在 JPA 中按截断日期分组

Question

我需要帮助！

我需要为

建立规范

SELECT date_trunc('day', start_time)
FROM Example
GROUP BY date_trunc('day', start_time)

(PostgreSQL)

我有代码：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Object[]> query = cb.createQuery(Object[].class);

Root<Example> root = query.from(Example.class);

Expression<Date> exp = cb.function("date_trunc", Date.class , cb.literal("day"), root.get("startTime"));

List<Object[]> result = entityManager.createQuery(query.select(exp).groupBy(exp)).getResultList();

但是我得到了例外：

Caused by: org.hibernate.exception.SQLGrammarException: could not extract ResultSet
    at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:106)
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:111)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:97)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:79)
    at org.hibernate.loader.Loader.getResultSet(Loader.java:2122)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1905)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1881)
    at org.hibernate.loader.Loader.doQuery(Loader.java:925)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:342)
    at org.hibernate.loader.Loader.doList(Loader.java:2622)
    at org.hibernate.loader.Loader.doList(Loader.java:2605)
    at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2434)
    at org.hibernate.loader.Loader.list(Loader.java:2429)
    at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:501)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:371)
    at org.hibernate.engine.query.spi.HQLQueryPlan.performList(HQLQueryPlan.java:216)
    at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1339)
    at org.hibernate.internal.QueryImpl.list(QueryImpl.java:87)
    at org.hibernate.jpa.internal.QueryImpl.list(QueryImpl.java:606)
    at org.hibernate.jpa.internal.QueryImpl.getResultList(QueryImpl.java:483)
    ... 122 common frames omitted
Caused by: org.postgresql.util.PSQLException: ERROR: column "examplei0_.start_time" must appear in the GROUP BY clause or be used in an aggregate function
  Position: 69

我做错了什么？

休眠sql:

select date_trunc(?, examplei0_.start_time) as col_1_0_ from examplei0_ group by date_trunc(?, examplei0_.start_time)

如果我在数据库控制台中执行它，它会起作用。

Answer 1

我猜 Postgres 看到了

select date_trunc(?, examplei0_.start_time) as col_1_0_ from examplei0_ group by date_trunc(?, examplei0_.start_time)

并拒绝，因为第一个 date_trunc(?, examplei0_.start_time) 不一定与第二个 date_trunc(?, examplei0_.start_time) 相同，而不查看传入的实际参数。

如果是这种情况，您需要休眠来生成 'day' 未参数化的查询。或者，在 postgres date_trunc_day(timestamp) 中创建一个调用 date_trunc('day', timestamp) 的函数并改为调用新函数。

在 JPA 中按截断日期分组

Group by trunced date in JPA

postgresql

specifications

jpa

hql