离合器中的弹出不起作用但在Codepen中运行良好

时间:2015-11-16 16:06:42

标签: angularjs ionic-framework

我正在尝试实现$ ionicPopover但由于某种原因它在我的项目中不起作用。但是当我在Codepen中运行时,它确实如此。有谁能告诉我为什么?

我的HTML代码:

<div ng-controller="SettingsCtrl">
<ion-view view-title="Instellingen">
    <ion-content >

        <div class="padding">
            <p class="textContact">Title</p>
            <p class="textContact">Title 2</p>
        </div>

        <div>   
           <p class="textContact" ng-click="openPopover($event)"><a href="#">Taal</a></p>
        </div>

    </ion-content>

    <script id="popover.html" type="text/ng-template">
      <ion-popover-view>
        <ion-content>
          <div class="list">
            <a class="item">
              Lorem Ipsum
            </a>
          </div>
        </ion-content>
      </ion-popover-view>
    </script>

</ion-view>
</div>

我的控制器代码:

angular.module('ChemicarSettings.controllers', [])

.controller('SettingsCtrl', function($scope, $ionicPopover){

    $scope.openPopover = function ($event) {
        $ionicPopover.fromTemplateUrl('popover.html', {
            scope: $scope
        }).then(function(popover) {
            $scope.popover = popover;
            $scope.popover.show($event);
        });
    }; 

})

1 个答案:

答案 0 :(得分:0)

检查app.js怎么样?您是否将ChemicarSettings.controllers添加到app.js中的angular.module行?比如..

angular.module('ChemicarSettings', ['ionic','ChemicarSettings.controllers'])

在ChemicarSettings.controllers的第一行添加“ionic”:

angular.module('ChemicarSettings.controllers', ['ionic'])