mysql - 1查询的SQL 3关系

我试图从不同的实体中获取3个表，当我使用eia.carreraHorizo​​ntal.kid = eig.carreraHorizo​​ntal.kid和eig.carreraHorizo​​ntal.kid = hch.carreraHorizo​​ntal.kid时，它一直在说我有错误。重复返回很多结果。
这是我的代码：

        List<EvaluacionIndividualAnualYGdp> lista;
        String jpql = "select DISTINCT new es.valencia.gp.sbch.entity.EvaluacionIndividualAnualYGdp(eia, eig, hch)"
                + " from EvaluacionIndividualAnual eia, EvaluacionIndividualGdp eig, HistoricoCarreraHorizontal hch"
                + " where eia.carreraHorizontal.kid = eig.carreraHorizontal.kid = eig.carreraHorizontal.kid"
                + " ";
                //GROUP BY eia ORDER BY eia.carreraHorizontal.empleado.apellido1 ASC, eia
        TypedQuery<EvaluacionIndividualAnualYGdp> query = em.createQuery(jpql, EvaluacionIndividualAnualYGdp.class);
        lista = query.getResultList();
        return lista;
    } ```

The error:
Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected AST node: = near line 1, column 302 [select DISTINCT new es.valencia.gp.sbch.entity.EvaluacionIndividualAnualYGdp(eia, eig, hch) from es.valencia.gp.sbch.entity.EvaluacionIndividualAnual eia, es.valencia.gp.sbch.entity.EvaluacionIndividualGdp eig, es.valencia.gp.sbch.entity.HistoricoCarreraHorizontal hch where eia.carreraHorizontal.kid = eig.carreraHorizontal.kid = eig.carreraHorizontal.kid ]

But i'm getting an exception. How i can get those values? Sorry for the question but i'm new.

最佳答案

您应该删除最后一个eig.carreraHorizo​​ntal.kid：

eia.carreraHorizontal.kid = eig.carreraHorizontal.kid = eig.carreraHorizontal.kid

应该

eia.carreraHorizontal.kid = eig.carreraHorizontal.kid

比较a = b = b无效。可能应该是a = b和b = c

    eia.carreraHorizontal.kid = eig.carreraHorizontal.kid
and eig.carreraHorizontal.kid = hch.carreraHorizontal.kid

而且我不知道在distinct之后的“ new”是否也是错误。