遍历树形结构，根据key值获取父级到子级的title

Question

  let counnt = [
        {
            title: 6,
            key: '/g',
        },
        {
            title: 1,
            key: '/a',
            subs: [
                {
                    title: 2,
                    key: '/a/b',
                    sybs: [
                        {
                            title: 3,
                            key: '/a/b/c',
                        },
                    ],
                },
            ],
        },
        {
            title: 4,
            key: '/d',
            subs: [
                {
                    title: 5,
                    key: '/d/e',
                },
            ],
        },
    ]

加入当前值是/a/b/c,需要拿到从父级到子级的title，/a/b/c,对应的就是[1,2,3]
/d/e 就是【4，5】 /g 就是【6】

Answer 1

function getTitleOfKey(list, key) {
  if(!list) return [];
  return list.reduce((res, v) => { 
    if(key.startsWith(v.key)) {
        res.push(v.title);
        // 相等即终止向下查找
        if(key !== v.key) res = res.concat(getTitleOfKey(v.subs, key))
    };
    return res;
  }, [])
}

Answer 2

function loop(arr, key, pks){
    arr = arr || []
    pks = pks || []
    arr.some(item => {
        //区分同一级有如 /a /ab的情况，不能简单使用indexOf
        //if(key.indexOf(item.key) === 0){
        if(key === item.key || key.indexOf(item.key + '/') === 0){
            pks.push(item.title)
            return key === item.key ? true : loop(item.subs, key, pks)
        }
    })

    return pks
}

//你的counnt格式不对
loop(counnt, '/d/e') //[4, 5]
loop(counnt, '/a/b') //[1, 2]

Answer 3

function doLoop(o, key) {
    if (o.key == key) {
        return [o.title]
    } else if (typeof o.subs !== 'undefined') {
        for (item in o.subs) {
            let r = doLoop(o.subs[item], key);
            if (r != null) {
                r.push(o.title);
                return r;
            }
        }
    }
    return null
}

counnt.forEach(i = >{
    let r = doLoop(i, "/a/b/c");
    if (r != null) {
        console.log(r)
    }
})

吐槽，楼主你自己看看你发出来的数据，到处都是问题。