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问题:angularJs列表中点击某条记录,跳转到新页面。当返回到列表后,仍然可以回到跳转前的页码处。 要做到上述效果,要把分页中的currentPage填写到url中,然后在新页面的url中也添加这个page信息。
$routeParams.page 获得url中的page $route.updateParams({page:$scope.page}) 修改url中的page信息 $routeProvider.when('/demo/:page',{ ... }) 路由写法
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问题:angularJs列表中点击某条记录,跳转到新页面。当返回到列表后,仍然可以回到跳转前的页码处。
要做到上述效果,要把分页中的currentPage填写到url中,然后在新页面的url中也添加这个page信息。
The text was updated successfully, but these errors were encountered: