org.hibernate.hql.internal.ast.QuerySyntaxException：

Question

我在 hql 中使用选择查询。但我不能在我的 API 中使用。

得到错误

org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token:

谁能告诉我是我的 HQL 中的错误

我的代码

Session session = SessionUtil.getSession();
//Query query = session.createQuery("from Login where email='" + email + "' and password='" + password + "'");
Query query = session.createQuery("SELECT CONCAT(p.FIRSTNAME, ' ', p.LASTNAME) as Name, a.mobile,  a.email, p.patientId\n"
        + "FROM ( from login where email= '\" + email + \"' and password= '\" + password + \"') a\n"
        + " INNER JOIN\n"
        + " patientprofile p ON a.loginId= p.loginId");
List<Login> logins = query.list();

session.close();
return logins;

Answer 1

首先：\n 在查询中未被接受，因此您必须删除它们。

Query query = session.createQuery("SELECT CONCAT(p.FIRSTNAME, ' ', p.LASTNAME) as Name, a.mobile,  a.email, p.patientId "
        + "FROM ( from login where email= '" + email + "' and password= '" + password + "') a "
        + " INNER JOIN "
        + " patientprofile p ON a.loginId= p.loginId");

第二：连接不安全，您可以使用setParameter 来避免语法错误和 SQL 注入：

Query query = session.createQuery("SELECT CONCAT(p.FIRSTNAME, ' ', p.LASTNAME) as Name,"
        + "a.mobile, a.email, p.patientId "
        + "FROM ( from login where email= :email and password= :password) a "
        //----------------------------------^----------------------^
        + "INNER JOIN "
        + "patientprofile p ON a.loginId= p.loginId");
query.setParameter("email", email);
query.setParameter("password", password);

第三：我真的不明白这部分查询( from login where email= :email and password= :password) 这可能会造成另一个问题，请确保您使用的是正确的查询！可以是：

Query query = session.createQuery("SELECT CONCAT(p.FIRSTNAME, ' ', p.LASTNAME) as Name,"
            + "a.mobile, a.email, p.patientId "
            + "FROM (SELECT l from login l where email= :email and password= :password) a "
            //---------------^-----------^
            + "INNER JOIN "
            + "patientprofile p ON a.loginId= p.loginId");
    query.setParameter("email", email);
    query.setParameter("password", password);

Forth：我想改用原生查询，因为 HQL 和 JPQL 都只在 SELECT、WHERE 或 HAVING 子句中接受 子查询，因此您可以使用：

Query query = session.createNativeQuery("SELECT a.mobile, a.email, p.patientId FROM "
        + "(SELECT * from login l where email= :email and password= :password) a "
        + "INNER JOIN patientprofile p ON a.loginId= p.loginId");
query.setParameter("email", email);
query.setParameter("password", password);

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在JPQL documentation了解更多信息