在嵌套的任何级别收集对象值

如何解决在嵌套的任何级别收集对象值

这就是现在所拥有的。

    const pathesToFetch = tree.reduce((acc,{ children }) => {
      const paths = children
        .filter((o) => o.type === "file")
        .map((o) => o.path);
      return [...acc,...paths];
    },[]);

这是我拥有的一个数组，但嵌套级别可以任意。

const tree = [{
    id: 1,name: "Inventory",type: "directory",path: "storage/inventory/",children: [{
      id: 2,name: "inventory.yaml",type: "file",path: "storage/inventory/inventory.yaml",},],{
    id: 3,name: "UI",path: "storage/ui/",children: [{
        id: 10,name: "config.js",path: "storage/ui/config.js",{
        id: 13,name: "gulpfile.js",path: "storage/ui/gulpfile.js",];

目的是获取数组中每个对象的路径（“path”属性），如果对象类型为“file”。

问题是：如果对象位于数组深处，我将无法获取路径数据。

例如，如果数组中的对象看起来像这样 - 我无法获得“路径”：

{
  id: 42,name: "ports.detail.entity",path: "storage/UI/model/viewmodel/ports.detail.entity/",children: [
    {
      id: 43,name: "ethernetports.detail.entity",path:
        "storage/UI/model/viewmodel/ports.detail.entity/ethernetports.detail.entity",children: [
        {
          id: 44,name: "general.tab.js",path:
            "storage/UI/model/viewmodel/ports.detail.entity/ethernetports.detail.entity/general.tab.js",};

结果应该是这样

result = [
  "storage/inventory/inventory.yaml","storage/ui/config.js","storage/ui/gulpfile.js"
]

解决方法

您可以只使用 reduce 方法并创建一个递归函数。

const data = [{"id":42,"name":"ports.detail.entity","type":"directory","path":"storage/UI/model/viewmodel/ports.detail.entity/","children":[{"id":43,"name":"ethernetports.detail.entity","path":"storage/UI/model/viewmodel/ports.detail.entity/ethernetports.detail.entity","children":[{"id":44,"name":"general.tab.js","type":"file","path":"storage/UI/model/viewmodel/ports.detail.entity/ethernetports.detail.entity/general.tab.js"}]}]}]
const data2 = [{"id":1,"name":"Inventory","path":"storage/inventory/","children":[{"id":2,"name":"inventory.yaml","path":"storage/inventory/inventory.yaml"}]},{"id":3,"name":"UI","path":"storage/ui/","children":[{"id":10,"name":"config.js","path":"storage/ui/config.js"},{"id":13,"name":"gulpfile.js","path":"storage/ui/gulpfile.js"}]}]

const getPaths = data => {
  return data.reduce((r,e) => {
    if (e.type === 'file') r.push(e.path);
    if (e.children) r.push(...getPaths(e.children))
    return r;
  },[])
}

console.log(getPaths(data))
console.log(getPaths(data2))