Postgresql 使用“case-when”时出现Spring Hibernate查询语法异常

该查询在PostgreSQL中用作本机查询。但在通过Hibernate将其传递给Spring数据JPA@Query时失败。哪种语法是编写查询的正确语法

@Repository
public interface ProcessoJudicialRepository extends JpaRepository<ProcessoJudicial, Long> {

  @Query("select p from ProcessoJudicial p where"
      + " case when :numero is not null then (p.numero = :numero) else true end"
      + " and case when :advogadoId is not null then (p.advogado_id = :advogadoId) else true end"
      + " and case when :assuntoId is not null then (p.assuntoId = :assuntoId) else true end"
      + " and case when :modalidadeId is not null then (p.modalidade_id = :modalidadeId) else true end")
  Page<ProcessoJudicial> find(@Param("numero") String numero, @Param("advogadoId") Long advogadoId,
      @Param("assuntoId") Long assuntoId, @Param("modalidadeId") Long modalidadeId,
      Pageable pageable);
}

启动此例外：

Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected AST node: case near line 1, column 77 [select p from com.rcsoyer.servicosjuridicos.domain.ProcessoJudicial p where case when :numero is not null then (p.numero = :numero) else true end and case when :advogadoId is not null then (p.advogado_id = :advogadoId) else true end and case when :assuntoId is not null then (p.assuntoId = :assuntoId) else true end and case when :modalidadeId is not null then (p.modalidade_id = :modalidadeId) else true end]
    at org.hibernate.hql.internal.ast.QuerySyntaxException.convert(QuerySyntaxException.java:74)
    at org.hibernate.hql.internal.ast.ErrorCounter.throwQueryException(ErrorCounter.java:91)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:272)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:189)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:141)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:115)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:77)
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:153)
    at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:553)
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:662)
    ... 125 common frames omitted

我试着解决另一个问题

---

尽管这在我的案例中不起作用。

该JPQL的第77列是开头的case关键字。如JPA规范中所示，需要一个简单的条件表达式，而case表达式不是JPA规范中的表达式

simple_cond_expression ::= 
    comparison_expression | 
    between_expression | 
    in_expression | 
    like_expression | 
    null_comparison_expression |
    empty_collection_comparison_expression | 
    collection_member_expression | 
    exists_expression

所以你可以有一个WHERE子句

(case_expression) = some_value

但不是

(case_expression) 

靠自己

---

请注意，并非所有JPA提供程序都有Hibernate所具有的这种限制，并且确实可以很好地对其进行评估，但这是JPA规范的一个扩展，例如DataNucleus JPA。

我没有找到一种简洁的方法来编写带有CASE-WHEN的查询，因此我改为使用and和OR。与我以前的效果完全相同，但现在在两个方面都有效，直接用于Postgres和Hibernate

最后的代码是：

  @Query("from #{#entityName} p where"
      + " ((:numero is not null and p.numero = :numero) or (:numero is null))"
      + " and ((:advogadoId is not null and p.advogado.id = :advogadoId) or (:advogadoId is null))"
      + " and ((:assuntoId is not null and p.assunto.id = :assuntoId) or (:assuntoId is null))"
      + " and ((:modalidadeId is not null and p.modalidade.id = :modalidadeId) or (:modalidadeId is null))")
  Page<ProcessoJudicial> query(@Param("numero") String numero, @Param("advogadoId") Long advogadoId,
      @Param("assuntoId") Long assuntoId, @Param("modalidadeId") Long modalidadeId, Pageable pageable);

---

尽管如此，如果有人知道正确的查询编写方法，可能更简单、更清晰，使用CASE-WHEN，甚至是上面查询的改进，我很乐意知道

实际上，当：advogadoId时，不能将参数用作case。你需要考虑我为什么不能使用它的任何可能的重复，@Nidhi257？@Nidhi257实际上JPA规范说你可以在CASE表达式中有一个参数。这是一种肮脏的方式，但你的答案对我的项目有帮助！谢谢分享