英文:
Could not extract ResultSet When i using Native Query in Spring Jpa
问题
我正试图使用Spring JPA本机查询获取带有比较条件的真或假。但我遇到了以下异常：
无法提取结果集；SQL [n/a]；嵌套异常是org.hibernate.exception.SQLGrammarException：无法提取结果集
我尝试的查询是：
SELECT case when ABS(am_savedagent.AGENT_VERSION) < 1.8 then 'true' else 'false' end as bool
FROM am_savedagent where am_savedagent.BOX_ID="ots-JIO6Yn0jZbxs";
我的存储库：
public interface SavedAgentRepository extends JpaRepository<SavedAgentDetails, String>, JpaSpecificationExecutor<SavedAgentDetails> {
@Query(value = "SELECT case when ABS(SavedAgentDetails.agentVersion) < ?1 then 'true' else 'false' end as bool FROM SavedAgentDetails where SavedAgentDetails.boxId=?2", nativeQuery = true)
public Optional<List<Object>> findByAgentVersionAndBoxId(String currentVersion, String boxid);
}
我找不到错误为什么会出现。请有人帮帮我。
英文:
I am trying to get true or false, with some compare condition using Spring JPA native query. But I got following Exception,
could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet
The query I am trying is:
SELECT case when ABS(am_savedagent.AGENT_VERSION)< 1.8 then 'true' else 'false' end as bool
FROM am_savedagent where am_savedagent.BOX_ID="ots-JIO6Yn0jZbxs";
My Repository
public interface SavedAgentRepository extends JpaRepository<SavedAgentDetails, String>, JpaSpecificationExecutor<SavedAgentDetails> {
@Query(value = "SELECT case when ABS(SavedAgentDetails.agentVersion)< ?1 then 'true' else 'false' end as bool FROM SavedAgentDetails where SavedAgentDetails.boxId=?2", nativeQuery = true)
public Optional<List<Object>> findByAgentVersionAndBoxId(String currentVersion, String boxid);
}
I can't find y the Errors Comming. please help me, someone.
答案1
得分: 1
public interface RegisterAgentRepository extends JpaRepository<RegisterAgentDetails, String>, JpaSpecificationExecutor<RegisterAgentDetails> {
@Query(value = "SELECT case when ABS(am_registeragent.AGENT_VERSION)< ?1 then 'true' else 'false' end as bool FROM am_registeragent where am_registeragent.SAASBOX_ID=?2", nativeQuery = true)
public Optional<List<Object>> findByAgentVersionAndSaasboxId(String currentVersion, String saasboxid);
}
英文:
Since you are using native query =true,
that's why you need to use the table name and not the entity name inside your query.
Try this -->
public interface RegisterAgentRepository extends JpaRepository<RegisterAgentDetails, String>, JpaSpecificationExecutor<RegisterAgentDetails> {
@Query(value = "SELECT case when ABS(am_registeragent.AGENT_VERSION)< ?1 then 'true' else 'false' end as bool FROM am_registeragent where am_registeragent.SAASBOX_ID=?2", nativeQuery = true)
public Optional<List<Object>> findByAgentVersionAndSaasboxId(String currentVersion, String saasboxid);
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论