php laravel - 如何将控制器输出传递给 iframe 源代码
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php laravel - how to pass controller output to view as iframe source
提问by Dython
I have a controller which gives output as URL , now I want to pass the output url to view and below is the code.But when I load the view I found that iframe is not loading and the url is showing in the body of the html.
我有一个控制器，它以 URL 的形式提供输出，现在我想将输出 url 传递给视图，下面是代码。但是当我加载视图时，我发现 iframe 没有加载并且 url 显示在 html 的正文中.
inside controller this is how I am returning the output
在控制器内部，这就是我返回输出的方式
return view('ifr',['name' => $url]);
And this is my view code
这是我的视图代码
<iframe src={{$name}}></iframe>
Could you please help me with the correct syntax
你能帮我正确的语法吗
回答by Andrius Rimkus
Instead of
<iframe src={{$name}}></iframe>
代替
<iframe src={{$name}}></iframe>
Try
<iframe src={!! $name !!}></iframe>
尝试
<iframe src={!! $name !!}></iframe>
More at: https://laravel.com/docs/5.4/blade/ Displaying Unescaped Data
更多信息：https: //laravel.com/docs/5.4/blade/ 显示未转义数据
回答by Rahul Chauhan
Pass the iframe src in single quotes
在单引号中传递 iframe src
Like
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$src = '"https://www.facebook.com/plugins/post.php?href=https%3A%2F%2Fwww.facebook.com%2Fhierensolanki%2Fposts%2F1264241026994313&width=500" width="500" height="608" style="border:none;overflow:hidden" scrolling="no" frameborder="0" allowTransparency="true"';
return view('welcome',compact('src'));
And in your view file.
并在您的视图文件中。
<iframe src=<?php echo $src; ?>></iframe>
It will work.
它会起作用。
