从没有 Twig 过滤器的函数中返回原始 HTML [英] Returning raw HTML from a function without a Twig filter

问题描述

我有一个具有返回 HTML 的函数的类.

I have a class that has a function that returns HTML.

class MyParser {
    public function getHTML() {
        return '<a href="#">Hello World</a>';
    }
}

然后在我的 Twig 模板中，我使用 raw 过滤器来输出文字 HTML，而不是让 Twig 为我转义它:

And then in my Twig template, I use the raw filter to output the literal HTML instead of having Twig escape it for me:

{{ myParserInstance.HTML | raw }}

有没有函数的方法(这不是 Twig 过滤器或函数) 返回原始 HTML 并按原样呈现?或者我将如何创建一个 Twig 过滤器或函数来为我无缝地执行此操作?

Is there a way for a function (that's not a Twig Filter or Function) to return raw HTML and render it as such? Or how would I go about creating a Twig Filter or Function that does this seamlessly for me?

例如，我不会想要这样的东西:

For example, I wouldn't want something like:

{{ render(myParserInstance) }}

相反，我只想能够使用 HTML 函数调用.这是可能的，还是我坚持使用 Twig 函数或使用 |原始?

Instead, I would like to just be able to use the HTML function call. Is this at all possible or am I stuck with a Twig function or using | raw?

推荐答案

您可以安全地返回和 Twig_Markup 对象对象...

You can return and Twig_Markup object object in safe way ...

class MyParser {
    public function getHTML() {
         $rawString = '<a href="#">Hello World</a>'
         return new \Twig_Markup( $rawString, 'UTF-8' );
    }
}

或

class MyParser {
    public function getHTML() {
         $rawString = '<a href="#">Hello World</a>'
         return new \Twig\Markup( $rawString, 'UTF-8' );
    }
}

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