Php 如何在Slim Framework中呈现404错误的自定义模板

我正在使用Slim 3进行渲染。我正在对渲染器进行如下初始化：

...

$container['view'] = new \Slim\Views\PhpRenderer("../mytemplatesfolder/");

$app = new \Slim\App();

$container = $app->getContainer();
$container['renderer'] = new PhpRenderer("templates");

$app->get('/someroute', function (Request $request, Response $response){
    return $this->renderer->render($response, "/onetemplate.phtml");
});

我可以在我的路线中呈现模板，没有任何问题，如下所示：

...

$container['view'] = new \Slim\Views\PhpRenderer("../mytemplatesfolder/");

$app = new \Slim\App();

$container = $app->getContainer();
$container['renderer'] = new PhpRenderer("templates");

$app->get('/someroute', function (Request $request, Response $response){
    return $this->renderer->render($response, "/onetemplate.phtml");
});

当404错误发生时，如何呈现自定义模板（使用PHP视图，而不是细枝）

---

我发现使用了Twig，但我不知道如何使用PHP视图来实现它。

考虑到您有这样一个composer.json：

{
    "require": {
        "slim/slim": "^3.0",
        "slim/php-view": "^2.2"
    }
}

这里是一个示例应用程序：

<?php
use \Psr\Http\Message\ServerRequestInterface as Request;
use \Psr\Http\Message\ResponseInterface as Response;
use \Slim\Views\PhpRenderer;

require '../vendor/autoload.php';

$app = new \Slim\App;
$container = $app->getContainer();
$container['renderer'] = new PhpRenderer("./templates");
$container['notFoundHandler'] = function ($container) {
    return function ($request, $response) use ($container) {
        return $container['renderer']->render($response, "/404.php");
    };
};

$app->get('/hello/{name}', function (Request $request, Response $response) {
    $name = $request->getAttribute('name');
    $response->getBody()->write("Hello, $name");

    return $response;
});

$app->run();

---

：）

工作起来很有魅力！非常感谢你的朋友！