java - 为什么我的查询或 jpa 查询在 spring 中不起作用

Question

当我在 Spring 项目中运行此查询时，它只返回一个错误。 我尝试了不同的方法，但没有任何结果，如果有人能帮助我，将非常感激。 我有 3 个表:User、Role 和 User_Role，并且嵌入了我的地址。也使用连接表。

所有其他类别为:User ,Role and User_Role spring security

    public List<User> getUsersByNameOrLogin(final String value){
    List<User> result = new ArrayList<User>(0);

    if(value ==null){
        return result;
    }

    final StringBuilder jpql = new StringBuilder("SELECT u FROM User u WHERE ");
    jpql.append(" u.lastname like  "+value);
   jpql.append(" OR u.firstname like  "+value);
    jpql.append(" OR u.email like  "+value);

    final String valueLike = "%"+value+"%";

    result =  entityManager.createQuery(jpql.toString(), User.class)
            .setParameter("lastname", valueLike)
           .setParameter("firstname", valueLike)
           .setParameter("email", valueLike)
            .getResultList();

    return result;
}  

错误是:

无法解析属性:姓氏:用户[SELECT u FROM User u WHERE u.lastname like pappo OR u.firstname like pappo OR u.email like pappo]

以及所有错误消息:

    Exception in thread "main" java.lang.IllegalArgumentException: org.hibernate.QueryException:      could not resolve property: lastname of: User [SELECT u FROM se.guard.User u WHERE  u.lastname like  pappo OR u.firstname like  pappo OR u.email like  pappo]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1222)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1168)
at             org.hibernate.ejb.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:292)
at se.datalayer.service.UserService.getUsersByNameOrLogin(UserService.java:189)
at se.datalayer.main.Main.getUserByParam(Main.java:131)
at se.datalayer.main.Main.main(Main.java:31)
     Caused by: org.hibernate.QueryException: could not resolve property: lastname of:      User [SELECT u FROM se.guard.User u WHERE  u.lastname like  pappo OR u.firstname like      pappo OR u.email like  pappo]
at       org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:81)
at        org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:75)
        at    org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:1451)
at org.hibernate.hql.ast.tree.FromElementType.getPropertyType(FromElementType.java:312)
at org.hibernate.hql.ast.tree.FromElement.getPropertyType(FromElement.java:487)
at org.hibernate.hql.ast.tree.DotNode.getDataType(DotNode.java:611)
at org.hibernate.hql.ast.tree.DotNode.prepareLhs(DotNode.java:263)
at org.hibernate.hql.ast.tree.DotNode.resolve(DotNode.java:210)
at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:117)
at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:113)
at org.hibernate.hql.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:868)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.expr(HqlSqlBaseWalker.java:1323)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.exprOrSubquery(HqlSqlBaseWalker.java:4387)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.comparisonExpr(HqlSqlBaseWalker.java:4004)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:1909)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:1859)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:1859)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.whereClause(HqlSqlBaseWalker.java:824)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:610)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:294)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:237)
at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:254)
at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:185)
at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:136)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:101)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:80)
at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:98)
at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:156)
at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:135)
at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1760)
at    org.hibernate.ejb.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:277)
... 3 more

Answer 1

如果您想使用命名参数，为什么要包含 value 变量？此外，属性名称区分大小写。更改为:

final String jpql = "SELECT u FROM User u WHERE "
  "    u.lastName  like :lastname" +;
  " OR u.firstName like :firstname" +;
  " OR u.email     like :email";

是的，StringBuilder在这里是不必要的。您可以仅使用一个参数来进一步简化它:

final String jpql = "SELECT u FROM User u WHERE "
  "    u.lastName  like :valueLike" +;
  " OR u.firstName like :valueLike" +;
  " OR u.email     like :valueLike";

result =  entityManager.createQuery(jpql, User.class)
        .setParameter("valueLike", valueLike)
        .getResultList();