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hibernate查询在java spring中不起作用_Java_Hibernate_Spring - Fatal编程技术网

hibernate查询在java spring中不起作用

hibernate查询在java spring中不起作用,java,hibernate,spring,Java,Hibernate,Spring,我有这个代码和它的工作良好 Registration person = (Registration) session.get(Registration.class, 8); person.setConfirmed(true); session.save(person); “但这不起作用,”映射错误说 String query = "FROM registration WHERE user_id = 8"; Query query2 = session.createQu

我有这个代码和它的工作良好

Registration person = (Registration) session.get(Registration.class, 8);
      person.setConfirmed(true);
      session.save(person);
“但这不起作用,”映射错误说

String query = "FROM registration WHERE user_id = 8";
 Query query2 = session.createQuery(query);
 Registration person = (Registration) query2.uniqueResult();
 person.setConfirmed(true);
 session.save(person);
这是我的注册课

@Column(name = "user_id")
    public Integer getUserId() {
        return userId;
    }

因为您使用的是非本机查询语言,所以可能需要

String query = "FROM Registration WHERE userId = 8";

因为您使用的是非本机查询语言,所以可能需要

String query = "FROM Registration WHERE userId = 8";

@下面的代码是SQL查询而不是HQL查询

String query = "FROM registration WHERE user_id = 8";
Query query2 = session.createQuery(query);
如果必须运行SQL查询,请改用以下命令:

String query = "FROM registration WHERE user_id = 8";
Query query2 = session.createSQLQuery(query); 
query2.executeUpdate();
要将SQL查询转换为HQL查询,假设注册类具有userId字段:


有关完整示例,请参阅以下指南:

@Pasha,以下代码是SQL查询而不是HQL查询

String query = "FROM registration WHERE user_id = 8";
Query query2 = session.createQuery(query);
如果必须运行SQL查询,请改用以下命令:

String query = "FROM registration WHERE user_id = 8";
Query query2 = session.createSQLQuery(query); 
query2.executeUpdate();
要将SQL查询转换为HQL查询,假设注册类具有userId字段:

有关完整示例,请参阅以下指南: