java递归获取父节点(java根据子节点获取它对应的所有父节点)
java 找到一节点的所有子节点 是不是得递归实现？
2L谁说必须用递归的，只是递归写起来简单罢了，迭代的方法一样OK；
深度优先或者广度优先都可以
LZ的问题用迭代的方法可以这样解决：
1。把当前节点（需要查找字节点的节点）压入一个堆栈，这步是初始化；
2。从堆栈中弹出一个节点，如果该节点是叶子节点，则这条路已经走不通了，如果是非叶子节点，那就把这个节点的所有子节点压入堆栈
3。重复第二步直到堆栈为空
上面三步就能遍历当前节点的所有字节点
递归的话：
f(node){
for(遍历node的所有子节点){
child=当前子节点
if(child为叶子节点){
..return
;}
else{
f(child)
}
}
}
java1.8使用mybaitis连接mysql递归查询所有父节点用到：和@报错Cause: java.util.NoSuchElementException
Java mysql mybatis批量更新数据库，采用以下写法即可执行，但是数据库连接必须配置：allowMultiQueries=true
例如：jdbc:mysql://192.168.1.236:3306/test?useUnicode=trueamp;characterEncoding=UTF-8allowMultiQueries=true
update?id="batchUpdate"??parameterType="java.util.List"
?????foreach?collection="list"?item="item"?index="index"?open=""?close=""?separator=";"
update?test?
set
?test=${item.test}+1
/set
where?id?=?${item.id}
/foreach
???/update
MyBatis是支持普通SQL查询，存储过程和高级映射的优秀持久层框架。MyBatis消除了几乎所有的JDBC代码和参数的手工设置以及结果集的检索。MyBatis使用简单的XML或注解用于配置和原始映射，将接口和Java的POJOs（Plan Old Java Objects，普通的Java对象）映射成数据库中的记录.
java Tree 怎么获得父节点的值。？SWT做的tree
TreeItem parentItem=item.getParentItem();得到父节点
Tree tree= item.getParent();得到树
parentItem.getText();节点显示的值
parentItem.getData();item上保存的数据
java的递归查询怎么写
原文在这里，写得不错，楼主可参考下，具体链接如下，我只是搬运工！
/**?
????*?说明方法描述：将list转为树tree结构?
????*??
????*?@param?allRrecords?
????*?@return?
????*?@time?2016年5月10日?下午6:00:35?
????*?@author?yangdong?
????*/??
???public?ListRecord?useListRecordToTree(ListRecord?allRrecords)?{??
??
???????ListRecord?listParentRecord?=?new?ArrayListRecord();??
???????ListRecord?listNotParentRecord?=?new?ArrayListRecord();??
???????//?第一步：遍历allRrecords保存所有数据的uuid用于判断是不是根节点??
???????MapString,?String?mapAllUuid?=?new?HashMapString,?String();??
???????MapString,?Record?allRecordMap?=?new?HashMapString,?Record();??
???????for?(Record?record?:?allRrecords)?{??
???????????mapAllUuid.put(record.getStr("uuid"),?record.getStr("uuid"));??
???????????allRecordMap.put(record.getStr("uuid"),?record);??
???????}??
???????//?第二步：遍历allRrecords找出所有的根节点和非根节点??
???????if?(allRrecords?!=?null??allRrecords.size()??0)?{??
???????????for?(Record?record?:?allRrecords)?{??
???????????????if?(StringUtil.isBlank(record.getStr("parent_uuid"))??
???????????????????||?!mapAllUuid.containsKey(record.getStr("parent_uuid")))?{??
???????????????????listParentRecord.add(record);??
???????????????}?else?{??
???????????????????listNotParentRecord.add(record);??
???????????????}??
???????????}??
???????}??
??
???????//?第三步：?递归获取所有子节点??
???????if?(listParentRecord.size()??0)?{??
???????????for?(Record?record?:?listParentRecord)?{??
???????????????//?添加所有子级??
???????????????record.set("childs",?this.getTreeChildRecord(listNotParentRecord,?record.getStr("uuid")));??
???????????}??
???????}??
???????return?listParentRecord;??
???}??
??
???/**?
????*?说明方法描述：使list转换为树并根据关键字和节点名称过滤?
????*??
????*?@param?allRecords?所有节点?
????*?@param?keywords?要过滤的关键字?
????*?@param?filterFields?要过滤的字段?
????*?@return?
????*?@time?2016年5月19日?下午3:27:32?
????*?@author?yangdong?
????*/??
???public?ListRecord?useListRecordToTreeByKeywords(ListRecord?allRecords,?String?keywords,?String...?filterFields)?{??
???????ListRecord?listRecord?=?new?ArrayListRecord();??
???????MapString,?Record?allRecordMap?=?new?HashMapString,?Record();??
???????for?(Record?record?:?allRecords)?{??
???????????allRecordMap.put(record.getStr("uuid"),?record);??
???????}??
???????//?遍历allRrecords找出所有的nodeName和关键字keywords相关的数据??
???????if?(allRecords?!=?null??allRecords.size()??0)?{??
???????????if?(filterFields.length??1)?{??
???????????????for?(Record?record?:?allRecords)?{??
???????????????????for?(String?field?:?filterFields)?{??
???????????????????????//?比较??
???????????????????????if?(record.getStr(field).toLowerCase().indexOf(keywords.toLowerCase())?!=?-1)?{??
???????????????????????????listRecord.add(record);??
???????????????????????}??
???????????????????}??
???????????????}??
???????????}?else?{??
???????????????for?(Record?record?:?allRecords)?{??
???????????????????//?比较??
???????????????????if?(record.getStr(filterFields[0]).toLowerCase().indexOf(keywords.toLowerCase())?!=?-1)?{??
???????????????????????listRecord.add(record);??
???????????????????}??
???????????????}??
???????????}??
???????}??
???????//?查找过滤出来的节点和他们的父节点??
???????listRecord?=?this.getSelfAndTheirParentRecord(listRecord,?new?ArrayListRecord(),??
?????????????????????????????????????????????????????new?HashMapString,?Record(),?allRecordMap);??
???????//?将过滤出来的数据变成树tree结构??
???????listRecord?=?this.useListRecordToTree(listRecord);??
???????return?listRecord;??
???}??
??
???/**?
????*?说明方法描述：递归查询子节点?
????*??
????*?@param?childList?子节点?
????*?@param?parentUuid?父节点id?
????*?@return?
????*?@time?2016年5月10日?下午3:29:35?
????*?@author?yangdong?
????*/??
???private?ListRecord?getTreeChildRecord(ListRecord?childList,?String?parentUuid)?{??
???????ListRecord?listParentRecord?=?new?ArrayListRecord();??
???????ListRecord?listNotParentRecord?=?new?ArrayListRecord();??
???????//?遍历tmpList，找出所有的根节点和非根节点??
???????if?(childList?!=?null??childList.size()??0)?{??
???????????for?(Record?record?:?childList)?{??
???????????????//?对比找出父节点??
???????????????if?(StringUtil.equals(record.getStr("parent_uuid"),?parentUuid))?{??
???????????????????listParentRecord.add(record);??
???????????????}?else?{??
???????????????????listNotParentRecord.add(record);??
???????????????}??
??
???????????}??
???????}??
???????//?查询子节点??
???????if?(listParentRecord.size()??0)?{??
???????????for?(Record?record?:?listParentRecord)?{??
???????????????//?递归查询子节点??
???????????????record.set("childs",?getTreeChildRecord(listNotParentRecord,?record.getStr("uuid")));??
???????????}??
???????}??
???????return?listParentRecord;??
???}??
??
???/**?
????*?说明方法描述：递归找出本节点和他们的父节点?
????*??
????*?@param?parentList?根据关键字过滤出来的相关节点的父节点?
????*?@param?resultList?返回的过滤出来的节点?
????*?@param?filterRecordMap?已经过滤出来的节点?
????*?@param?allRecordMap?所有节点?
????*?@return?
????*?@time?2016年5月19日?上午9:53:56?
????*?@author?yangdong?
????*/??
???private?ListRecord?getSelfAndTheirParentRecord(ListRecord?parentList,?ListRecord?resultList,??
????????????????????????????????????????????????????MapString,?Record?filterRecordMap,??
????????????????????????????????????????????????????MapString,?Record?allRecordMap)?{??
???????//?当父节点为null或者节点数量为0时返回结果，退出递归??
???????if?(parentList?==?null?||?parentList.size()?==?0)?{??
???????????return?resultList;??
???????}??
???????//?重新创建父节点集合??
???????ListRecord?listParentRecord?=?new?ArrayListRecord();??
???????//?遍历已经过滤出来的节点??
???????for?(Record?record?:?parentList)?{??
??
???????????String?uuid?=?record.getStr("uuid");??
???????????String?parent_uuid?=?record.getStr("parent_uuid");??
??
???????????//?如果已经过滤出来的节点不存在则添加到list中??
???????????if?(!filterRecordMap.containsKey(uuid))?{??
???????????????listParentRecord.add(record);//?添加到父节点中??
???????????????filterRecordMap.put(uuid,?record);//?添加到已过滤的map中??
???????????????allRecordMap.remove(uuid);//?移除集合中相应的元素??
???????????????resultList.add(record);//?添加到结果集中??
???????????}??
??
???????????//?找出本节点的父节点并添加到listParentRecord父节点集合中，并移除集合中相应的元素??
???????????if?(StringUtil.isNotBlank(parent_uuid))?{??
???????????????Record?parentRecord?=?allRecordMap.get(parent_uuid);??
???????????????if?(parentRecord?!=?null)?{??
???????????????????listParentRecord.add(parentRecord);??
???????????????????allRecordMap.remove(parent_uuid);??
???????????????}??
???????????}??
??
???????}??
???????//?递归调用??
???????getSelfAndTheirParentRecord(listParentRecord,?resultList,?filterRecordMap,?allRecordMap);??
??
???????return?resultList;??
???}??
[java]?view?plain?copy
//示例??
[java]?view?plain?copy
/**?
?????*?说明方法描述：递归查询所有权限?
?????*??
?????*?@param?keyword?
?????*?@param?is_deleted?
?????*?@return?
?????*?@time?2016年5月10日?下午3:47:50?
?????*?@author?yangdong?
?????*/??
????public?ListRecord?getRecordByKeywordRecursive(String?keyword,?String?is_deleted)?{??
????????//?第一步：查询所有的数据??
????????StringBuffer?sql?=?new?StringBuffer(??
????????????????????????????????????????????"?select?pa.uuid,pa.parent_uuid,pa.author_code,pa.author_name,pa.is_menu,pa.sort_number,pa.is_enable,pa.menu_icon?");??
????????sql.append("??from?s_author?pa");??
????????ListObject?params?=?new?ArrayListObject();??
????????sql.append("?where??pa.is_deleted=??");??
????????params.add(is_deleted);??
????????sql.append("?order?by?pa.sort_number?asc?");??
??
??
????????ListRecord?allRrecords?=?Db.use(AppConst.DB_DATASOURCE_MAIN).find(sql.toString(),?ParamUtil.listToArray(params));??
[java]?view?plain?copy
????????????//第二步：将list变为树tree结构??
????????if?(StringUtil.isNotBlank(keyword))?{??
????????????return?super.useListRecordToTreeByKeywords(allRrecords,?keyword,?"author_name");??
????????}?else?{??
????????????return?super.useListRecordToTree(allRrecords);??
????????}??
????}
java 递归数据库生成 树形结构问题
1、准备表结构及对应的表数据
a、表结构：
create table TB_TREE
(
CID NUMBER not null,
CNAME VARCHAR2(50),
PID NUMBER //父节点
)
b、表数据：
insert into tb_tree (CID, CNAME, PID) values (1, '中国', 0);
insert into tb_tree (CID, CNAME, PID) values (2, '北京市', 1);
insert into tb_tree (CID, CNAME, PID) values (3, '广东省', 1);
insert into tb_tree (CID, CNAME, PID) values (4, '上海市', 1);
insert into tb_tree (CID, CNAME, PID) values (5, '广州市', 3);
insert into tb_tree (CID, CNAME, PID) values (6, '深圳市', 3);
insert into tb_tree (CID, CNAME, PID) values (7, '海珠区', 5);
insert into tb_tree (CID, CNAME, PID) values (8, '天河区', 5);
insert into tb_tree (CID, CNAME, PID) values (9, '福田区', 6);
insert into tb_tree (CID, CNAME, PID) values (10, '南山区', 6);
insert into tb_tree (CID, CNAME, PID) values (11, '密云县', 2);
insert into tb_tree (CID, CNAME, PID) values (12, '浦东', 4);
2、TreeNode对象，对应tb_tree
public class TreeNode implements Serializable {
private Integer cid;
private String cname;
private Integer pid;
private List nodes = new ArrayList();
public TreeNode() {
}
//getter、setter省略
}
3、测试数据
public class TreeNodeTest {
@Test
public void loadTree() throws Exception{
System.out.println(JsonUtils.javaToJson(recursiveTree(1)));
}
/**
* 递归算法解析成树形结构
*
* @param cid
* @return
* @author jiqinlin
*/
public TreeNode recursiveTree(int cid) {
//根据cid获取节点对象(SELECT * FROM tb_tree t WHERE t.cid=?)
TreeNode node = personService.getreeNode(cid);
//查询cid下的所有子节点(SELECT * FROM tb_tree t WHERE t.pid=?)
List childTreeNodes = personService.queryTreeNode(cid);
//遍历子节点
for(TreeNode child : childTreeNodes){
TreeNode n = recursiveTree(child.getCid()); //递归
node.getNodes().add(n);
}
return node;
}
}
输出的json格式如下：
{
"cid": 1,
"nodes": [
{
"cid": 2,
"nodes": [
{
"cid": 11,
"nodes": [
],
"cname": "密云县",
"pid": 2
}
],
"cname": "北京市",
"pid": 1
},
{
"cid": 3,
"nodes": [
{
"cid": 5,
"nodes": [
{
"cid": 7,
"nodes": [
],
"cname": "海珠区",
"pid": 5
},
{
"cid": 8,
"nodes": [
],
"cname": "天河区",
"pid": 5
}
],
"cname": "广州市",
"pid": 3
},
{
"cid": 6,
"nodes": [
{
"cid": 9,
"nodes": [
],
"cname": "福田区",
"pid": 6
},
{
"cid": 10,
"nodes": [
],
"cname": "南山区",
"pid": 6
}
],
"cname": "深圳市",
"pid": 3
}
],
"cname": "广东省",
"pid": 1
},
{
"cid": 4,
"nodes": [
{
"cid": 12,
"nodes": [
],
"cname": "浦东",
"pid": 4
}
],
"cname": "上海市",
"pid": 1
}
],
"cname": "中国",
"pid": 0
}