我有以下一系列深层嵌套的对象:
const data = [
{
name: "foo",
children:[
{
count: 1,
name: "A"
},
{
count: 2,
name: "B"
}
]
},
{
name: "bar",
children: [
{
count: 3,
name: "C",
children: [
{
count: 4,
name: "D"
}
]
}
]
}
]
我要转换的方式如下:
const expectedStructure = [
{
count: 1,
name: "A",
label: "foo = A"
},
{
count: 2,
name: "B",
label: "foo = B"
},
{
count: 3,
name: "C",
label: "bar = C"
},
{
count: 4,
name: "D",
label: "bar = D"
}
]
我创建了将嵌套数组转换为平面对象数组的递归函数。
这是我的代码:
function getChildren(array, result=[]) {
array.forEach(({children, ...rest}) => {
result.push(rest);
if(children) {
getChildren(children, result);
}
});
return result;
}
这是我得到的输出:
[ { name: 'foo' },
{ count: 1, name: 'A' },
{ count: 2, name: 'B' },
{ name: 'bar' },
{ count: 3, name: 'C' },
{ count: 4, name: 'D' } ]
问题是我需要向输出数组中的每个对象添加label
字段,而且如果不对最终数组进行多次迭代以进行所需的转换,就无法找到解决方案。如何在不大大增加函数复杂性的情况下正确插入label
字段?
答案 0 :(得分:2)
检查每次迭代是否当前项目是“父”项目,如果是,则重新分配label
。
const data = [{name:"foo",children:[{count:1,name:"A"},{count:2,name:"B"}]},{name:"bar",children:[{count:3,name:"C",children:[{count:4,name:"D"}]}]}];
function getChildren(array, result = [], label = "") {
array.forEach(({ children, name, count }) => {
if (!label || name[1]) {
label = `${name} = `;
}
if (count) {
result.push({ count, name, label: label + name });
}
if (children) {
getChildren(children, result, label);
}
});
return result;
}
const res = getChildren(data);
console.log(res);
答案 1 :(得分:1)
您可以对嵌套级别使用其他函数,因此可以将顶级name
属性向下传递到所有这些递归级别。
function getTopChildren(array, result = []) {
array.forEach(({
name,
children
}) => {
if (children) {
getChildren(children, name, result);
}
});
return result;
}
function getChildren(array, name, result) {
array.forEach(({
children,
...rest
}) => {
rest.label = `${name} = ${rest.name}`;
result.push(rest);
if (children) {
getChildren(children, name, result);
}
});
}
const data = [{
name: "foo",
children: [{
count: 1,
name: "A"
},
{
count: 2,
name: "B"
}
]
},
{
name: "bar",
children: [{
count: 3,
name: "C",
children: [{
count: 4,
name: "D"
}]
}]
}
]
console.log(getTopChildren(data));
答案 2 :(得分:1)
您还可以根据是否已将parent
传递到递归调用中,使用flatMap
进行递归操作:
const data = [{
name: "foo",
children: [{
count: 1,
name: "A"
},
{
count: 2,
name: "B"
}
]
},
{
name: "bar",
children: [{
count: 3,
name: "C",
children: [{
count: 4,
name: "D"
}]
}]
}
];
function flatten(arr, parent = null) {
return parent
? arr.flatMap(({name, count, children}) => [
{name, count, label: `${parent} = ${name}`},
...flatten(children || [], parent)
])
: arr.flatMap(({name, children}) => flatten(children || [], name));
}
console.log(flatten(data));
答案 3 :(得分:1)
有时候,对代码进行推理并使用生成器清楚地编写代码要容易一些。您可以从递归调用中yield*
:
const data = [{name: "foo",children:[{count: 1,name: "A"},{ count: 2,name: "B"}]},{name: "bar",children: [{count: 3,name: "C",children: [{count: 4,name: "D"}]}]}]
function* flat(input, n){
if (!input) return
if (Array.isArray(input)) {
for (let item of input)
yield* flat(item, n)
}
let _name = n || input.name
if ('count' in input) {
yield { count:input.count, name:input.name, label:`${_name} = ${input.name}`}
}
yield* flat(input.children, _name)
}
let g = [...flat(data)]
console.log(g)
该函数返回一个生成器,因此,如果需要列表,则需要将其扩展到列表[...flat(data)]
中;如果不需要存储列表,则需要对其进行迭代。