我需要将$scope从我的父窗口传递到新窗口.我在我的
javascript函数中为新窗口编写了$window.open().但是在新窗口中,我没有获得$scope.下面我给出了代码.
1)JS文件
$scope.mainPageNewsArchive = data.newsArchiveList;
var myWindow1 = $window.open("resources/partials/NewFile.html","News Archive",'toolbar=0,scrollbars=1,location=0,status=0,menubar=0,resizable=1,width=650,height=550,left = 300,top=100');
2)NewFile.html
<div>
<table border="0" cellspacing=0 cellpadding=2 width="100%" id="tkm">
<tbody>
<tr x-ng-repeat="newsArchive in mainPageNewsArchive track by $index">
<td align=left >{{newsArchive.createDate}}</td>
<td> </td>
<td align=left><a href="" ng-click="showNewsItem(newsArchive.id)" title="Click for more detail">{{newsArchive.title}}</a>
</td>
</tr>
</tbody>
</table>
</div>
解决方法
由主窗口打开的窗口在架构上将无法与其父窗口共享相同的范围.因为角度范围是根据DOM元素定义的.但是您可以使用window.open()方法返回的窗口对象来传递数据
angular.module('originalModule').service('popupService',function(someOtherDataService) {
var popupWindow = window.open('popupWindow.html');
popupWindow.mySharedData = someOtherDataService.importantData;
});
一旦初始化了辅助“弹出”角度应用程序,它就可以通过读取window.mySharedData来引用共享数据.
要么
您可以引用新生成的窗口的rootscope,然后通过父窗口的子窗口范围传递事件.
家长：
var newWindowRef = $window.open("","New Window","width=1280,height=890,resizable=1");
var newWindowRootScope = newWindowRef.angular.element("#MY_ROOT_APP_ELEMENT_ID").scope();
newWindowRootScope.$broadcast("INTER_WINDOW_DATA_TRANSFER",{someData : "I'm data"});
儿童：
$rootScope.$on("INTER_WINDOW_DATA_TRANSFER",function (data,args) {
console.log("DATA RECEIVED: " + args.someData);
});
