[英]Convert SQL statement 'LIKE' to JPQL statement
I want to write this SQL statement in eclipse JPQL query. 我想在Eclipse JPQL查询中编写此SQL语句。 It works in SQL but I'm not sure how to write it in eclipse.
它可以在SQL中工作,但是我不确定如何在Eclipse中编写它。
SELECT *
FROM hardware h
WHERE h.`Staffid` LIKE '%150%'
My staffid in hardware table is a foreign key to staff table's staffid primary key. 我在硬件表中的人员ID是人员表的人员ID主键的外键。 So the staff in the hardware table is
所以硬件表中的人员是
private Staff staff;
This is what I write to run my search: 这是我写来运行搜索的内容:
@SuppressWarnings("unchecked")
public List<Hardware> searchstaff(Staff a) {
try {
entityManager.getTransaction().begin();
Query query = entityManager
.createQuery("SELECT st from Hardware st where st.staff LIKE :x");
query.setParameter("x", a);
List<Hardware> list = query.getResultList();
entityManager.getTransaction().commit();
return list;
} catch (Exception e) {
return null;
}
}
But it shows 但这表明
javax.servlet.ServletException: java.lang.IllegalStateException:
Exception Description: Transaction is currently active
when I run the search. 当我运行搜索时。
Can anyone help me correct? 谁能帮我纠正?
The like operator work only with string. like运算符仅适用于字符串。 So do something like that:
所以做这样的事情:
Query query = entityManager
.createQuery("SELECT st from Hardware st where st.staff.id LIKE :x");
query.setParameter("x", '%' + a.getId() + '%');
Link : 连结:
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