使用 ui-sref 的站点导航，不可用时如何删除 ui-sref 属性 [英] Site navigation using ui-sref, how to remove ui-sref attribute when not available

问题描述

我使用 ng-repeat 设置了如下导航，效果很好

I've setup navigation as follows, using ng-repeat, which works very well

<a ui-sref="{{link.Route}}" ng-click="clickLink(link)">
    <span class="title"> {{link.Text}} </span><span class="selected"></span>
</a>

但是，我的导航项经常有子链接，这意味着父链接并不是真正的导航链接，它只是用于展开和查看子链接.但有时它是一个链接，没有子链接可显示.

However, my navigation items frequently have sublinks, which means the parent link isn't really a navigation link, it's just used to expand and view the sublinks. But sometime it is a link, and has no sublinks to display.

问题是对于那些特殊情况，当没有可用状态时，我需要一起删除 ui-sref，因为根本不应该有链接.有了它就会抛出错误:无效状态引用"

The problem is for those particular cases, when there is no state available, I need to remove the ui-sref all together, because there shouldn't be a link at all. Having it there is throwing 'Error: Invalid state ref '''

当状态不可用时如何删除 ui-sref?

How do I remove the ui-sref when a state isn't available?

推荐答案

You can use {{}} with expression

You could use {{}} with expression

标记

ui-sref="{{expression ? '.childState' : '.'}}"

. 会创建自己的状态路由，所以当点击它时，它不会重定向到任何地方.

. will create own state route, so while click on it, it will redirect no where.

希望能帮到你，谢谢.

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