AngularJS - 在NG-重复未定义的属性过滤器？ [英] AngularJS - filter for undefined properties in ng-repeat?

问题描述

有关我AngularJS项目（V1.2.3），我的路线列表，我试图建立从对象的导航栏。我想要做的是显示与任何对象未定义 isRight 物业于一体的风格，并在财产是另一个定义。

For my AngularJS project (v1.2.3), I have a list of routes and am trying to build a navigation bar from the object. What I want to do is display any object with an undefined isRight property in one style, and where the property is defined in another.

在一个 NG-重复我想用一个未定义的 isRight 属性过滤这些对象。我怎样才能做到这一点的 NG-重复属性中，而无需借助于创建自定义过滤器功能？

In one ng-repeat I would like to filter those objects with an undefined isRight property. How can I accomplish this inside the ng-repeat attribute, without having to resort to creating a custom filter function?

$scope.nav = [
    { path: '/', title: 'Home' },
    { path: '/blog', title: 'Blog' },
    { path: '/about', title: 'About' },
    { path: '/login', title: 'Login', isRight: true }
];

我知道我可以只添加属性 isRight：假来的每个对象，或对右侧和左侧的链接，以及其他类似简单的解决办法单独的导航对象，但我很好奇，如果有与当前的结构来实现这一点的方式，使用的东西线沿线的：

I realize I could just add the attribute isRight: false to each object, or have separate nav objects for right and left side links, and other such simple workarounds, but I am curious if there is a way to accomplish this with the current structure, using something along the lines of:

<li ng-repeat="link in nav | filter:{isRight:undefined}">

这是一个多需要的好奇心，但我AP preciate任何建议。

This is more a curiosity than a need, but I appreciate any suggestions.

推荐答案

您可以否定一个过滤器前pression。因此，而不是处理未定义你可以过滤掉任何东西，其中 isRight 不是（！ ）真。像这样的：

You can negate a filter expression. So instead of dealing with undefined you can just filter out anything where isRight is not (!) true. Like this:

<li ng-repeat="link in nav | filter:{isRight:'!true'} ">

而对于相反的就可以了，当然，这样做：

And for the opposite you can, of course, do:

<li ng-repeat="link in nav | filter:{isRight:'true'} ">

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