Angularjs过滤器不为空 [英] Angularjs filter not null

问题描述

试图筛选出具有一定的属性，不是空的项目因此，对于：

Trying to filter out items with a certain property that is not null So for:

var details = [{name:'Bill', shortDescription: null}, {name:'Sally', shortDescription: 'A girl'}]

我想只显示一华里;在一个突破口。这是我没有成功尝试

I would like to only show one li; the one for sally. This is what I have tried with no success

<ul>
<li ng-repeat="detail in details | filter:{shortDescription:'!'}">
<p>{{detail.shortDescription}}</p>
</li>
</ul>

任何想法，我怎么能做到这一点，而无需创建自定义过滤器？或者，即使如此，什么自定义过滤器会是什么样子？

Any idea how I can do this without creating a custom filter? Or even so, what the custom filter would look like?

推荐答案

据<一个href=\"https://github.com/angular/angular.js/issues/11573\">https://github.com/angular/angular.js/issues/11573对角> = 1.4，则建议使用''它匹配任何原始除空/未定义。

According to https://github.com/angular/angular.js/issues/11573 for Angular >= 1.4, the recommendation is to use '' which matches any primitive except null/undefined.

<ul>
    <li ng-repeat="detail in details | filter:{shortDescription: ''}">
        <p>{{detail.shortDescription}}</p>
    </li>
</ul>

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