如何在angularJs上使用onClick？

Question

我写了条件

onclick="window.open({{video_call_url}}, '_system', 'location=yes'); return false;"

此处video_call_url在myController中定义为$scope.video_call_url = 'http://www.google.com/';

但是当我点击按钮时，我收到错误video_call_url is not defined.

Answer 1

您可以在控制器中执行逻辑：

function myController($scope, $window) {
    $scope.openVideoCallUrl = function() {
        $window.open($scope.video_call_url, "_system", "location=yes");
        return false;
    }
}

在你看来

<a ng-click="openVideoCallUrl()">Open!</a>

Answer 2

您可以使用ng-click，而不是使用onclick

ng-click="open(video_call_url)"

$scope.open = function(url) {
  //inject $window inside controller.
  $window.open(url, '_system', 'location=yes');
  return false;
}