作者:我-是二毛控控控_ | 来源:互联网 | 2023-01-06 19:11
假设我们有以下实体:项目:
class Item {
...
@Index(unique=true)
private String guid;
...
@ToMany
@JoinEntity(entity = JoinItemsWithTags.class, sourceProperty = "itemGuid", targetProperty = "tagName")
private List tagsWithThisItem;
...
}
标签:
class Tag {
@Id
private Long localId;
@Index(unique = true)
private String name;
...
}
我们需要加入他们.这是我的连接实体类:
@Entity(nameInDb = "item_tag_relations")
class JoinItemsWithTags {
@Id
private Long id;
private String itemGuid;
private String tagName;
...
}
我想使用标记名称作为连接属性而不是Long id,因为在与服务器同步时更容易支持一致性.但是当前Item类中的标签getter总是返回一个空列表.我查看了日志并找到了在getter中内部使用的生成查询:
SELECT * <<-- there were a long sequence of fields
FROM "tags" T JOIN item_tag_relations J1
ON T."_id"=J1."TAG_NAME" <<-- here is the problem, must be `T."NAME"=J1."TAG_NAME"`
WHERE J1."ITEM_GUID"=?
所以问题是join是基于tag的_id
字段.生成的List _queryItem_TagsWithThisItem(String itemGuid)
方法隐式使用该id进行连接:
// this `join` nethod is overloaded and pass tag's id as source property
queryBuilder.join(JoinItemsWithTags.class, JoinItemsWithTagsDao.Properties.TagName)
.where(JoinItemsWithTagsDao.Properties.ItemGuid.eq(itemGuid));
我认为正确的方法可能是以下情况:
// source property is passed explicitly
queryBuilder.join(/* Desired first parameter -->> */ TagDao.Properties.Name,
JoinItemsWithTags.class, JoinItemsWithTagsDao.Properties.TagName)
.where(JoinItemsWithTagsDao.Properties.ItemGuid.eq(itemGuid));
但是这个代码是生成的dao,我不知道如何用它做任何事情.有没有办法解决这个问题?