javascript - 如何在 javascript 中编写递归平面 map ？

Question

我有一个嵌套路由的对象。

任何路线可以包含路线列表childRoutes。

我想获取包含键 menu 的所有路由的列表。

const routes = [{
        "name": "userManagement",
        "childRoutes": [
          {
            "name": "blogManagement",
            "childRoutes": [
              {
                "name": "blog",  // <=== I want to have this route
                "menu": {
                  "role": 1020
                }
              }
            ],
          },
          {
            "name": "organizationList", // <=== and this one
            "menu": {
              "role": 1004
            }
          }

        ],
      }, { 
      	"name": "test", 
  	"menu": { "role": 4667 }
  	}];

const deepFlatten = arr => [].concat(...arr.map(v => (Array.isArray(v) ? deepFlatten(v) : v)));

// Should handle nesting of route 
const links = deepFlatten(routes).filter((r) => !!r.menu); 

console.log('it should have a length of 3:', links.length === 3);
console.log('it should be blog:', links[0].name === 'blog');
console.log('it should be organizationList:', links[1].name === 'organizationList');
console.log('it should be test:', links[2].name === 'test');

上面的代码片段还不能递归地工作。

在没有任何第三方库的情况下，我如何递归执行此操作？

Answer 1

@yBrodsky 的回答可以用来隔离和展示通用的 flatMap 操作——在这里，你会看到路线被扁平化了很多 reduce-map-concat 为程序员开路。

// polyfill if you don't have it
Array.prototype.flatMap = function (f)
{
  return this.reduce ((acc, x) =>
    acc.concat (f (x)), [])
}

// your data   
const routes =
    [ { name : "userManagement"
      , childRoutes :
            [ { name : "blogManagement"
              , childRoutes :
                    [ { name : "blog"
                      , menu : { role : 1020 }
                      }
                    ]
              }
            , { name : "organizationList"
              , menu : { role : 1004 }
              }
            ]
      }
    , { name : "test"
      , menu : { role : 4667 }
      }
    ]

// flat-mapped routes
const allChildRoutes =
  routes.flatMap (function loop (node) {
    if (node.childRoutes)
      return node.childRoutes.flatMap (loop)
    else
      return [node]
  })     
  
console.log (allChildRoutes)