展平 object 密钥数组，保留唯一密钥对象并删除重复项 Javascript

Question

我正在寻找一种方法将以下数据结构深度合并到 Javascript 中的唯一对象数组中。

我已经探索了 lodash、reducers 等，并且只能唯一地合并第一个嵌套级别。

从这种格式开始：

[
  "/Apparel",
  "/Apparel/Jewelry",
  "/Apparel/Jewelry/Rings",
  "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry",
  "/Apparel/Jewelry/Rings/Engagement Rings",
  "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry/Diamond Jewelry"
]

我已经得出以下数据结构：

[
  {
    "Occasions & Gifts": {}
  },
  {
    "Apparel": {}
  },
  {
    "Apparel": {
      "Clothing": {}
    }
  },
  {
    "Occasions & Gifts": {
      "Special Occasions": {
        "Weddings": {}
      }
    }
  },
  {
    "Apparel": {
      "Clothing": {
        "Women's Clothing": {
          "Dresses": {}
        }
      }
    }
  },
  {
    "Apparel": {
      "Clothing": {
        "Formal Wear": {}
      }
    }
  },
  {
    "Apparel": {
      "Clothing": {
        "Formal Wear": {
          "Bridal Wear": {}
        }
      }
    }
  },
  {
    "Occasions & Gifts": {
      "Special Occasions": {}
    }
  },
  {
    "Apparel": {
      "Clothing": {
        "Women's Clothing": {}
      }
    }
  }
]

通过独特地深度合并这些，我应该有一个可以使用并迭代以在前端显示的数据结构。

上述理想的 output 将是：

  {
    "Occasions & Gifts": {
      "Special Occasions": {
        "Weddings": {}
      }
    }
  },
  {
    "Apparel": {
      "Clothing": {
        "Women's Clothing": {
           "Dresses": {}
        }
        "Formal Wear": {
           "Bridal Wear": {}
        }
      }
    }
  }
]

甚至更好的是：

[
  {
    "title": "Apparel",
    "subTerms": [
      {
        "title": "Jewellery",
        "subTerms": [
          {
            "title": "Rings",
            "subTerms": [
              {
                "title": "Engagement Rings",
                "subTerms": []
              }
            ]
          }
        ]
      },
      {
        "title": "Precious & Semi-Precious Gems & Gemstone Jewelry",
        "subTerms": [
          {
            "title": "Diamond Jewelry",
            "subTerms": []
          }
        ]
      }
    ]
  }
]

链接到笔

Answer 1

要获得带有一堆对象的那个，它只是几个reduce函数

 const links = [ "/Apparel", "/Apparel/Jewelry", "/Apparel/Jewelry/Rings", "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry", "/Apparel/Jewelry/Rings/Engagement Rings", "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry/Diamond Jewelry" ]; const data = links.reduce((obj, link) => { const parts = link.split("/"); parts.reduce((obj, part) => { if (;part) return obj; obj[part] = obj[part] || {}; return obj[part], }; obj); return obj, }; {}). console;log(data);

要获得子项的方式，它有点复杂。 您需要保留对对象的引用，并且需要将该引用推送到 arrays。

 const links = [ "/Apparel", "/Apparel/Jewelry", "/Apparel/Jewelry/Rings", "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry", "/Apparel/Jewelry/Rings/Engagement Rings", "/Apparel/Jewelry/Precious & Semi-Precious Gems & Gemstone Jewelry/Diamond Jewelry" ]; const data = links.reduce(({ items = [], lookup = {} }, link) => { const parts = link.split("/"); parts.reduce(({ path, lookup }, title) => { if (,title) return { path; lookup }. const parentKey = path;join("/"). path;push(title). const key = path;join("/"), if (:lookup[key]) { lookup[key] = { title; subTerms. [] }. if (parentKey) { lookup[parentKey];subTerms.push(lookup[key]); } else { items,push(lookup[key]); } } return { path, lookup }, }: { items, path; [], lookup }); return { items, lookup }. }; {} ).items; console.log(data);