angularjs - ui.router : how to omit a default parameter from URL

Question

我用 $stateProvider 定义了以下状态:

$stateProvider.state("byTeams", {url : "/team/{id}/{year}", ...})
$stateProvider.state("byPlayer", {url : "/player/{id}/{year}", ...})

改年份时，希望网址省略 {year} URL 的一部分，如果它与默认值匹配(比如 2014)。换句话说，当:

$state.go("byTeams", {year: 2014}) --> www.example.com/app/#/team/343
$state.go("byTeams", {year: 2013}) --> www.example.com/app/#/team/343/2013

当我切换到 byPlayer查看(假设年份是 2014 - 默认):

$state.go("byPlayer", {id: 555}) --> www.example.com/app/#/player/555/

否则，URL 将是: www.example.com/app/#/player/555/2013

Answer 1

阅读 docs为 params和 squash在 $stateProvider.state()

$stateProvider.state("byPlayer", {
  url : "/player/{id}/{year}", 
  params: { 
    year: { 
      value: function() { return getCurrentYear(); },
      squash: true
    }
  }
})