在 Laravel 中更改查询参数时重定向到当前 URL
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Redirect to current URL while changing a query parameter in Laravel
提问by Martti Laine
Is there a built-in way to do something like this?
有没有内置的方法来做这样的事情？
Let's say I have a search-page that has a few parameters in the URL:
假设我有一个搜索页面，其 URL 中有几个参数：
example.com/search?term=foo&type=user
A link on that page would redirect to an URL where typeis link. I'm looking for a method to do this without manually constructing the URL.
该页面上的链接将重定向到一个 URL，其中type是link。我正在寻找一种无需手动构建 URL 的方法来执行此操作。
Edit:
I could build the URL manually like so:
编辑：
我可以像这样手动构建 URL：
$qs = http_build_query(array(
'term' => Input::get('term'),
'type' => Input::get('type')
));
$url = URL::to('search?'.$qs);
However, what I wanted to know is if there is a nicer, built-in way of doing this in Laravel, because the code gets messier when I want to change one of those values.
但是，我想知道的是，在 Laravel 中是否有更好的内置方法来执行此操作，因为当我想更改其中一个值时，代码会变得更加混乱。
Giving the URL generator a second argument ($parameters) adds them to the URL as segments, not in the query string.
为 URL 生成器提供第二个参数 ( $parameters) 将它们作为段添加到 URL 中，而不是在查询字符串中。
回答by clarkf
You can use the URL Generator to accomplish this. Assuming that search is a named route:
您可以使用 URL Generator 来完成此操作。假设搜索是一个命名路由：
$queryToAdd = array('type' => 'user');
$currentQuery = Input::query();
// Merge our new query parameters into the current query string
$query = array_merge($queryToAdd, $currentQuery);
// Redirect to our route with the new query string
return Redirect::route('search', $query);
Laravel will take the positional parameters out of the passed array (which doesn't seem to apply to this scenario), and append the rest as a query string to the generated URL.
Laravel 将从传递的数组中取出位置参数（这似乎不适用于这种情况），并将其余部分作为查询字符串附加到生成的 URL 中。
See: URLGenerator::route(),
URLGenerator::replaceRouteParameters()URLGenerator::getRouteQueryString()
见：URLGenerator::route()，
URLGenerator::replaceRouteParameters()URLGenerator::getRouteQueryString()
回答by Bouke Versteegh
I prefer native PHP array merging to override some parameters:
我更喜欢原生 PHP 数组合并来覆盖一些参数：
['type' => 'link'] + \Request::all()
To add or override the typeparameter and remove another the term:
要添加或覆盖type参数并删除另一个参数term：
['type' => 'link'] + \Request::except('term')
Usage when generating routes:
生成路由时的用法：
route('movie::category.show', ['type' => 'link'] + \Request::all())
回答by TonyArra
You can do it with Laravel's URLGenerator
URL::route('search', array(
'term' => Input::get('term'),
'link' => Input::get('type')
));
Edit: be sure to name the route in your routes.php file:
编辑：一定要在你的 routes.php 文件中命名路由：
Route::get('search', array('as' => 'search'));
That will work even if you're using a Route::controller()
即使您使用的是 Route::controller()
回答by J.C. Gras
From Laravel documentation:
if your route has parameters, you may pass them as the second argument to the route method.
如果你的路由有参数，你可以将它们作为第二个参数传递给路由方法。
In this case, for return an URI like example.com/search?term=foo&type=user, you can use redirect function like this:
在这种情况下，要返回像example.com/search?term=foo&type=user这样的 URI ，您可以使用如下重定向功能：
return redirect()->route('search', ['term' => 'foo', 'type' => 'user']);
回答by daylerees
The Input component should also contain query parameters.
Input 组件还应包含查询参数。
i.e Input::get('foo');
IE Input::get('foo');
