清理过滤路径数组的方法

Question

如何筛选以下数组

[ '/pageA/pageAA.html',
  '/pageB.html',
  '/pageC.html',
  '/pageC/',
  '/pageC/pageW/pageC.html',
  '/' ]

要得到

[ '/pageA/pageAA.html',
  '/pageB.html',
  '/pageC/pageW/pageC.html' ]

规则是：

• 路径不以/结束
• 路径不应该类似于以/ ex结尾的另一条路径：'/pageC/'和'/pageC.html'相似

我做了以下但是它不干净也许有一个更清洁的方式与正则表达式或类似的东西

const routes = ['/pageA/pageAA.html',
  '/pageB.html',
  '/pageC.html',
  '/pageC/',
  '/pageC/pageW/pageC.html',
  '/'
]

const filterRoutes = routes => {
  var unWontedRoutes = []
  for (let i = 0; i < routes.length - 1; i++) {
    for (let j = i + 1; j < routes.length; j++) {
      if (Math.abs(routes[i].length - routes[j].length) == 4) {
        var a = routes[i].length < routes[j].length ? routes[j] : routes[i]
        var b = routes[i].length > routes[j].length ? routes[j] : routes[i]
        if (a.replace(b.slice(0, b.length - 1), '') === '.html') {
          unWontedRoutes.push(a)
        }
      }
    }
  }
  return unWontedRoutes
}

const unWontedRoutes = filterRoutes(routes)
const newRoutes = routes.filter(route => (!(unWontedRoutes.indexOf(route) > -1) && !route.endsWith('/')))

console.log(newRoutes)

Answer 1

你可以使用filter()和some()来做到这一点

const routes = ['/pageA/pageAA.html',
  '/pageB.html',
  '/pageC.html',
  '/pageC/',
  '/pageC/pageW/pageC.html',
  '/'
]


const res = routes.filter(x => !x.endsWith('/') && !routes.some(a => a.endsWith('/') && x.split('.')[0] === a.replace(/\/$/,'')))

console.log(res)