在线等，挺急的！js 数组重组 根据配置嵌套结构 形成 父子级数据跪求！

Question

原数组如下[[{"name":"VKORG","value":"1000"},{"name":"VTWEG","value":"100"},{"name":"SPART","value":"10"},{"name":"VKORGTEXT","value":"1000text"},{"name":"VTWEGTEXT","value":"100text"},{"name":"SPARTTEXT","value":"10text"}],[{"name":"VKORG","value":"1000"},{"name":"VTWEG","value":"100"},{"name":"SPART","value":"9"},{"name":"VKORGTEXT","value":"1000text"},{"name":"VTWEGTEXT","value":"100text"},{"name":"SPARTTEXT","value":"9text"}],[{"name":"VKORG","value":"1252"},{"name":"VTWEG","value":"100"},{"name":"SPART","value":"10"},{"name":"VKORGTEXT","value":"1252text"},{"name":"VTWEGTEXT","value":"100text"},{"name":"SPARTTEXT","value":"10text"}]]父子关系是动态的可配置规则展示二级或者三级嵌套如['VKORG','VTWEG','SPART']三级或者['VKORG','VTWEG']二级如果形成三级父子级数组如下[{'value':'1000','label':'1000text','children':[{'value':'100','label':'100text','children':[{'value':'10','label':'10text'},{'value':'9','label':'9text'}]}]},{'value':'1252','label':'1252text','children':[{'value':'100','label':'100text','children':[{'value':'10','label':'10text'}]}]}]如果形成二级嵌套数组如下[{'value':'1000','label':'1000text','children':[{'value':'100','label':'100text'}]},{'value':'1252','label':'1252text','children':[{'value':'100','label':'100text'}]}]请问下怎么样根据设置的嵌套层数得到以上结果###题目描述题目来源及自己的思路相关代码//请把代码文本粘贴到下方（请勿用图片代替代码）你期待的结果是什么？实际看到的错误信息又是什么？

Answer 1

consttransform=(()=>{
constunique=items=>{
consthash=newMap();
for(constitemofitems){
constid=String([item.label,item.value]);
if(hash.has(id)){
hash.get(id).children.push(...item.children);
}else{
hash.set(id,item);
}
}
constcollection=[];
for(const[,item]ofhash){
collection.push(item);
if(item.children)item.children=unique(item.children);
}
returncollection;
};
returnitems=>{
items=items.map(item=>{
consthash={};
for(const{name,value}ofitem){
hash[name]=value;
}
returnhash;
});
returnkeys=>{
constcollection=items.map(item=>{
constcollection=[];
for(
leti=0,key=keys[i],p=collection;
key;
key=keys[(i+=1)],p=p[0].children
){
p.push({children:[],label:item[key+"TEXT"],value:item[key]});
}
returncollection[0];
});
returnunique(collection);
};
};
})();
constitems=[
[
{name:"VKORG",value:"1000"},
{name:"VTWEG",value:"100"},
{name:"SPART",value:"10"},
{name:"VKORGTEXT",value:"1000text"},
{name:"VTWEGTEXT",value:"100text"},
{name:"SPARTTEXT",value:"10text"}
],
[
{name:"VKORG",value:"1000"},
{name:"VTWEG",value:"100"},
{name:"SPART",value:"9"},
{name:"VKORGTEXT",value:"1000text"},
{name:"VTWEGTEXT",value:"100text"},
{name:"SPARTTEXT",value:"9text"}
],
[
{name:"VKORG",value:"1252"},
{name:"VTWEG",value:"100"},
{name:"SPART",value:"10"},
{name:"VKORGTEXT",value:"1252text"},
{name:"VTWEGTEXT",value:"100text"},
{name:"SPARTTEXT",value:"10text"}
]
];
consttrans=transform(items);
console.log(trans(["VKORG","VTWEG","SPART"]));
console.log(trans(["VKORG","VTWEG"]));