如何點擊ionicActionSheet選項打開ionicPopup？

Question

我試圖打開彈出式窗口，當用戶點擊操作表上的「自定義」按鈕時，我無法弄清楚如何在兩者之間進行交互。我最好的猜測是在下面，當我在視圖中調用ng-click =「showPrompt（）」時，彈出窗口被觸發，但是當我嘗試從操作表的buttonClicked事件中執行它時，它是一個禁止行爲。

.controller('TablesCtrl', function($scope, $ionicPopup, $ionicActionSheet) { 
    $scope.tables = []; 

    /* Choose Number of Guests */ 
    $scope.showActionsheet = function($ionicPopup) { 
     $ionicActionSheet.show({ 
      titleText: 'How many guests?', 
      buttons: [ 
      { text: '1' }, 
      { text: '2' }, 
      { text: '3' }, 
      { text: '4' }, 
      { text: '5' }, 
      { text: '6' }, 
      { text: 'Custom' } 
      ], 
      cancelText: 'Cancel', 
      cancel: function() { 
      console.log('CANCELLED'); 
      }, 
      buttonClicked: function(index, $ionicPopup) { 
      console.log('BUTTON CLICKED', index); 
      if(index==6){showPrompt();} 
      return true; 
      } 
     }); 
     }; 
    /* CUSTOM Number of Guests */ 
    $scope.showPrompt = function() { 
     var myPopup = $ionicPopup.show({ 
     template: '<input type="password" ng-model="data.wifi">', 
     title: 'Enter Wi-Fi Password', 
     subTitle: 'Please use normal things', 
     scope: $scope, 
     buttons: [ 
      { text: 'Cancel' }, 
      { 
      text: '<b>Save</b>', 
      type: 'button-positive', 
      onTap: function(e) { 
       if (!$scope.data.wifi) { 
       //don't allow the user to close unless he enters wifi password 
       e.preventDefault(); 
       } else { 
       return $scope.data.wifi; 
       } 
      } 
      }, 
     ] 
     }); 
    }; 
}) 

Answer 1

嘗試使用$scope因爲該功能是在這方面不確定，但$範圍通過封閉定義：

if(index==6){$scope.showPrompt();}