如何从弹出框选项中显示Ionic弹出窗口?

时间:2015-04-27 21:14:17

标签: javascript angularjs ionic-framework ionic

我的Ionic框架应用程序中有一个popover,带有选项:共享和删除。我需要在选择删除选项时显示确认弹出窗口,但我不知道如何。

如何做到这一点?我是否需要为弹出窗口创建单独的控制器?我已经从ActionSheet做了一个弹出窗口,但这有点不同。

这是控制器:

select colname from tablename where (select precision from sys.columns 
where name='<colname>') > 4;

这是popover模板:

$ionicPopover.fromTemplateUrl('templates/popover.html', {
    scope: $scope
}).then(function(popover) {
    $scope.popover = popover;
});

// Triggered on a button click, or some other target
$scope.openPopover = function($event) {

    $scope.popover.show($event);
}; 

1 个答案:

答案 0 :(得分:12)

您可以在删除时放置ng-click(我认为您的模板中有Eliminar吗?)

<ion-popover-view style="height: 120px">
  <ion-content>
    <div class="list">
      <a class="item">
        Compartir
      </a>
      <a class="item" ng-click="showConfirm()">
        Eliminar
      </a>
    </div>
  </ion-content>
</ion-popover-view>
$ionicPopover.fromTemplateUrl('templates/popover.html', {
    scope: $scope
}).then(function(popover) {
    $scope.popover = popover;
});

// Triggered on a button click, or some other target
$scope.openPopover = function($event) {

    $scope.popover.show($event);
}; 

$scope.showConfirm = function() {
   var confirmPopup = $ionicPopup.confirm({
     title: 'Are you sure?',
     template: 'Are you sure you want to delete?'
   });
   confirmPopup.then(function(res) {
     if(res) {
       console.log('You are sure');
     } else {
       console.log('You are not sure');
     }
   });
 };