Hibernate Criteria 多层次查询关联有关问题(转)

弈蒌，收藏一下！

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http://terrencemail.javaeye.com/blog/197821

hibernate 存在如下表及关联:

Java代码

1. class ?House{ ??
2. ?? private ?string?houseid; ??
3. ?? private ?string?housename; ??
4. } ??
5. class ?Shelf{ ??
6. ?? private ?string?shelfid; ??
7. ?? private ?House?house; ??
8. } ??
9. class ?Position{ ??
10. ?? private ?string?positionid; ??
11. ?? private ?Shelf?shelf; ??
12. } ??
13. class ?Warehouse{ ??
14. ?? private ?string?warehouseid; ??
15. ?? private ?Position?position; ??
16. }??

class House{ private string houseid; private string housename; } class Shelf{ private string shelfid; private House house; } class Position{ private string positionid; private Shelf shelf; } class Warehouse{ private string warehouseid; private Position position; }

Warehouse外键关联Pisition, Position外键关联Shelf, Shelf外键关联houseid。
现在要对Warehouse使用Criteria查询, 查询House的id，可使用如下方法:

Java代码

1. Criteria?criteria?=?BaseUtil.getCriteria(Warehouse. class ); ??
2. criteria.createAlias( "position" ,? "position" ); ??
3. criteria.createAlias( "position.shelf" ,? "shelf" ); ??
4. criteria.add(expression_r.eq( "shelf.house.houseid" ,? "111" ));??

Criteria criteria = BaseUtil.getCriteria(Warehouse.class); criteria.createAlias("position", "position"); criteria.createAlias("position.shelf", "shelf"); criteria.add(expression_r.eq("shelf.house.houseid", "111"));

如果要查询House的name, 必须再关联House表:

Java代码

1. Criteria?criteria?=?BaseUtil.getCriteria(Warehouse. class ); ??
2. criteria.createAlias( "position" ,? "position" ); ??
3. criteria.createAlias( "position.shelf" ,? "shelf" ); ??
4. criteria.createAlias( "shelf.house" ,? "house" ); ??
5. criteria.add(expression_r.eq( "house.housename" ,? "name?A" ));??

Criteria criteria = BaseUtil.getCriteria(Warehouse.class); criteria.createAlias("position", "position"); criteria.createAlias("position.shelf", "shelf"); criteria.createAlias("shelf.house", "house"); criteria.add(expression_r.eq("house.housename", "name A"));

以上代码就可以实现了。

可是还会出现一个问题:如果表路径position已经被关联过了(比如此criteria是传入的参数,已经关联了Alias路径position, 并查询过了),则会报错。解决方法：
检查criteria里面关联的路径，如果已经关联，则不再关联，直接使用就行了。

Java代码

1. class ?CriteriaUtil{ ??
2. public ? static ?Criteria?addAlias(CriteriaImpl?criteriaImpl,?String?path,?String?name){ ??
3. ?? if (path?==? null )? return ?criteriaImpl; ??
4. ?? for (Iterator?iter?=?criteriaImpl.iterateSubcriteria();iter.hasNext();){ ??
5. ????Subcriteria?subCriteria?=?(Subcriteria)iter.next(); ??
6. ???? if (path.equals(subCriteria.getPath())) ??
7. ?????? return ?criteriaImpl; ??
8. ??} ??
9. ?? return ?criteriaImpl.createAlias(path,?name); ??
10. } ??
11. }??

class CriteriaUtil{ public static Criteria addAlias(CriteriaImpl criteriaImpl, String path, String name){ if(path == null) return criteriaImpl; for(Iterator iter = criteriaImpl.iterateSubcriteria();iter.hasNext();){ Subcriteria subCriteria = (Subcriteria)iter.next(); if(path.equals(subCriteria.getPath()))
                  

(注：这段代码有时是存在问题的，不是所有的Criteria 都能转换为CriteriaImpl，可能会是Subcriteria。因此，需要确保传入的参数为CriteriaImpl)

调用代码就变成：

Java代码

1. Criteria?criteria?=?BaseUtil.getCriteria(Warehouse. class ); ??
2. CriteriaUtil.addAlias((CriteriaImpl)criteria,? "position" ,? "position" ); ??
3. CriteriaUtil.addAlias((CriteriaImpl)criteria,? "position.shelf" ,? "shelf" ); ??
4. criteria.add(expression_r.eq( "shelf.house.houseid" ,? "111" ));?