AngularJS:过滤没有特定值的 ng-options [英] AngularJS: Filter ng-options not having specific values

问题描述

我想像这样过滤一个选择:

I want to filter on a select like so :

<select ng-model="test" ng-options="c as c.label group by c.type for c in columns
| filter:{c.type:'!field'} | filter:{c.type:'!map'}"></select>

添加列模型:

Columns = [
{
    name: "name",
    label: "Label",
    info: "Information displayed in help",
    type: "type",
    view: "html template",
    style: "min-width: 10em;",
    show: true
},
{
 ...
}
];

列用于多种用途并优化我的代码我需要它也在选择中，但没有类型为字段"或地图"的条目

Columns is used for several things and to optimize my code I need it to be also in a Select, but without the entries whose type are 'field' nor 'map'

然而，我可以从所有内容中进行选择，甚至是字段"和地图"类型的条目.有没有干净的方法来做到这一点?

Yet, I get to choose from everything, even the entries which types are 'field' and 'map'. Is there a clean way to do it ?

推荐答案

AngularJS NOT Filter

<select ng-model="test" ng-options="c as c.label group by c.type for c in columns
    | filter:{ type : '!field' }
    | filter:{ type : '!map' }">
</select>

小提琴

来自文档:

From the docs:

"...可以通过在字符串前加上 ! 来否定谓词."

"模式对象可用于过滤数组包含的对象的特定属性.例如，{name:M", phone:1"} 谓词将返回具有包含M"的属性名称的项的数组.和财产电话包含1"...

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