Angularjs 过滤器不为空 [英] Angularjs filter not null
本文介绍了Angularjs 过滤器不为空的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
试图过滤掉具有不为空的特定属性的项目因此:
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'}]
我只想展示一里;莎莉的那个.这是我尝试过但没有成功的方法
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>
知道如何在不创建自定义过滤器的情况下执行此操作吗?或者即便如此,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
根据 https://github.com/angular/angular.js/issues/11573 对于 Angular >= 1.4,建议使用 '' 匹配除 null/undefined 之外的任何原语.
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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