1. In actual development, we will encounter an array containing chidren in the array. If we want to obtain the corresponding value of a specified key in the array to form a new key and value, how can we change the simple and convenient operation at this time? Woolen cloth?
1: How to get all the id values inside, including those in children
[
{
id: 1,
age: 23,
name: '测试1',
children: [
{
id: 2,
age: 24,
name: '测试11',
children: [
{
id: 3,
age: 244,
name: '测试111'
},
]
}
]
},
{
id: 4,
age: 26,
name: '测试2',
children: [],
}
]
Method 1... The operator is used to solve:
let a = [
{
id: 1,
age: 23,
name: '测试1',
children: [
{
id: 2,
age: 24,
name: '测试11',
children: [
{
id: 3,
age: 244,
name: '测试111'
},
]
}
]
},
{
id: 4,
age: 26,
name: '测试2',
children: [],
}
]
// (1)
function func(arr){
return arr.reduce((a, b) => {
let res = [...a,b.id];
if (b.children) res = [...res, ...func(b.children)];
return res;
}, []);
}
// (2)
function func(arr) {
return arr.reduce((a, b) => {
return b.children
? [...a, b.id, ...func(b.children)]
: [...a, b.id];
}, []);
}
console.log(func(a))
// 结果： [1,2,3,4]
Method 2: Use for loop to traverse, and then recurse
// （1）
let arr= [
{
id: 1, children: [{
id: 2 },{
id: 3 }] },
{
id: 4, children: [] }
];
// 获取对应id
const getIds = function(arr) {
const res = arr.reduce((prev, next) => {
if (next.id) prev.push(next.id);
if (next.children && next.children.length) return prev.concat(getIds(next.children));
return prev;
}, []);
return res;
};
const res = getIds(arr);
console.log(res);
// 获取对应需要的key
function getKeys(list, key) {
return list.reduce((res, v) => {
if(v.children && v.children.length) res = res.concat(getKeys(v.children,key))
res.push(v[key]);
return res;
}, [])
}
console.log(getKeys(arr, 'id'))
// （2）
const getIdsArry = (arr) => {
const ans = [];
const dfs = (root, ans) => {
let neighbor;
if (Array.isArray(root)) {
neighbor = root;
} else {
ans.push(root.id);
neighbor = root.children;
}
if (!neighbor) return;
for (let i = 0; i < neighbor.length; ++i) {
dfs(neighbor[i], ans);
}
};
dfs(arr, ans);
return ans;
};
console.log(getIds (arr))
2: Same as the borrowing structure above, get the corresponding id [2,3] in children
let a = [
{
id: 1, children: [{
id: 2 },{
id: 3 }] },
{
id: 4, children: [] }
];
function func(arr) {
return arr.reduce((a, b) => {
return b.children ? [...a, ...func(b.children)] : [...a, b.id];
}, []);
}
console.log(func(a));
// 结果 [2,3]
3. Form a new value and store it in the array
var aa = [
{
id: 1,
age: 23,
name: '测试1',
children: [
{
id: 2,
age: 24,
name: '测试11',
children: [
{
id: 3,
age: 244,
name: '测试111'
},
]
}
]
},
{
id: 4,
age: 26,
name: '测试2',
children: [],
}
]
for (var itemproject = 0; itemproject < aa.length; itemproject++) {
if (aa[itemproject].children && aa[itemproject].children.length > 0) {
aa[itemproject].selectedFields = getidarry(aa[itemproject].children)
} else {
aa[itemproject].selectedFields = []
}
}
function getidarry (arr) {
return arr.reduce((a, b) => {
let res = [...a,b.id];
if (b.children) res = [...res, ...getidarry(b.children)];
return res;
}, []);
}
console.log('新数组', aa)