遍历树结构，比对id值，返回对应name 

 1  filterType(id, tree) {
 2       function filterOurType(id,tree){
 3           for(let l of tree){
 4             if(id===l.id){
 5               return l.name;
 6             }else{
 7               if(l.children&&l.children.length>0){
 8                 let res=filterOurType(id,l.children);
 9                 if(res){
10                   return res
11                 }
12               }
13             }                      
14           }
15       }
16       return filterOurType(id,tree);
17   },

 

posted @ 2023-05-31 09:36  鹿俊俊  阅读(56)  评论(0)    收藏  举报