現(xiàn)有一個(gè)數(shù)組
[
{id: 1, pk: null, name: '學(xué)校'},
{id: 2, pk: 1, name: '一年級(jí)'},
{id: 3, pk: 1, name: '二年級(jí)'},
{id: 4, pk: 1, name: '三年級(jí)'},
{id: 6, pk: 2, name: '一班'},
{id: 7, pk: 2, name: '二班'},
{id: 8, pk: 2, name: '三班'},
{id: 9, pk: 2, name: '四班'},
{id: 10, pk: 3, name: '一班'},
{id: 11, pk: 3, name: '二班'},
{id: 12, pk: 3, name: '三班'},
{id: 13, pk: 3, name: '四班'},
{id: 14, pk: 4, name: '一班'},
{id: 15, pk: 4, name: '二班'},
{id: 16, pk: 4, name: '三班'},
{id: 17, pk: 4, name: '四班'}
]
pk為null的放在最外層, 子pk對(duì)應(yīng)父id,則當(dāng)作對(duì)應(yīng)id的children, id轉(zhuǎn)為對(duì)應(yīng)的key,
如 id為17這一條, pk為4,則對(duì)應(yīng)為三年級(jí)的children
.
最終轉(zhuǎn)換結(jié)果
[{
title: '學(xué)校',
key: 1,
children: [{
title: '一年級(jí)',
key: 2,
children: [{
title: '一班',
key: 6,
},{
title: '二班',
key: 7,
},{
title: '三班',
key: 8,
},{
title: '四班',
key: 9,
}]
},{
title: '二年級(jí)',
key: 3,
children: [{
title: '一班',
key: 10,
},{
title: '二班',
key: 11,
},{
title: '三班',
key: 12,
},{
title: '四班',
key: 13,
}]
},{
title: '三年級(jí)',
key: 4,
children: [{
title: '一班',
key: 14,
},{
title: '二班',
key: 15,
},{
title: '三班',
key: 16,
},{
title: '四班',
key: 17,
}]
}]
}]
真實(shí)數(shù)據(jù)可能沒(méi)那么整齊, 但是轉(zhuǎn)換規(guī)則是這樣的, 有什么好的轉(zhuǎn)換寫(xiě)法么?
無(wú)限樹(shù)結(jié)構(gòu)可以用遞歸
function buildChildren(parents, source) {
return parents.map(parent => {
let children = source.find(item => item.pk === parent.id);
parent.children = buildChildren(children, source);
});
}
let items = [ ... ];
let result = buildChildren(items.find(item=>!item.pk), items);
隨手寫(xiě)的,生產(chǎn)環(huán)境可以再優(yōu)化一下。。
let result = [];
let key_map = data.reduce((obj, item) => {
return ((pk, data) => pk
? (obj[pk] = obj[pk] || []).push(data)
: result.push(data))(item.pk, {
title: item.name,
key: item.id
}), obj
}, {});
let key_walker = item => (key => key_map[key]
&& (item.children = key_map[key]).forEach(key_walker))(item.key)
result.forEach(key_walker)
console.dir(result);
其實(shí)思路跟前邊幾位差不多,替換了一些 if
結(jié)構(gòu)而已。
let result = [];
const setData = (list, d) => {
for(let i = 0; i < list.length; i++) {
if(list[i].key === d.pk) {
if(!list[i].children) {
list[i].children = [{key: d.id, title: d.name}];
} else {
list[i].children.push({key: d.id, title: d.name});
}
return list;
} else if(list[i].children) {
setData(list[i].children, d);
} else {
return list;
}
}
};
const data = [
{id: 1, pk: null, name: '學(xué)校'},
{id: 2, pk: 1, name: '一年級(jí)'},
{id: 3, pk: 1, name: '二年級(jí)'},
{id: 4, pk: 1, name: '三年級(jí)'},
{id: 6, pk: 2, name: '一班'},
{id: 7, pk: 2, name: '二班'},
{id: 8, pk: 2, name: '三班'},
{id: 9, pk: 2, name: '四班'},
{id: 10, pk: 3, name: '一班'},
{id: 11, pk: 3, name: '二班'},
{id: 12, pk: 3, name: '三班'},
{id: 13, pk: 3, name: '四班'},
{id: 14, pk: 4, name: '一班'},
{id: 15,pk: 4, name: '二班'},
{id: 16, pk: 4, name: '三班'},
{id: 17, pk: 4, name: '四班'}
];
data.forEach(d => {
if(!d.pk) {
result.push({key: d.id, title: d.name});
} else {
setData(result, d);
}
});
console.log(result);
// 結(jié)果
let result = []
// hash 用于快速查找數(shù)據(jù)
let tempList = []
arr.forEach(item => {
// 生成待保存結(jié)構(gòu)數(shù)據(jù)
let temp = {
title: item.name,
key: item.id
}
// 存在hash上
tempList[item.id] = temp
if (item.pk === null) {
// 放置最外層
result.push(temp)
} else if (tempList[item.pk] !== undefined) {
// 存在則添加至children中
if (tempList[item.pk].children === undefined) {
tempList[item.pk].children = []
}
tempList[item.pk].children.push(temp)
} else {
// 否則提示錯(cuò)誤
console.log(item + '未找到對(duì)應(yīng)父級(jí)')
}
})
var items = [
{id: 1, pk: null, name: '學(xué)校'},
{id: 2, pk: 1, name: '一年級(jí)'},
{id: 3, pk: 1, name: '二年級(jí)'},
{id: 4, pk: 1, name: '三年級(jí)'},
{id: 6, pk: 2, name: '一班'},
{id: 7, pk: 2, name: '二班'},
{id: 8, pk: 2, name: '三班'},
{id: 9, pk: 2, name: '四班'},
{id: 10, pk: 3, name: '一班'},
{id: 11, pk: 3, name: '二班'},
{id: 12, pk: 3, name: '三班'},
{id: 13, pk: 3, name: '四班'},
{id: 14, pk: 4, name: '一班'},
{id: 15,pk: 4, name: '二班'},
{id: 16, pk: 4, name: '三班'},
{id: 17, pk: 4, name: '四班'}
]
var ret = []
var mids = {}
items.forEach(item => {
var data = {
title: item.name,
key: item.id
}
if (!mids[item.pk]) {
mids[item.id] = data
ret.push(data)
} else {
mids[item.id] = data
if (!mids[item.pk].children) {
mids[item.pk].children = []
}
mids[item.pk].children.push(data)
}
})
console.log(ret)
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