Laravel URL helper：如何使用查询参数和哈希生成完美的 URL
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Laravel URL helper: How to generate a perfect URL with query parameters and hash
提问by Debiprasad
Suppose the route is like this:
假设路线是这样的：
Route::get('messages/{messages}', ['as' => 'messages.show', 'uses' => 'MessagesController@show']);
So, when we will create an URL using URL helper of Laravel,
所以，当我们使用 Laravel 的 URL helper 创建一个 URL 时，
{{ route('messages.show', 12) }}
will display example.com/messages/12.
将显示example.com/messages/12.
This is correct. Let's have some hash in the URL.
这是对的。让我们在 URL 中添加一些哈希值。
{{ route('messages.show', [12, '#reply_23']) }}
This will display example.com/messages/12#reply_23.
这将显示example.com/messages/12#reply_23.
This looks good. Now let's add some query strings instead of the hash.
这看起来不错。现在让我们添加一些查询字符串而不是哈希。
{{ route('messages.show', [12, 'ref=email']) }}
This will display example.com/messages/12?ref=email. This looks cool.
这将显示example.com/messages/12?ref=email. 这看起来很酷。
Now add both query string and hash.
现在添加查询字符串和哈希。
{{ route('messages.show', [12, 'ref=email', '#reply_23']) }}
Now this will display example.com/messages/12?ref=email&#reply_23. This looks little ugly because of the &in the URL. However it's not creating a lot of problem, I would like to get a clean URL like example.com/messages/12?ref=email#reply_23. Is there a way to get rid of the unnecessary &in the URL?
现在这将显示example.com/messages/12?ref=email&#reply_23. 由于&URL中的，这看起来有点难看。但是它并没有造成很多问题，我想获得一个干净的 URL，如example.com/messages/12?ref=email#reply_23. 有没有办法去掉&URL 中不必要的内容？
Edit:There is a workaround, but I am looking for a solid answer.
编辑：有一种解决方法，但我正在寻找可靠的答案。
<a href="{{ route('messages.show', [12, 'ref=email']) }}#reply_23">Link to view on website</a>
回答by Bogdan
The Laravel UrlGeneratorclass does not support specifying the #fragmentpart of the URL. The code responsible for building the URL is the following, and you can see it just appends the query string parameters and nothing else:
LaravelUrlGenerator类不支持指定#fragmentURL的一部分。负责构建 URL 的代码如下，您可以看到它只附加了查询字符串参数，没有其他任何内容：
$uri = strtr(rawurlencode($this->trimUrl(
$root = $this->replaceRoot($route, $domain, $parameters),
$this->replaceRouteParameters($route->uri(), $parameters)
)), $this->dontEncode).$this->getRouteQueryString($parameters);
A quick test of your code reveals that the second example you posted:
对您的代码的快速测试显示您发布的第二个示例：
{{ route('messages.show', [12, '#reply_23']) }}
Actually generates:
实际生成：
/messages/12?#reply_23 // notice the "?" before "#reply_23"
So it treats #reply_23as a parameter rather than as a fragment.
所以它被视为#reply_23一个参数而不是一个片段。
An alternative to this shortcoming would be to write a custom helper function that allows to pass the fragment as a third parameter. You could create a file app/helpers.phpwith your custom function:
这个缺点的替代方法是编写一个自定义帮助函数，允许将片段作为第三个参数传递。您可以app/helpers.php使用自定义函数创建一个文件：
function route_with_fragment($name, $parameters = array(), $fragment = '', $absolute = true, $route = null)
{
return route($name, $parameters, $absolute, $route) . $fragment;
}
Then add the following line at the end of your app/start/global.phpfile:
然后在app/start/global.php文件末尾添加以下行：
require app_path().'/helpers.php';
You can then use it like this:
然后你可以像这样使用它：
{{ route_with_fragment('messages.show', [12, 'ref=email'], '#reply_23') }}
Of course you can name the function whatever you want, if you feel the name I gave it is too long.
当然你可以随意命名函数，如果你觉得我给它起的名字太长了。
