JQuery方法不能在指令链接函数中工作

Question

我有三个下拉菜单： -

<select clas="col-md-3" ng-model="a.userList"
        ng-options="p as p.users for p in List"
        ng-change="selectValues(a.userList)">
        <option value="">Select</option>
</select>
<select clas="col-md-3" ng-model="a.userCode" id="dropdown1">
        <option>A</option>
        <option>B</option>
        <option>C</option>
        <option>D</option>
</select>
<select clas="col-md-3" ng-model="a.userId" id="dropdown2">
        <option>X</option>
        <option>F</option>
        <option>M</option>
</select>

我想要的是当用户选择AX然后从其他两个下拉列表中分别选择A和X，并从这两个下拉列表中删除所有其他值。

我的指令代码： -

(function () {
    'use strict';
    angular.module('myApp.components')
        .directive('user', user);

    user.$inject = ['$http', '$timeout', 'ApiServices'];

    function user($http, $timeout, ApiServices) {
        return {
            restrict: 'EA',
            scope: {

            },
            link: function (scope, el, attrs) {
                 scope.a = {};

                $('#user').on('hide.bs.modal', function () {
                    scope.a = {};
                });

                $('#user').on('shown.bs.modal', function () {

                    scope.getAllUserList = function(){
                        scope.List = [];
                        ApiServices.getAllUserList().then(function (response) {
                            scope.List = response.data;
                        });
                    }
                        scope.getAllUserList();

                });

                scope.selectValues = function(userList){
                    switch(userList){
                        case "user1" : scope.a.userCode = "A";
                                            $('dropdown1').find('option[value!="'+scope.a.userCode+'"]').remove();
                                            scope.a.userCode = "X";
                                            $('dropdown2').find('option[value!="'+scope.a.userId+'"]').remove();
                                    break;
                        case "user1" : scope.a.userCode = "B";
                                            $('dropdown1').find('option[value!="'+scope.a.userCode+'"]').remove();
                                            scope.mlmScrip.segment = "X";
                                            $('dropdown2').find('option[value!="'+scope.a.userId+'"]').remove();
                                    break;
                        case "user1" : scope.a.userCode = "C";
                                            $('dropdown1').find('option[value!="'+scope.a.userCode+'"]').remove();
                                            scope.mlmScrip.segment = "X";
                                            $('dropdown2').find('option[value!="'+scope.a.userId+'"]').remove();
                                    break;
      //and so on

                    }         
                }


            //Link function ends below
            },
            templateUrl: 'js/components/folder/info.html'
        };
    }

})();

在上面的代码中，我能够选择我想要选择的特定值，但是无法从两个下拉列表中删除所有其他值。我的jquery代码无效。我知道代码是正确的，但我认为我把它弄错了。任何人都可以指出错误。