求java递归算法，帮我把模块表里面的数据遍历，然后转成json形式传到前台，形成树形事后加分+,非常感谢

private void putModule(Module module, List<Module> list)

{

Integer moduleId = modulegetModuleId();

for (Module m : list)

{

if (mgetModuleParentId()equals(moduleId)){

if (modulegetmodules() == null ){

List<Module> modules = new ArrayList<Module>():

modulesetModules(modules);

}

modulegetModules()add(m);

}

}

}

private void parseModule(List<Module> modules, List<Module> sources)

{

for (Module m : modules)

{

putModule(m, sources);

if (mgetModules()size() > 0)

{

parseModule(mgetModules(), sources);

}

}

}

// 调用这个方法即可，插入查询回来的列表

public Module dealTopLevel(List<Module> modules){

Module module = null;

for (Module m : modules)

{

if (mgetModuleLevel() == 0){

module = m;

break;

}

}

putModule(module, modules);

if (modulegetModules()size() > 0)

{

parseModule(modulegetModules(), modules);

}

return module；

}

注意在实体类中加入 List<Module> modules；的属性

Spring直接返回这个Module即可得到结构

辛苦写的，请采纳

首先要遍历json,JSONObject的keySet类似于map可以遍历下面所有的key,

但是如果节点下面还有对象,那就要用到递归了

可以看到对象有好几层,并且有可能包含数组及不规则对象,如果我们想寻找所有节点下的value并找出包含"wav"的字符串并替换掉,做法是什么呢

可以看出所有包含"wav"的value都被替换了

注:文中会使用hutool工具类,需要在maven中引入

//是类似这种吗

//控制层使用JSONArray jsonObject=JSONArrayfromObject(）;转换

Map<String,Object> map = new HashMap<String, Object>();

     mapput("id","1");

     mapput("text","实验外国语学校");

     List<Map<String,Object>> fatherList=new ArrayList<Map<String,Object>>();

     List<Map<String,Object>> list=new ArrayList<Map<String,Object>>();

     for(Bean bean:list){

         if("1"equals(listgetParent_level())){

    

         Map<String,Object> map2=new HashMap<String, Object>();

    

         map2put("id",listgetId());

    

         map2put("text",listgetName());

    

         listadd(map2);

    

         }

    

     }

     mapput("children",list);

可以，就是需要遍历所有的数据。

def print_dict(k, v):

    if isinstance(v, dict):

        print k, v

        for kk in vkeys():

            print_dict(kk, v[kk])

    else:

        print k, v

for k in dkeys():

    print_dict(k, d[k])

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{

Systemoutprintln(JsonUtilsjavaToJson(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 tcid=)

TreeNode node = personServicegetreeNode(cid);

//查询cid下的所有子节点(SELECT FROM tb_tree t WHERE tpid=)

List childTreeNodes = personServicequeryTreeNode(cid);

//遍历子节点

for(TreeNode child : childTreeNodes){

TreeNode n = recursiveTree(childgetCid()); //递归

nodegetNodes()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

}

1、方法思路使用js数组自带的filter()方法；数据格式要求，父子节点通过，Id，ParentId进行关联。默认父结节id为0。

样例数据：var jsonData = [{"id":"1","pid":"0","name":"家用电器"}, {"id":"4","pid":"1","name":"大家电"}, {"id":"5","pid":"1","name":"生活电器"}, {"id":"2","pid":"0","name":"服饰"}, {"id":"3","pid":"0","name":"化妆"}, {"id":"7","pid":"4","name":"空调"}, {"id":"8","pid":"4","name":"冰箱"}, {"id":"9","pid":"4","name":"洗衣机"}, {"id":"10","pid":"4","name":"热水器"}, {"id":"11","pid":"3","name":"面部护理"}, {"id":"12","pid":"3","name":"口腔护理"}, {"id":"13","pid":"2","name":"男装"}, {"id":"14","pid":"2","name":"女装"}, {"id":"15","pid":"7","name":"海尔空调"}, {"id":"16","pid":"7","name":"美的空调"}, {"id":"19","pid":"5","name":"加湿器"}, {"id":"20","pid":"5","name":"电熨斗"}];

2、实现方法：

function creatTreeData(list){

var clonelist = JSONparse(JSONstringify(list));

var result =clonelistfilter(function(father){

var childlist = clonelistfilter(function(child){

return fatherid= childpid;

});

if(childlistlength>0){

fatherchildren=childlist;

}

return fatherpid ==0;

});

return result;

}

3、方法封装

function treeData(list,id,pid,children){

var clone = JSONparse(JSONstringify(list));

return clonefilter(function(father){

var childlist =clonefilter(function(child){

return father[id]==child[pid];

});

if(childlistlength>0){

father[children]=childlist;

}

return father[pid]==0;

});

}

以上就是关于求java递归算法，帮我把模块表里面的数据遍历，然后转成json形式传到前台，形成树形事后加分+,非常感谢全部的内容，包括:求java递归算法，帮我把模块表里面的数据遍历，然后转成json形式传到前台，形成树形事后加分+,非常感谢、递归解析json节点并替换、如何用Java拼接JSON方式遍历整个树形节点等相关内容解答，如果想了解更多相关内容，可以关注我们，你们的支持是我们更新的动力！