如何将原始html传递给Play框架视图？ [英] How to pass raw html to Play framework view?

问题描述

我正在尝试将一个简单的URL传递给play框架应用程序中的视图，但是当作为字符串传递时，url中的& 将更改为& amp; 导致网址无效。

I'm trying to pass a simple URL to a view within a play framework app, however when passed as a string, the & in the url is changed to & which causes the URL to not work.

如果我将参数更改为Html，即 @（url：Srting）到 @（url：Html），当我尝试传递时出现错误url as string to view.render（） method。

If I change the argument to Html, i.e @(url: Srting) to @(url: Html), then I get an error when I try to pass the url as string to view.render() method.

如何将url转换为Html并传递给它那个？

How can I convert the url into Html and pass it as that?

推荐答案

为了防止视图上的动态内容发生默认转义，你需要包装 String with @Html（String） function：

To prevent the default escaping that happens for dynamic content on views you need to wrap the String with @Html(String) function:

查看：

@(url: String)
<div class="myLink">
   Go to: @Html(url) <br>
   not to: @url
</div>

控制器：

public static Result displayLink(){
   return ok(view.render("<a href='http://stackoverflow.com/'>Stack Overflow</a>"));
}

参见模板引擎页面上的更多信息（特别是最底部的Escaping部分）。

See The template engine page on the documentation for more info (specifically the "Escaping" section at the very bottom).

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