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我想使用一条路线来获取完整的集合,如果有的话,还有一个过滤的集合。

所以我的路线:

$app->get("/companies", \App\Handler\CompanyPageHandler::class, 'companies');

我的这条路线的处理程序:

use App\Entity\Company;
use App\Entity\ExposeableCollection;
use Psr\Http\Message\ResponseInterface;
use Psr\Http\Message\ServerRequestInterface;

class CompanyPageHandler extends AbstractHandler
{
    public function handle(ServerRequestInterface $request): ResponseInterface
    {
        $categories = new ExposeableCollection();
        foreach (['test', 'test1', 'test3'] as $name) {
            $category = new Company();
            $category->setName($name);

            $categories->addToCollection($category);
        }

        return $this->publish($categories);
    }
}

获得这条路线/companies时,我得到了预期的收藏

[{"name":"test"},{"name":"test1"},{"name":"test3"}]

所以现在我改变路线

$app->get("/companies[/:search]", \App\Handler\CompanyPageHandler::class, 'companies');

当我浏览到/companies. 但是如果我尝试可选参数/companies/test1,那么我会得到一个错误

无法获取http://localhost:8080/companies/test1

我的作曲家需要部分:

"require": {
    "php": "^7.1",
    "zendframework/zend-component-installer": "^2.1.1",
    "zendframework/zend-config-aggregator": "^1.0",
    "zendframework/zend-diactoros": "^1.7.1 || ^2.0",
    "zendframework/zend-expressive": "^3.0.1",
    "zendframework/zend-expressive-helpers": "^5.0",
    "zendframework/zend-stdlib": "^3.1",
    "zendframework/zend-servicemanager": "^3.3",
    "zendframework/zend-expressive-fastroute": "^3.0"
},

在 Zend Framework 2 和 Symfony4 中,这个路由定义工作得很好。所以我很困惑。为什么我的可选参数不起作用?

4

1 回答 1

3

那是因为您使用的是https://github.com/nikic/FastRoute路由器,正确的语法是:

$app->get("/companies[/{search}]", \App\Handler\CompanyPageHandler::class, 'companies');

或者更严格并验证搜索参数,如下所示:

$app->get("/companies[/{search:[\w\d]+}]", \App\Handler\CompanyPageHandler::class, 'companies');
于 2019-01-16T08:10:39.653 回答