使用in ng-options排除值

Question

我一直在尝试找出如何从使用过滤器的ng-options生成的数组中排除一个值。

下面的代码片段生成的数组生成了下面的数组["publicExtra","public","private"]，我想从选项中排除"public“。提前谢谢。

<select ng-model="elb_instance.use_vpc"
                    ng-options="('Subnet: ' + type) for type in elb_instance.vpc_names_with_subnet_types_for_elb[elb_instance.vpc_name]"
                    ng-disabled="!elb_instance['new_record?']"
                    ng-show="elb_instance.vpc_name"
                    id="use_vpc"
                    class="input-medium">

Answer 1

您可以很容易地使用filter和在搜索字符串之前使用!来否定谓词，如下所示：

ng-options="('Subnet: ' + type) for type in types | filter: '!public'">

但请注意，这会忽略public和publicExtra，因为基本筛选器不会进行精确匹配。为此，我们还需要传递comparator的true，如下所示：

ng-options="('Subnet: ' + type) for type in types | filter: '!public' : true">

​

var app = angular.module('myApp', []);
app.controller('AppCtrl', function($scope) {
  $scope.selected = "";
  $scope.types = ["publicExtra","public","private"];
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.7.5/angular.min.js"></script>
<section ng-app="myApp">
  <div ng-controller="AppCtrl">
    <select ng-model="selected" 
      ng-options="('Subnet: ' + type) for type in types | filter: '!public' : true">
    </select>
  </div>
</section>

​