如何在Angular JS中刪除QueryString

Question

我有一個網址http://sitename.com/#/products/manage-options/31?form_id=32&type=create-template

在我的網址上方，我想刪除type但在我的代碼下方，則無法正常工作。 請檢查

我當前頁面的網址-http: http://sitename.com/#/products/manage-options/31?type=create-template create- http://sitename.com/#/products/manage-options/31?type=create-template

<a href="" ng-click="clickMe();">Click Here</a>

$scope.clickMe = function(){
  $location.search('type', null);
  $state.go('products.manage_options', ({form_id: $stateParams.form_id}));
};

頁面重定向后，我看到type QueryString沒有刪除。

我的輸出是http://sitename.com/#/products/manage-options/31?form_id=32&type=create-template //sitename.com/#/products/manage-options/31? http://sitename.com/#/products/manage-options/31?form_id=32 - http://sitename.com/#/products/manage-options/31?form_id=32&type=create-template我想要http://sitename.com/#/products/manage-options/31?form_id=32

請幫我。

編輯：

路由器

.state('products.manage_options', {
        url: '/manage-options/:product_id?form_id=&type',
        templateUrl: "views/product/manage_options.html",
        controller: "manageAttributeCtrl",
        data: {pageTitle: 'Manage Options'},
})

Answer 1

去掉

$location.search('type', null);

更改以下行：

$state.go('products.manage_options', ({form_id: $stateParams.form_id, type:null}));

Answer 2

參見文檔https://docs.angularjs.org/api/ng/service/ $ location＃search

     $state.go('products.manage_options', ({form_id: $stateParams.form_id, type:null}));