存在某些查询字符串时定义Laravel路由 [英] Defining Laravel routes when certain query strings are present

问题描述

说我希望能够在将?s=附加到当前URL的任何页面上提交搜索表单，但返回SERP:https://example.com/my-page?s=foobar. (我见过一些站点这样做而不是指向/search?s=.*-不是我这样做的方法，但这说明了我的问题.)

Say I want to be able to submit a search form on any page that will append a ?s= to the current url but return a SERP: https://example.com/my-page?s=foobar. (I've seen a few sites do this instead of pointing to /search?s=.* - not the way I'd do it, but it illustrates my question.)

在我的Laravel web.php路线中，除了可能使用不规则的正则表达式外，目前还有其他方法可以注册吗?

In my Laravel web.php routes, is there currently a way to register this other than maybe using a wonky regex?

<?php
Route::get('.+?\?.+?\bs={search}', 'SearchController@search');
// This regex is probably wrong, but you get what I was going for,
// and that kinda highlights why this is not an ideal way to do it
?>

例如，当您要捕获多个$ _GET参数时，这变得更难看:

This gets uglier when, say, you want to capture multiple $_GET params:

https://example.com/my-page?s=foobar&dept=clothing

我没有在Laravel文档中找到任何可让您在Route $uri参数上定义查询字符串选项的内容.推荐的选项是仅使用漂亮的URL(https://example.com/search/foobar)，但是肯定有保留查询字符串的有效用例.

I haven't found anything in Laravel docs that let you define query string options on the Route $uri parameter. The recommended option is to just use pretty URLs (https://example.com/search/foobar), but there are definitely valid use cases for keeping query strings.

推荐答案

1:创建中间件( https ://laravel.com/docs/5.7/middleware ) 2:使用$ request-> query('s')验证传入请求 3:如果验证成功，则返回下一个，否则重定向到任何地方或显示错误响应

1: Create a middleware (https://laravel.com/docs/5.7/middleware) 2: Validate incoming request using $request->query('s') 3: If validation successfull return to next else redirect to wherever or display an error response

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