java将list转为树形结构的⽅法⽬录
1、通过转化成json封装数据
原始数据如下
[
{
"name":"⽢肃省",
"pid":0,
"id":1
},
{
"name":"天⽔市",
"pid":1,
"id":2
},
{
"name":"秦州区",
"pid":2,
"id":3
},
{
"name":"北京市",
"pid":0,
"id":4
},
{
"name":"昌平区",
"pid":4,
"id":5
}
]
现需要是使⽤java将以上数据转为树形结构，转化后下的结构如下
[
{
"children":[
{
"children":[
{
"name":"秦州区",
"pid":2,
"id":3
}
],
"name":"天⽔市",
"pid":1,
"id":2
}
],
"name":"⽢肃省",
"pid":0,
"id":1
},
{
"children":[
{
"name":"昌平区",
"pid":4,
"id":5
}
],
"name":"北京市",
"pid":0,
"id":4
}
]
代码如下
/**
-
listToTree
- <p>⽅法说明<p>
- 将JSONArray数组转为树状结构
- @param arr 需要转化的数据
- @param id 数据唯⼀的标识键值
- @param pid ⽗id唯⼀标识键值
- @param child ⼦节点键值
- @return JSONArray
*/
public static JSONArray listToTree(JSONArray arr,String id,String pid,String child){
JSONArray r = new JSONArray();
JSONObject hash = new JSONObject();
//将数组转为Object的形式，key为数组中的id
for(int i=0;i<arr.size();i++){
JSONObject json = (JSONObject) (i);
hash.String(id), json);
}
//遍历结果集
for(int j=0;j<arr.size();j++){
//单条记录
JSONObject aVal = (JSONObject) (j);
//在hash中取出key为单条记录中pid的值
JSONObject hashVP = (JSONObject) ((pid).toString());
//如果记录的pid存在，则说明它有⽗节点，将她添加到孩⼦节点的集合中
if(hashVP!=null){
//检查是否有child属性
(child)!=null){
JSONArray ch = (JSONArray) (child);
ch.add(aVal);
hashVP.put(child, ch);
}else{
JSONArray ch = new JSONArray();
ch.add(aVal);
hashVP.put(child, ch);
}
}else{
r.add(aVal);
}
}
return r;
}
测试代码如下
public static void main(String[] args){
List<Map<String,Object>> data = new ArrayList<>();
Map<String,Object> map = new HashMap<>();
map.put("id",1);
map.put("pid",0);
map.put("name","⽢肃省");
data.add(map);
Map<String,Object> map2 = new HashMap<>();
map2.put("id",2);
map2.put("pid",1);
map2.put("name","天⽔市");
data.add(map2);
Map<String,Object> map3 = new HashMap<>();
map3.put("id",3);
map3.put("pid",2);
map3.put("name","秦州区");
data.add(map3);
Map<String,Object> map4 = new HashMap<>();
map4.put("id",4);
map4.put("pid",0);
map4.put("name","北京市");
data.add(map4);
Map<String,Object> map5 = new HashMap<>();
map5.put("id",5);
map5.put("pid",4);
map5.put("name","昌平区");
data.add(map5);
System.out.JSONString(data));
JSONArray result = listToTree(JSONArray.JSONString(data)),"id","pid","children");
System.out.JSONString(result));
}
2、通过java8 stream转换
我在⽹上了很多⽅法，⾃⼰写的这个思路清晰，代码量少，希望能到志同道合的朋友，看看还有没有优化的地⽅。import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
@Data
@AllArgsConstructor
@NoArgsConstructor
public class ZhField {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private String id;
/**
* 上级领域id
*/
@Column(name = "parent_id")
private String parentId;
/**
* 领域名称
*/
private String name;
/**
* 排序
*/
private Integer sort;
}
import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import java.io.Serializable;
import java.util.List;
import java.util.Map;
@Data
@AllArgsConstructor
@NoArgsConstructor
public class TreeMenuNode implements Serializable {
private String id;
private String parentId;
private String name;
private Integer sort;
private List<TreeMenuNode> children;
private Boolean isAble;
/**20180929zhw添加树的额外属性(⾄少含有⽗节点ID:"parentId")**/
private Map<String,Object> attributes;sortedlist
}
aoqi.service.implpany;
ity.ZhField;
aoqi.util.TreeMenuNode;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
/**
* @author zhizhao
* @description
* @create 2018-11-14 9:07
*/
public class toModel {
private static void forEach(Map<String, List<TreeMenuNode>> collect, TreeMenuNode treeMenuNode) {
List<TreeMenuNode> treeMenuNodes = (Id());
(Id())!=null){
//排序
treeMenuNodes.sort((u1, u2) -> u1.getSort()Sort()));
treeMenuNodes.stream().sorted(Comparatorparing(TreeMenuNode::getSort)).List());            treeMenuNode.setChildren(treeMenuNodes);
forEach(collect,t);
});
}
}
public static void main(String[] args) {
List<ZhField> zhFields = new ArrayList<>();
List<TreeMenuNode> treeNodeList = new ArrayList<>();
//转换数据，这个是前端需要的格式。
zhFields.forEach(t->{
TreeMenuNode treeMenuNode = new TreeMenuNode();
treeMenuNode.Id());
treeMenuNode.ParentId());
treeMenuNode.Name());
treeMenuNode.Sort());
treeNodeList.add(treeMenuNode);
});
/
/分组
Map<String, List<TreeMenuNode>> collect = treeNodeList.stream().upingBy(TreeMenuNode::getParentId));        //树形结构肯定有⼀个根部，我的这个根部的就是parentId.euqal("0"),⽽且只有⼀个就get（"0"）
TreeMenuNode treeMenuNode = ("0").get(0);
//拼接数据
forEach(collect, treeMenuNode);
}
}

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