0

我有这个 jpa 查询

@Query(nativeQuery = true, value = "with\n"
            + "    validBonansasAssignation as (\n"
            + "        select aga.autorisation_bonansas_id, aga.autorisation_bonansa_assign_id as iBonansaAffectation_id,\n"
            + "               s.site_id as ISiteNoTypeBaseData_id, s.desc_court as iSiteNoTypeBaseData_nom,\n"
            + "               aga.dt_debut as iBonansaAffectation_dt_debut, aga.dt_fin as iBonansaAffectation_dt_fin\n"
            + "        from AUTORISATION_BONANSA_ASSIGNATION aga\n"
            + "        inner join site s on aga.site_id = s.site_id\n"
            + "        where ?1 between aga.dt_debut and aga.dt_fin\n"
            + "    )\n"
            + "select ag.AUTORISATION_BONANSAS_ID, ag.NOM, ag.PRENOM, ag.EULOGIN, ag.dt_Debut, ag.dt_Fin\n"
            + "    , aga.iBonansaAffectation_id, aga.iSiteNoTypeBaseData_id, aga.iSiteNoTypeBaseData_nom\n"
            + "    , aga.iBonansaAffectation_dt_debut, aga.iBonansaAffectation_dt_fin\n"
            + "from autorisation_bonansa ag\n"
            + "left join validBonansasAssignation aga on ag.autorisation_bonansas_id = aga.autorisation_bonansas_id\n"
            + "where ?1 between ag.dt_debut and ag.dt_fin\n"
            + "and ag.organisation_id = ?2")
List<AutoUsers> find(Date date, Long id);

但是当我运行测试时出现此错误：

org.springframework.orm.jpa.JpaSystemException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.7.2.v20180622-f627448): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: org.h2.jdbc.JdbcSQLException: Syntax error in SQL statement "WITH
    VALIDBONANSASASSIGNATION AS[*] (
        SELECT AGA.AUTORISATION_BONANSAS_ID, AGA.AUTORISATION_BONANSA_ASSIGN_ID AS IBONANSAAFFECTATION_ID,
                  S.SITE_ID AS ISITENOTYPEBASEDATA_ID, S.DESC_COURT AS ISITENOTYPEBASEDATA_NOM,
               AGA.DT_DEBUT AS IBONANSAAFFECTATION_DT_DEBUT, AGA.DT_FIN AS IBONANSAAFFECTATION_DT_FIN
        FROM AUTORISATION_BONANSA_ASSIGNATION AGA
        INNER JOIN SITE S ON AGA.SITE_ID = S.SITE_ID
        WHERE ? BETWEEN AGA.DT_DEBUT AND AGA.DT_FIN
    )
SELECT AG.AUTORISATION_BONANSAS_ID, AG.NOM, AG.PRENOM, AG.EULOGIN, AG.DT_DEBUT, AG.DT_FIN
    , AGA.IBONANSAAFFECTATION_ID, AGA.ISITENOTYPEBASEDATA_ID, AGA.ISITENOTYPEBASEDATA_NOM
    , AGA.IBONANSAAFFECTATION_DT_DEBUT, AGA.IBONANSAAFFECTATION_DT_FIN
FROM AUTORISATION_BONANSA AG
LEFT JOIN VALIDBONANSASASSIGNATION AGA ON AG.AUTORISATION_BONANSAS_ID = AGA.AUTORISATION_BONANSAS_ID
WHERE ? BETWEEN AG.DT_DEBUT AND AG.DT_FIN
AND AG.ORGANISATION_ID = ? "; expected "., ("; SQL statement:
4

2 回答 2

1

你可以nativeQuery = true像下面这样设置，看看这是否有效。

@Query(
  value = "SELECT * FROM USERS u WHERE u.status = 1", 
  nativeQuery = true)
于 2020-09-29T18:30:55.923 回答
0

通过JPA形成错误堆栈生成原始查询的方式，我建议如下更改并尝试，

我现在没有 IDE 以正确的格式实际修改查询，因此提供了 SQL 格式的查询，但目的是将列定义到WITH子句中，因为正如您在错误堆栈中看到的那样，它是WITH VALIDBONANSASASSIGNATION AS[*] (在[*]之后生成的AS是 Oracle 无法识别并抛出语法错误的东西。

WITH VALIDBONANSASASSIGNATION 
(
  AUTORISATION_BONANSAS_ID
, IBONANSAAFFECTATION_ID
, ISITENOTYPEBASEDATA_ID
, ISITENOTYPEBASEDATA_NOM
, IBONANSAAFFECTATION_DT_DEBUT
, IBONANSAAFFECTATION_DT_FIN
)
AS
(
  SELECT AGA.AUTORISATION_BONANSAS_ID
       , AGA.AUTORISATION_BONANSA_ASSIGN_ID AS IBONANSAAFFECTATION_ID
       , S.SITE_ID AS ISITENOTYPEBASEDATA_ID
       , S.DESC_COURT AS ISITENOTYPEBASEDATA_NOM
       , AGA.DT_DEBUT AS IBONANSAAFFECTATION_DT_DEBUT
       , AGA.DT_FIN AS IBONANSAAFFECTATION_DT_FIN
    FROM AUTORISATION_BONANSA_ASSIGNATION AGA
  INNER JOIN SITE S 
     ON AGA.SITE_ID = S.SITE_ID
  WHERE ? BETWEEN AGA.DT_DEBUT 
    AND AGA.DT_FIN
)
SELECT AG.AUTORISATION_BONANSAS_ID
     , AG.NOM
     , AG.PRENOM
     , AG.EULOGIN
     , AG.DT_DEBUT
     , AG.DT_FIN
     , AGA.IBONANSAAFFECTATION_ID
     , AGA.ISITENOTYPEBASEDATA_ID
     , AGA.ISITENOTYPEBASEDATA_NOM
     , AGA.IBONANSAAFFECTATION_DT_DEBUT
     , AGA.IBONANSAAFFECTATION_DT_FIN
FROM AUTORISATION_BONANSA AG
LEFT JOIN VALIDBONANSASASSIGNATION AGA 
  ON AG.AUTORISATION_BONANSAS_ID 
   = AGA.AUTORISATION_BONANSAS_ID
WHERE ? BETWEEN AG.DT_DEBUT AND AG.DT_FIN
  AND AG.ORGANISATION_ID = ?

我试了一下，看看它是否有效。

于 2020-09-29T18:36:42.453 回答