有一次要做一个行业分类的菜单,获取级联关系,由于是个新手,没有这方便的经验,做了很久才整出来一个蹩脚的方法,下面展示一下:
数据库:
CREATE TABLE `enterprise_type` (
`c_id` int(11) NOT NULL auto_increment COMMENT '企业类型编号',
`c_name` varchar(100) NOT NULL COMMENT '企业类型名称',
`c_parentid` int(11) default NULL COMMENT '父类型编号',
PRIMARY KEY (`c_id`),
KEY `FK_enterprise_type` (`c_parentid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
用Myeclipse的工具生成的实体类为:
// default package import java.util.ArrayList; import java.util.HashSet; import java.util.List; import java.util.Set; import javax.persistence.CascadeType; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.FetchType; import javax.persistence.GeneratedValue; import static javax.persistence.GenerationType.IDENTITY; import javax.persistence.Id; import javax.persistence.JoinColumn; import javax.persistence.ManyToOne; import javax.persistence.OneToMany; import javax.persistence.Table; /** * EnterpriseCategory entity. @author MyEclipse Persistence Tools */ @Entity @Table(name = "enterprise_type", catalog = "") public class EnterpriseCategory implements java.io.Serializable { // Fields private Integer CId; private EnterpriseCategory parentCategory; private String CName; private List<EnterpriseCategory> childCategory = new ArrayList<EnterpriseCategory>(); // Property accessors @Id @GeneratedValue(strategy = IDENTITY) @Column(name = "c_id", unique = true, nullable = false) public Integer getCId() { return this.CId; } public void setCId(Integer CId) { this.CId = CId; } @ManyToOne(fetch = FetchType.LAZY) @JoinColumn(name = "c_parentid") public EnterpriseCategory getParentCategory() { return this.parentCategory; } public void setParentCategory(EnterpriseCategory parentCategory) { this.parentCategory = parentCategory; } @Column(name = "c_name", nullable = false, length = 100) public String getCName() { return this.CName; } public void setCName(String CName) { this.CName = CName; } @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "EnterpriseCategory") public List<EnterpriseCategory> getChildCategory() { return this.childCategory; } public void setChildCategory( List<EnterpriseCategory> childCategory) { this.childCategory = childCategory; } }
其他的不展示了,直接展示实现方法:
import org.apache.log4j.Category; import org.hibernate.Criteria; import org.hibernate.FetchMode; import org.hibernate.criterion.Restrictions; import org.springframework.stereotype.Repository; import java.util.ArrayList; import java.util.Iterator; import java.util.List; /** * Created with IntelliJ IDEA. * User: Administrator * Date: 12-7-13 * Time: 上午11:35 * To change this template use File | Settings | File Templates. */ @Repository("enterpriseCategoryDao") public class EnterpriseCategoryDaoHobernate extends EnterpriseAutoInjectGenericHibernateDAO<EnterpriseCategory, Integer> implements EnterpriseCategoryDao { /** * 根据给定的类别获取所有级联的cid * * @param id * @return */ public List<Integer> getList(Integer id) { List<Integer> result = new ArrayList<Integer>(); //存放最终结果 if (id == null) { return result; } List list = this.getSession().createCriteria(EnterpriseCategory.class) .setFetchMode("parentCategory", FetchMode.JOIN) .add(Restrictions.eq("parentCategory.CId", id)) .list(); if (list.size() == 0) { result.add(id); } else { result.add(id); result.addAll(add(list)); } return result; } /** * @param list 得到的enterpriseCategory对象的集合 * @return */ public List<Integer> add(List list) { List<Integer> result = new ArrayList<Integer>(); //存放最终结果 if (list.size() == 0) { return result; } try { for (int i = 0; i < list.size(); i++) { EnterpriseCategory childCategory = (EnterpriseCategory) list.get(i); result.add(childCategory.getCId()); List<EnterpriseCategory> childList = childCategory.getChildCategory(); if (childList.size() != 0) { result.addAll(add(childList)); } } } catch (Exception e) { e.printStackTrace(); } return result; } }
这样就能获取到级联关系的类型Tree,这个方法可以实现N层级联关系
下面是其他高手写的
实体类:
import javax.persistence.*; import static javax.persistence.GenerationType.IDENTITY; @Entity @Table(name = "enterprise_type", catalog = "") public class EnterpriseCategory implements java.io.Serializable { // Fields private Integer CId; private String CName; private Integer parentId; // Property accessors @Id @GeneratedValue(strategy = IDENTITY) @Column(name = "c_id", unique = true, nullable = false) public Integer getCId() { return this.CId; } public void setCId(Integer CId) { this.CId = CId; } @Column(name="c_parentid") public Integer getParentId() { return parentId; } public void setParentId(Integer parentId) { this.parentId = parentId; } @Column(name = "c_name", nullable = false, length = 100) public String getCName() { return this.CName; } public void setCName(String CName) { this.CName = CName; } }
实现方法:
private List<Integer> getCidList(Integer catId, boolean includeSelf) { List<Integer> ret = new ArrayList<Integer>(); List<Integer> list = getSubCatList(catId); ret.addAll(list); if (!includeSelf) ret.remove(catId); return ret; } /** * 指定分类ID的所有子节点 * * @param catId * @return */ private List<Integer> getSubCatList(Integer catId) { List<EnterpriseCategory> allCats = enterpriseCategoryDao.loadAll(); return getSubCatList(catId, allCats); } private List<Integer> getSubCatList(Integer cid, List<EnterpriseCategory> allCats) { Map<Integer, List<Integer>> FK_Parent_Child = new HashMap<Integer, List<Integer>>(); for (EnterpriseCategory cat : allCats) { // 建立所有带有子节点的节点对应关系 List<Integer> list = FK_Parent_Child.get(cat.getParentId()); if (list == null) { list = new ArrayList<Integer>(); list.add(cat.getCId()); FK_Parent_Child.put(cat.getParentId(), list); } else { list.add(cat.getCId()); } } return getSubCats(cid, FK_Parent_Child); } private List<Integer> getSubCats(Integer cid, Map<Integer, List<Integer>> FK_Parent_Child) { List<Integer> nodes = new ArrayList<Integer>(); if (FK_Parent_Child.containsKey(cid)) { // 非叶子节点 nodes.add(cid); List<Integer> childNodes = FK_Parent_Child.get(cid); for (Integer childId : childNodes) { nodes.addAll(getSubCats(childId, FK_Parent_Child)); } } else { // 叶子节点 nodes.add(cid); } return nodes; }
可以根据指定的ID获取下面的所有的子节点的id。
记下来,说不定以后会用到。虽然自己写的不怎么的,单也是辛苦的成果,记下来
相关推荐
hibernate面向对象树型结果开放实例
Struts+Hibernate完成的Tree(公司机构) lib里面的hibernate jar包被我删除了,因为文件太大。
hibernate 资料hibernate 资料hibernate 资料hibernate 资料
Struts_Spring_Hibernate_CRUD操作案例_-分页查询
hibernate3hibernate3hibernate3hibernate3hibernate3hibernate3hibernate3hibernate3
自己整合的例子。spring3+hibernate3+jquerytree 没放jar包 里边有oracle的建表语句,全标注形式
hibernate教程hibernate教程hibernate教程
hibernate annotation hibernate3
Hibernate.jar包,Hibernate可以应用在任何使用JDBC的场合,包含 hibernate-commons-annotations-4.0.1.Final.jar hibernate-core-4.1.12.Final.jar hibernate-ehcache-4.1.12.Final.jar hibernate-entitymanager-...
hibernate5 hibernate PDF 讲义 动力 hibernate ssh hibernate5 节点
hibernate 5.2.15 hibernate 5.2.15 hibernate 5.2.15 hibernate 5.2.15 hibernate 5.2.15hibernate 5.2.15
hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,hibernate,包含4个说明文档,分别详细解说了hibernate...
Hibernate面试题专栏 - 最全的Hibernate面试题, Hibernate笔试题, Hibernate问题 Hibernate教程
hibernate 实战 hibernate 实战 hibernate 实战 hibernate 实战
DmDialect-for-hibernate2.0、DmDialect-for-hibernate2.1、DmDialect-for-hibernate3.0、DmDialect-for-hibernate3.1、DmDialect-for-hibernate3.6、DmDialect-for-hibernate4.0、DmDialect-for-hibernate5.0、...
hibernate第一个hibernate
HIBERNATE HIBERNATE HIBERNATE HIBERNATE
Hibernate连接SQLite配置步骤: 1、添加SQLite库: 1.1、将两个jar包:com.zy.hibernate.dialect.SQLiteDialect.jar、sqlite-jdbc-3.7.2.jar拷贝到“\WEB-INF\lib”文件夹下; 2、配置hibernate: 2.1、将...
包含hibernate所有所需jar包还有一些其他包日志包、jpa支持包等: 列如:hibernate-core-5.1.0.Final.jar hibernate-ehcache-5.1.0.Final.jar hibernate-entitymanager-5.1.0.Final.jar hibernate-envers-5.1.0....
hibernate aiphibernate aiphibernate aip