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我试图打开一个whats'app链接,我得到这样的错误:离子2:获得不安全网址错误
WARNING: sanitizing unsafe URL value SafeValue must use
(see http://g.co/ng/security#xss)
这里是我的代码:
<a href="{{surl}}">
<img src="assets/imgs/whatsapp.png" height=35px/></a>
和TS文件这里是什么SURL认为:
this.url = 'whatsapp://send?text=Hello World!&phone=+966'+this.phone
this.surl = this.dom.bypassSecurityTrustUrl(this.url);
,你可以告诉的问题是,我传递一个变量的url,但科尔多瓦不会信任它! 任何想法如何解决这个问题?
看看这个主题https://stackoverflow.com/questions/38593515/angular2-warning-sanitizing-unsafe-style-value-urlsafevalue-must-use-proper –
@ChristianBenseler他们说我应该包装整个网址,这正是我所做的:( –
如果你尝试把它直接放在视图中,如下所示: '' Where 'url' is set in the component code like 'this.url = 'whatsapp://send?text=Hello World!&phone=+966' + this.phone;' – sebaferreras