iScroll in angularjs指令

我正在尝试使用角度指令实现水平iScroll。iScroll in angularjs指令

这是我的指令代码。

link: function(scope, elem, attrs) { 
     scope.winWidth = window.innerWidth; 
     scope.iScrollWidth = scope.winWidth *8.5 +'px'; 
     jQuery(angular.element(attrs.thelist)).css({'width':scope.iScrollWidth}); 

     var k_videoScroll = new iScroll(angular.element(attrs.scrollwrap),{ 
      snap: true, 
      bounce: false, 
      checkDOMChanges: false, 
      momentum: false, 
      hScrollbar: false, 
      vScrollbar: false, 
      overflow: false 
     });

我有id为我的HTML页面相关的一个div标签的attrs.scrollwrap

<div id="videolist" class="videoScrollWrap" options="#thelist" wrapper="#deal-wrap" scrollwrap="#videoReviewsWrap">  
    <div id="videoReviewsWrap"> 
     <ul id="thelist"> 
     <li id='deal-wrap' ng-repeat="video in videoList" videoList='video'></li> 
     </ul> 
    </div> 
</div>

我坚持是“无法设置属性‘’未定义的”溢出的问题。

TypeError: Cannot set property 'overflow' of undefined 
at Object.iScroll (http://localhost/swordfish/web/js/lib/iscroll.js:89:31) 
at setScroll (http://localhost/swordfish/web/js/directives/d-product-details.js:395:33) 
at link (http://localhost/swordfish/web/js/directives/d-product-details.js:405:13) 
at o (http://localhost/swordfish/web/js/lib/angular/angular.min.js:42:187) 
at e (http://localhost/swordfish/web/js/lib/angular/angular.min.js:38:28) 
at http://localhost/swordfish/web/js/lib/angular/angular.min.js:37:118 
at <error: illegal access> 
at Object.e.$broadcast (http://localhost/swordfish/web/js/lib/angular/angular.min.js:88:517) 
at http://localhost/swordfish/web/js/lib/angular/angular.min.js:81:85 
at i (http://localhost/swordfish/web/js/lib/angular/angular.min.js:76:207) <div id="videolist" class="videoScrollWrap ng-isolate-scope ng-scope" options="#thelist" wrapper="#deal-wrap" scrollwrap="#videoReviewsWrap"> angular.min.js:60

（匿名函数）

可能是什么问题？看来定义新的iScroll的语法是不正确的。什么是正确的方法来做到这一点？

在此先感谢！

来源

2013-04-24 ayimos

你可以发表评论作为答案，这样的问题将不会出现在未解答的问题列表，并变得更有助于他人？:) – 2013-04-27 20:36:52

我想出了这个问题。这是因为当我尝试获取（attrs.scrollwrap）时，它实际上是重新调整对象。但iScroll库试图直接获取对象子元素，并设置该对象的style属性。一旦我尝试使用（attrs.scrollwrap [0]）访问div元素，问题就解决了。

来源

2013-04-29 14:41:13 ayimos

iScroll in angularjs指令

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