Laravel中HTML :: link的URLEncode [英] URLEncode for HTML::link in Laravel

问题描述

我正在Laravel中建立一个博客，似乎无法弄清楚如何使用HTML::link(..)生成转义的URL.例如，我在Blog中有指向不同类别的链接，例如Department News，我试图获得这样的链接-http://localhost/blog/category/department+news.，其中Department News由$post->category生成

I am setting up a blog in Laravel and cannot seem to figure out how to generate escaped URLs with HTML::link(..). For example, I have links to different categories in the Blog such as Department News, I am trying to get a link formatted like so - http://localhost/blog/category/department+news., where Department News is generated by $post->category

我尝试了以下代码，并产生了http://localhost/blog/Department News

I have tried the following code and it produces http://localhost/blog/Department News

{{ HTML::link('admin/blog/category/' . $post->category, $post->category) }}

我该如何转义并生成所需的URL?

How can I escape this and generate the desired URL?

推荐答案

通常，您将使每个帖子类别在数据库中使用slug列作为URL片段，然后使用类似以下内容的东西:

Usually, you'd have each post category use a slug column in the database for the URL fragment, and then use something like:

HTML::link('blog/category/'.$post->category->slug, $post->category->name)

Laravel似乎无法自动对URL的某些部分进行自动编码，因此您必须自己做:

There's appears to be no way with Laravel to automatically encode only certain parts of the URL, so you'd have to do it yourself:

HTML::link(
    'admin/blog/category/'.urlencode(strtolower($post->category)),
    $post->category
)

您可能要考虑使用塞子"方法.您不必将其存储在数据库中，您可以让您的类动态生成它:

You might want to consider using the "slug" approach. You don't necessarily have to store it in the database, you could have your class generate it on the fly:

class Category {
    function slug() {
        return urlencode(strtolower($this->name));
    }
}

我不确定您正在使用什么，但希望您能理解.

I'm not sure what you're working with exactly, but hopefully you get the idea.

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