ui.router:如何从 URL 中省略默认参数 [英] ui.router: how to omit a default parameter from URL

问题描述

我使用 $stateProvider:

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

更改年份时，如果 URL 与默认值匹配(例如 2014)，我希望 URL 省略 URL 的 {year} 部分.换句话说，当:

When changing a year, I would like the URL to omit the {year} part of the URL if it matches the default (say 2014). In other words, when:

$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 视图时(假设年份是 2014 - 默认):

And when I switch to a byPlayer view (assuming the year is 2014 - default):

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

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

推荐答案

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

Read the docs for params and squash in $stateProvider.state()

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

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