Javascript 在selectbox angularjs中捕获正确的选项数据

Javascript 在selectbox angularjs中捕获正确的选项数据,javascript,angularjs,Javascript,Angularjs,我有一个选择框,其中包含来自后端的选项: $http.get($rootScope.appUrl + '/nao/abb/getDataOptionsForAbb/' + $rootScope.abbForm) .then(function(response) { $scope.abbOptions = response.data; //console.log($scope.abbOptions);

我有一个选择框,其中包含来自后端的选项:

  $http.get($rootScope.appUrl + '/nao/abb/getDataOptionsForAbb/' + $rootScope.abbForm)
            .then(function(response) {
                $scope.abbOptions = response.data;
                //console.log($scope.abbOptions);

        });

$scope.onChangeSuperCustomer = function() {
            console.log($scope.selectedSupercustomer);
        }
我正在使用ng选项:

<td><select class="form-control input-sm2" ng-model="selectedSupercustomer" ng-options="item.superkund_id as item.namn for item in abbOptions" ng-change="onChangeSuperCustomer()" ><option value=''>Select</option></select></td>
选择
my console.log($scope.selectedSupercustomer)的输出错误。这不是我在选择框中选择的值。

请尝试以下操作:

<td><select class="form-control input-sm2" ng-model="selectedSupercustomer" ng-options="item.superkund_id as item.namn for item in abbOptions" ng-change="onChangeSuperCustomer(selectedSupercustomer)" ><option value=''>Select</option></select></td>
或:

   $scope.onChangeSuperCustomer = function() {
        console.log($scope.selectedSupercustomer);
    }

@Anant:我不知道如何使用JSFIDLE:/你的代码应该可以用。。在任何情况下,您都可以使用
$scope重现Plunkr的问题。控制器函数内部的selectedSupercustomer
也应该工作,为什么要将scope值作为函数参数传递
$scope.onChangeSuperCustomer = function(selectedSupercustomer) {
        console.log(selectedSupercustomer);
    }
   $scope.onChangeSuperCustomer = function() {
        console.log($scope.selectedSupercustomer);
    }