需求:数组转成树状结构

data: [{label: '张大大',children: [{label: '小亮',children: [{ label: '小丽' }, { label: '大光' }]}]}],
数据:
const arr = [{ id: '01', name: '张大大', pid: '', job: '项目经理' },{ id: '02', name: '小亮', pid: '01', job: '产品leader' },{ id: '03', name: '小美', pid: '01', job: 'UIleader' },{ id: '04', name: '老马', pid: '01', job: '技术leader' },{ id: '05', name: '老王', pid: '01', job: '测试leader' },{ id: '06', name: '老李', pid: '01', job: '运维leader' },{ id: '07', name: '小丽', pid: '02', job: '产品经理' },{ id: '08', name: '大光', pid: '02', job: '产品经理' },{ id: '09', name: '小高', pid: '03', job: 'UI设计师' },{ id: '10', name: '小刘', pid: '04', job: '前端工程师' },{ id: '11', name: '小华', pid: '04', job: '后端工程师' },{ id: '12', name: '小李', pid: '04', job: '后端工程师' },{ id: '13', name: '小赵', pid: '05', job: '测试工程师' },{ id: '14', name: '小强', pid: '05', job: '测试工程师' },{ id: '15', name: '小涛', pid: '06', job: '运维工程师' },{ id: '16', name: '小才', pid: '07', job: '运维工程师' },]
答案:
方法一
arr.forEach(item => {// item.pid 为" "时 返回underfinedlet parent = obj[item.pid]if (parent) {(parent.children || (parent.children = [])).push(item)} else {// 这里push的item是pid为undefined的数据result.push(item)}})console.log(result)
方法二:用递归函数
function filterArray(data, pid) {var tree = []var tempfor (var i = 0; i < data.length; i++) {if (data[i].pid == pid) {var obj = data[i]temp = filterArray(data, data[i].id)if (temp.length > 0) {obj.children = temp}tree.push(obj)}}return tree}console.log(filterArray(arr, ''))
