js----递归获取父节点

话不多说,直接上代码:

var list = [
  {id:1,pid:0,name:'中国'},
  {id:2,pid:0,name:'美国'},
  {id:3,pid:1,name:'云南'},
  {id:4,pid:3,name:'昆明'},
  {id:5,pid:4,name:'盘龙区'},
  {id:6,pid:5,name:'北辰小区'},
];
function findAncestry (arr, id) {
  var temp = []
  var forFn = function (list, id) {
    for (var i = 0; i < list.length; i++) {
      var item = list[i]
      if (item.id === id) {
        temp.push(item)
        forFn(list, item.pid)
        break
      } else {
        if (item.children) {
          forFn(item.children, id)
        }
      }
    }
  }
  forFn(arr, id)
  return temp
}
var allParent = findAncestry(list,6);
console.log(allParent);

打完收工。

posted @ 2020-06-30 17:59  帅到要去报警  阅读(795)  评论(0编辑  收藏  举报