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Angularjs 如何访问状态视图的自定义属性?_Angularjs_Ionic Framework - Fatal编程技术网

Angularjs 如何访问状态视图的自定义属性?

Angularjs 如何访问状态视图的自定义属性?,angularjs,ionic-framework,Angularjs,Ionic Framework,我将该州定义为: .state('app.feedback', { url: '/feedback', views: { 'appContent': { templateUrl: 'templates/feedback.html', controller: 'feedbackCtrl', customParameter: 'This is my

我将该州定义为:

.state('app.feedback', {
        url: '/feedback',
        views: {
            'appContent': {
                templateUrl: 'templates/feedback.html',
                controller: 'feedbackCtrl',
                customParameter: 'This is my Name'
            }
        }
    })
我试图在
appContent
上显示
customParameter

这是我的控制器:

.controller('feedbackCtrl',
    ['$scope', '$log', '$state',
 function($scope, $log, $state) {

    $scope.customParameter = $state.current.customParameter;

}]);
这是我的观点:

<ion-view view-title="{{customParameter}}">
    <ion-content>
        <h1>{{customParameter}}</h1>
    </ion-content>
</ion-view>

{{customParameter}}
当我在状态声明中使用没有视图对象的自定义参数时,视图上不会显示任何内容,但我需要在这个项目中有几个视图。如何访问该自定义参数


Ps:我正在使用Ionic 1.3.1

使用状态参数将自定义数据传递到状态

.state('app.feedback', {
    url: '/feedback',
    views: {
        'appContent': {
            templateUrl: 'templates/feedback.html',
            controller: 'feedbackCtrl'
        }
    },
    params:{
     customParameter: 'This is my Name'
    }
})
在控制器中,使用$stateparms获取customParameter

.controller('feedbackCtrl',
    ['$scope', '$log', '$state', $stateParams
 function($scope, $log, $state, $stateParams) {

    $scope.customParameter = $stateParams.customParameter;

}]);
而这一观点是正确的

<ion-view view-title="{{customParameter}}">
    <ion-content>
        <h1>{{customParameter}}</h1>
    </ion-content>
</ion-view>

{{customParameter}}

使用状态参数将自定义数据传递到状态

.state('app.feedback', {
    url: '/feedback',
    views: {
        'appContent': {
            templateUrl: 'templates/feedback.html',
            controller: 'feedbackCtrl'
        }
    },
    params:{
     customParameter: 'This is my Name'
    }
})
在控制器中,使用$stateparms获取customParameter

.controller('feedbackCtrl',
    ['$scope', '$log', '$state', $stateParams
 function($scope, $log, $state, $stateParams) {

    $scope.customParameter = $stateParams.customParameter;

}]);
而这一观点是正确的

<ion-view view-title="{{customParameter}}">
    <ion-content>
        <h1>{{customParameter}}</h1>
    </ion-content>
</ion-view>

{{customParameter}}

我用来获取参数的rute错误。我需要在
$state
对象中进一步挖掘,以找到正确的方法

所有参数都在
$state.当前

A以一个示例为例(基于ionic starter侧菜单)

国家:

 .state('app.playlists', {
      url: "/playlists",
      views: {
        'menuContent' :{
          templateUrl: "playlists.html",
          controller: 'PlaylistsCtrl',
          menuContentParam:'menuContentParam'
        },
        viewsParam:'viewsParam'
      },
      stateParam:'stateParam'
    });
控制器

.controller('PlaylistsCtrl', function($scope, $state) {
 $scope.stateParam = $state.current.stateParam;
    $scope.viewsParam = $state.current.views.viewsParam;
    $scope.menuContentParam = $state.current.views.menuContent.menuContentParam;
})
以及以下观点:

<ion-view title="Playlists">
  <ion-nav-buttons side="left">
    <button menu-toggle="left" class="button button-icon icon ion-navicon"></button>
  </ion-nav-buttons>
  <ion-content class="has-header">
    <ion-list>
      <ion-item>{{stateParam}}</ion-item>
      <ion-item>{{viewsParam}}</ion-item>
      <ion-item>{{menuContentParam}}</ion-item>
    </ion-list>
  </ion-content>
</ion-view>

{{stateParam}}
{{viewsParam}}
{{MENUCONTENTPRAM}}

我用来获取参数的rute错误。我需要在
$state
对象中进一步挖掘,以找到正确的方法

所有参数都在
$state.当前

A以一个示例为例(基于ionic starter侧菜单)

国家:

 .state('app.playlists', {
      url: "/playlists",
      views: {
        'menuContent' :{
          templateUrl: "playlists.html",
          controller: 'PlaylistsCtrl',
          menuContentParam:'menuContentParam'
        },
        viewsParam:'viewsParam'
      },
      stateParam:'stateParam'
    });
控制器

.controller('PlaylistsCtrl', function($scope, $state) {
 $scope.stateParam = $state.current.stateParam;
    $scope.viewsParam = $state.current.views.viewsParam;
    $scope.menuContentParam = $state.current.views.menuContent.menuContentParam;
})
以及以下观点:

<ion-view title="Playlists">
  <ion-nav-buttons side="left">
    <button menu-toggle="left" class="button button-icon icon ion-navicon"></button>
  </ion-nav-buttons>
  <ion-content class="has-header">
    <ion-list>
      <ion-item>{{stateParam}}</ion-item>
      <ion-item>{{viewsParam}}</ion-item>
      <ion-item>{{menuContentParam}}</ion-item>
    </ion-list>
  </ion-content>
</ion-view>

{{stateParam}}
{{viewsParam}}
{{MENUCONTENTPRAM}}

实际上,当我看到$stateparms对象时,它是空的,我认为它只适用于url生成的参数。实际上,当我看到$stateparms对象时,它是空的,我认为它只适用于url生成的参数。