JPQL并使用Hibernate选择1：n关系(JPQL and selecting 1:n relationships using Hibernate)

 我有一个条目MenuItem有一种语言。 我还有一个设备需要所有具有设备选择语言的MenuItem。 （然后还有一个AccessGroup，用于确定设备是否可以访问MenuItem） 
 
 我有一个适用于AccessGroup选择的查询，但无法考虑如何制定查询以包含语言选择。 
 
 简化： 
 
MenuItem
- id
- published
- Language
- AccessGroup
- App

Device
- id
- List<Languages>
- User

AccessGroup
- id
- universal
- <List>AccessGroupLink

AccessGroupLink
- id
- Device
- User
- AccessGroup
- perpetual
- validTo
 
 我如何在JPQL中执行此操作。 目前我已经/尝试过（使用Play！Framework）： 
 
return find("SELECT DISTINCT m FROM MenuItem m " +
                    "INNER JOIN m.accessGroups ag " +
                    "LEFT OUTER JOIN ag.accessGroupLinks agl " +
                    "WHERE m.app = ? AND m.language IN ? AND (ag.universal = TRUE OR ((agl.device = ? OR agl.user = ?) AND (agl.validTo >= ? OR agl.perpetual = TRUE))) AND m.published = true", device.app, device.languages, device, device.user, new Date()).fetch();
 
 我添加了m.language IN device.languages，但这不起作用。 所以我想我必须添加一个JOIN，但由于某种原因无法看到。 所以任何帮助都非常感谢。 
 
 编辑：我上面的查询得到的错误是： 
 
 已捕获到play.db.jpa.JPABase $ JPAQueryException，执行查询时出错SELECT DISTINCT m FROM MenuItem m INNER JOIN m.accessGroups ag LEFT OUTER JOIN ag.accessGroupLinks agl WHERE m.app =？ 和m.language IN？ AND（ag.universal = TRUE OR（（agl.device =？OR agl.user =？）AND（agl.validTo> =？OR agl.perpetual = TRUE）））和m.published = true ：org.hibernate。 PropertyAccessException：无法通过models.Language.id的反射getter获取字段值 
 
 但我不明白错误，因为model.Language.id是实体的一部分。 

I have a an entry MenuItem which has a language. I also have a device which needs all MenuItems which have the language selected by the device. (Then there is also an AccessGroup which determines if a device has access to a MenuItem)
 
I have a query which works for the AccessGroup selection, but cannot think how to formulate the query to include the language selection.
 
Simplified:
 
MenuItem
- id
- published
- Language
- AccessGroup
- App

Device
- id
- List<Languages>
- User

AccessGroup
- id
- universal
- <List>AccessGroupLink

AccessGroupLink
- id
- Device
- User
- AccessGroup
- perpetual
- validTo
 
How do I do this in JPQL. Currently I have/tried (using the Play! Framework):
 
return find("SELECT DISTINCT m FROM MenuItem m " +
                    "INNER JOIN m.accessGroups ag " +
                    "LEFT OUTER JOIN ag.accessGroupLinks agl " +
                    "WHERE m.app = ? AND m.language IN ? AND (ag.universal = TRUE OR ((agl.device = ? OR agl.user = ?) AND (agl.validTo >= ? OR agl.perpetual = TRUE))) AND m.published = true", device.app, device.languages, device, device.user, new Date()).fetch();
 
I added m.language IN device.languages, but that doesn't work. So I guess I have to add a JOIN, but cannot see how for some reason. So any help is really appreciated.
 
EDIT: The error I get with the query above is:
 
A play.db.jpa.JPABase$JPAQueryException has been caught, Error while executing query SELECT DISTINCT m FROM MenuItem m INNER JOIN m.accessGroups ag LEFT OUTER JOIN ag.accessGroupLinks agl WHERE m.app = ? AND m.language IN ? AND (ag.universal = TRUE OR ((agl.device = ? OR agl.user = ?) AND (agl.validTo >= ? OR agl.perpetual = TRUE)))AND m.published = true: org.hibernate.PropertyAccessException: could not get a field value by reflection getter of models.Language.id
 
But I don't understand the error, as model.Language.id is part of the Entity.

原文：https://stackoverflow.com/questions/17652312