【发布时间】:2014-11-12 10:37:24
【问题描述】:
我创建了控制器来打开 $modal 。现在如何制作打开 $modal 的常用方法。我还需要将一些对象传递给 $modal 模板。
下面我创建了控制器,我可以将其作为常用方法。但是,我需要在打开 $modal 时传递对象数据 [不使用工厂]
HTML:
<a class="view-all" href="#" ng-controller="MoreDetailControllerbasic"
ng-click="get_form('login')">+ add more</a>
Javascript:
viewmoreApp.controller('MoreDetailControllerbasic', ['$scope', '$http', '$compile', '$modal',
function($scope, $http, $compile, $modal) {
$scope.data="i am trying to send this Dataaaaaaaaaaaaaaaaaaaaaa";
$scope.get_form = function(form) {
debugger;
$scope.form = form;
templateUrl = "/form/" + form + ' ';
modalInstance = $modal.open({
templateUrl : '/view_more',
controller : 'MoreDetailController',
//backdrop : 'static'
resolve: {
GalleryData: function () {
return $scope.data;
},
}
});
modalInstance.result.then(function() {
//Get triggers when modal is closed
}, function() {
//gets triggers when modal is dismissed.
});
};
}]);
viewmoreApp.controller('MoreDetailController', ['$scope', '$http', '$compile', '$modalInstance','GalleryData',
function($scope, $http, $compile, $modalInstance,GalleryData) {
$scope.data_list=GalleryData;
$scope.greetings="Welcome to wiki ";
$scope.closeModal = function() { debugger;
$modalInstance.dismiss('cancel');
};
}]);
【问题讨论】:
-
也许您需要
scope属性? -
@dfsq 我正在为此目的制定指令
标签: javascript angularjs angular-ui-bootstrap