使用Like和％％运算符的Hibernate命名查询? [英] Hibernate Named Query Using Like and % % operators?

问题描述

在我的Hibernate JPA示例代码中.

In my Hibernate JPA Sample code..

public List<AttendeesVO> addAttendees(String searchKeyword) {
    TypedQuery<AttendeesVO> query = entityManager.createQuery(" select at from AttendeesVO at where at.user.firstName LIKE :searchKeyword",AttendeesVO.class);
    query.setParameter("searchKeyword", searchKeyword+"%");
    return query.getResultList();
}

在给出整个字符串firstName=Narasimham

但是当我们给任何字符Narasimham即a或n

But it is not working when we give any character of Narasimham i.e a or n

实际上我的想法是我给Like运算符加上% %以便它可以处理给定String的任何字符.

Actually my thinking is i'm giving Like operator with % % so it working any character of given String..

推荐答案

您正在使用query.setParameter("searchKeyword", searchKeyword+"%");

而不是query.setParameter("searchKeyword", "%"+searchKeyword+"%");

第一个将返回Narasimham N Na Nar Nara等的行.

first one will return rows for Narasimham N Na Nar Nara etc.

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