[英]Can I remove a redundant data structure wrapper from my nested array to nested json script?
我编写了这个脚本,将具有下面结构的嵌套数组转换为具有父子关系的嵌套对象。
list = [
['lvl-1 item-1', 'lvl-2 item-1'],
['lvl-1 item-1', 'lvl-2 item-1', 'lvl-3 item-1'],
['lvl-1 item-1', 'lvl-2 item-1', 'lvl-3 item-2'],
['lvl-1 item-2', 'lvl-2 item-1', 'lvl-3 item-1'],
['lvl-1 item-2', 'lvl-2 item-2', 'lvl-3 item-2', 'lvl-4 item-1'],
];
它似乎可以解决问题,但是为了data.children
脚本,我必须在初始数据结构周围添加data.children
包装器。 我不相信这是必要的,尽管我还没有能够锻炼如何摆脱它。
任何人都能看到我失踪的东西吗?
console.log(nestedArrayToJson(list));
function nestedArrayToJson(structure) {
const top_item = '0';
// This was added to behave like the child data structure.
let data = {
children: [
{
name: top_item,
parent: null,
children: [],
}],
};
for(let i = 0; i < structure.length; i++) {
let parents = [top_item];
for(let j = 0; j < structure[i].length; j++) {
let obj = data;
for(parent of parents) {
obj = obj.children.find(o => o.name === parent);
}
const name = structure[i][j];
if(!obj.children.find(o => o.name === name)) {
obj.children.push({
name,
parent,
children: [],
});
}
parents.push(structure[i][j]);
}
}
return data.children[0];
}
样本输出
{
"name": "0",
"parent": null,
"children": [
{
"name": "lvl-1 item-1",
"parent": "0",
"children": [
{
"name": "lvl-2 item-1",
"parent": "lvl-1 item-1",
"children": [
{
"name": "lvl-3 item-1",
"parent": "lvl-2 item-1",
"children": []
},
{
"name": "lvl-3 item-2",
"parent": "lvl-2 item-1",
"children": []
}
]
}
]
},
{
"name": "lvl-1 item-2",
"parent": "0",
"children": [
{
"name": "lvl-2 item-1",
"parent": "lvl-1 item-2",
"children": [
{
"name": "lvl-3 item-1",
"parent": "lvl-2 item-1",
"children": []
}
]
},
{
"name": "lvl-2 item-2",
"parent": "lvl-1 item-2",
"children": [
{
"name": "lvl-3 item-2",
"parent": "lvl-2 item-2",
"children": [
{
"name": "lvl-4 item-1",
"parent": "lvl-3 item-2",
"children": []
}
]
}
]
}
]
}
]
}
可以通过向命名函数提取一些功能来清除for
循环。
const node = (name, parent = null) => ({name, parent, children: []})
处理创建节点。
然后可以使用addNode()
添加节点
搜索当前下一个父节点findNamedNode()
如果找到具有current
名称的node
,则它将向下移动到下一个node
。 如果不存在具有current
名称的node
,则创建该node
。
function createTree(arr, topItem = 'Top') {
const node = (name, parent = null) => ({name, parent, children: []});
const addNode = (parent, child) => {
parent.children.push(child);
return child;
};
const findNamedNode = (name, parent) => {
for(const child of parent.children) {
if(child.name === name) { return child; }
const found = findNamedNode(name, child);
if(found) { return found; }
}
};
const top = node(topItem);
let current;
for(const children of arr) {
current = top;
for(const name of children) {
const found = findNamedNode(name, current);
current = found ? found : addNode(current,
node(name, current.name));
}
}
return top;
}
感谢@ Blindman67对Code Review的帮助。
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