JPA - 加入對同桌

我得到了這個請求：JPA - 加入對同桌

TypedQuery<ParamGenerauxExternes> q = entityMgr 
      .createQuery("SELECT p FROM ParamGenerauxExternes p " 
        + "WHERE EXISTS " 
        + "(SELECT q FROM ParamGenerauxExternes q " 
        + "WHERE q.key.origine = :pOrigine " 
        + "AND q.key.typeParam LIKE :pTypeParametreBis " 
        + "AND p.key.sousType LIKE CONCAT('%',q.libelleParam) " 
        + "AND q.actif = 'Y') " 
        + "ORDER BY p.libelleParam", ParamGenerauxExternes.class) 
      //.setParameter("pTypeParametre", "REL_TO_HOUSEHOLD") 
      .setParameter("pOrigine", pOrigineGrc) 
      .setParameter("pTypeParametreBis", "GRC_HOUSEHOLD_TYPE_P%");

但它不工作，我得到這個錯誤消息告訴prenthesis丟失：

GRAVE: EJB Exception: : javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: ORA-00907: missing right parenthesis

生成的SQL通過JPA有後的第二個括號中進行選擇：

select 
    paramgener0_.CODE_PARAM as CODE1_2823_, 
    paramgener0_.ORIGINE as ORIGINE2823_, 
    paramgener0_.SOUS_TYPE as SOUS3_2823_, 
    paramgener0_.TYPE_PARAM as TYPE4_2823_, 
    paramgener0_.ACTIF as ACTIF2823_, 
    paramgener0_.LIBELLE_PARAM as LIBELLE6_2823_ 
from 
    FOA_PARAM_GEN_EXTERNE paramgener0_ 
where 
    exists (
     select 
      (paramgener1_.CODE_PARAM, 
      paramgener1_.ORIGINE, 
      paramgener1_.SOUS_TYPE, 
      paramgener1_.TYPE_PARAM) 
     from 
      FOA_PARAM_GEN_EXTERNE paramgener1_ 
     where 
      paramgener1_.ORIGINE=? 
      and (
       paramgener1_.TYPE_PARAM like ? 
      ) 
      and (
       paramgener0_.SOUS_TYPE like '%'||paramgener1_.LIBELLE_PARAM 
      ) 
      and paramgener1_.ACTIF='Y' 
    ) 
order by 
    paramgener0_.LIBELLE_PARAM

我不知道爲什麼冬眠第二後加括號？請如果你知道該怎麼辦...

來源

2014-11-03 user3469203

該查詢在Oracle上失敗，但傳遞，例如，Postgres。如果您正在使用Oracle數據庫，請嘗試將Oracle SQL方言設置爲您的JPA提供程序。
Hibernate SQL Dialects

來源

2014-11-10 13:37:13

JPA - 加入對同桌

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