernate.cache.region.factory_class: io.github.jhipster.config.jcache.NoDefaultJCacheRegionFactory
spring.jpa.properties.hibernate.naming-strategy: org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl

        #hibernate.naming_strategy: org.hibernate.cfg.EJB3NamingStrategy
        #hibernate.naming.implicit-strategy: org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl
        #hibernate.dialect: org.hibernate.dialect.PostgreSQLDialect

Answer 1

我以这种方式复制了你需要的东西：

CREATE TABLE temp."SOME_UPPERCASE_TABLE" (
  dt      TIMESTAMP NOT NULL DEFAULT now(),
  column1 VARCHAR(10) PRIMARY KEY
);

实体：

@Entity
@Table(name = "SOME_UPPERCASE_TABLE", schema = "tmp", catalog = "cram")
public class SomeUppercaseTableEntity {
    private Timestamp dt;
    private String column1;
    // getters, setters...

一切都运作良好。

使用Spring JPA为Postgres生成大写表名

1 个答案: