hql

Spring boot jpa hibernate HQL query with group by gives Validation failed for query

由 僤鯓⒐⒋嵵緔 提交于 2021-01-07 06:58:51
问题 In spring boot app I have an ElementEntity: .... @Entity public class ElementEntity { @Id int id; Double quantity; String form; String color; String type; String description; @ManyToOne @JoinColumn(name = "form") private FormEntity form; } And I have here FormEntity: @Entity public class FormEntity { ..... @OneToMany @JoinColumn(name = "links") private List<LinksEntity> links; } I want using HQL to generate a query that will sum the values of quantity field, group by color and and to return

How to convert PSQLs ::json @> ::json to a jpa/jpql-predicate

由 北战南征 提交于 2020-12-31 06:21:59
问题 Say i have a db-table looking like this: CREATE TABLE myTable( id BIGINT, date TIMESTAMP, user_ids JSONB ); user_ids are a JSONB-ARRAY Let a record of this table look like this: { "id":13, "date":"2019-01-25 11:03:57", "user_ids":[25, 661, 88] }; I need to query all records where user_ids contain 25. In SQL i can achieve it with the following select-statement: SELECT * FROM myTable where user_ids::jsonb @> '[25]'::jsonb; Now i need to write a JPA-Predicate that renders "user_ids::jsonb @> '

How to convert PSQLs ::json @> ::json to a jpa/jpql-predicate

由 岁酱吖の 提交于 2020-12-31 06:20:16
问题 Say i have a db-table looking like this: CREATE TABLE myTable( id BIGINT, date TIMESTAMP, user_ids JSONB ); user_ids are a JSONB-ARRAY Let a record of this table look like this: { "id":13, "date":"2019-01-25 11:03:57", "user_ids":[25, 661, 88] }; I need to query all records where user_ids contain 25. In SQL i can achieve it with the following select-statement: SELECT * FROM myTable where user_ids::jsonb @> '[25]'::jsonb; Now i need to write a JPA-Predicate that renders "user_ids::jsonb @> '

Hibernate Bug

由 这一生的挚爱 提交于 2020-12-18 07:56:22
Hibernate对于用原生的Sql查询有bug，当使用sql语句查询时，要注意进行标量查询或者是实体查询（SQLQuery）query.addEntity(Entity.Class) ，但是有时候即使是配置了实体的id，也即表的主键，也会发生错误Column notFound ‘’id“ ；所以不建议使用原生sql语句在hibernate中； hql查询单表部分字段： 在hibernate中，用hql语句查询实体类，采用list方法的返回结果为一个List，该List中封装的对象分为以下三种情况： 1.查询全部字段的情况下，如"from 实体类"，list中封装的对象为实体类本身，各属性都将得到填充。 2.只查询一个字段，默认情况下，list中封装的是Object对象。 3.查询两个或两个以上的字段，默认情况下，list中封装的是Object[],长度与所查询的字段数一致。 对于后两种情况，用标签遍历时不太方便，因为无法直接转换成实体类的对象。比较简单的解决方法是： 在hql中使用 select new 包名.类名(属性1，属性2……) from实体类，同时 在实体类中添加带参的构造方法 ，参数的个数和顺序与（属性1，属性2……)保持一致，这样我们得到的list中存放的依然是实体类的 对象，所查询到的属性得到了填充，使用起来更为方便。 附hql查询多表语句： String hql

How can I use mysql null safe equality operator <=> in HQL?

由 好久不见. 提交于 2020-12-13 11:27:15
问题 I wanted to get records of Patient (POJO class) who's contact number is not null. So, I referred this post. In the answer two ways are specified SELECT * FROM table WHERE YourColumn IS NOT NULL; SELECT * FROM table WHERE NOT (YourColumn <=> NULL); From above I wrote below hql which runs successfully from Patient p where p.contactNo is not null But, for 2nd type of hql from Patient p where not (p.contactNo <=> null) throws Exception org.hibernate.hql.internal.ast.QuerySyntaxException:

How to INSERT using a SELECT in Hibernate

由 丶灬走出姿态 提交于 2020-12-08 08:47:18
问题 I need to implement the following request in hibernate: insert into my_table(....,max_column) values(...,(select max(id) from special_table where ....)) How to do that in hibernate, using annotations? special_table may be not a child or dependency of my_table, just a subselect. 回答1: You can use the INSERT INTO ... SELECT ... feature: int updateCount = session.createQuery(""" insert into MyEntity( ..., max_column ) select ..., max(id) from SpecialEntity """) .executeUpdate(); 来源： https:/

How to INSERT using a SELECT in Hibernate

由 ぐ巨炮叔叔 提交于 2020-12-08 08:46:52
问题 I need to implement the following request in hibernate: insert into my_table(....,max_column) values(...,(select max(id) from special_table where ....)) How to do that in hibernate, using annotations? special_table may be not a child or dependency of my_table, just a subselect. 回答1: You can use the INSERT INTO ... SELECT ... feature: int updateCount = session.createQuery(""" insert into MyEntity( ..., max_column ) select ..., max(id) from SpecialEntity """) .executeUpdate(); 来源： https:/
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