Symfony - 如何执行另一个 php 脚本并获取结果？

发布于 2025-01-02 17:32:54 字数 899 浏览 4 评论 0原文

我有一个脚本，它生成一些图形并返回它作为结果，它还缓存这个图形等。我

使用 symfony2 并且在控制器中我需要调用这个脚本，现在我使用这个函数来调用我的 php 脚本

        private function http_post($url, $data)
            {
                $data_url = http_build_query ($data);
                $data_len = strlen ($data_url);

                return array ('content'=>file_get_contents ($url, false, stream_context_create (array ('http'=>array ('method'=>'POST'
                        , 'header'=>"Connection: close\r\nContent-Length: $data_len\r\nContent-Type: application/x-www-form-urlencoded\r\n"
                        , 'content'=>$data_url
                        ))))
                    , 'headers'=>$http_response_header
                    );
            }

：认为这种方式不是最好的，而且我记得 file_get_contents 非常慢？所以我的问题是：通过“http_post”将 POST 发送到该脚本是好方法吗？如果没有的话，那什么会更好呢？

编辑：我不想在控制器中包含此脚本，因此请不要包含解决方案:)。

原文

I have got script which generates some graphic and returns it as a result, also it caches this graphics etc.

I use symfony2 and in a controller I need to invoke this script, for now I use this function, to invoke my php script:

        private function http_post($url, $data)
            {
                $data_url = http_build_query ($data);
                $data_len = strlen ($data_url);

                return array ('content'=>file_get_contents ($url, false, stream_context_create (array ('http'=>array ('method'=>'POST'
                        , 'header'=>"Connection: close\r\nContent-Length: $data_len\r\nContent-Type: application/x-www-form-urlencoded\r\n"
                        , 'content'=>$data_url
                        ))))
                    , 'headers'=>$http_response_header
                    );
            }

I think this way isn't the best and as I remember well file_get_contents is pretty slow?
So my question: is it good way to send POST to this script via "http_post"? If not, than what would be better?

EDIT: I don't want to have this script in a controller, so no include solutions please :).

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李不 2025-01-09 17:32:54

如果您希望将图像生成器实现为控制器，则只需照常返回响应：return new Response($ generated_image);。

所以这可能是您的控制器：

namespace Acme\MyBundle\Controller;
use Symfony\Component\HttpFoundation\Response;

class ImageGeneratorController
{
    public function generateAction($parameters)
    {
      //Generate an image using parameters and store it in $image
      $image = ....

      return new Response($image);
    }
}

然后，您可以从任何控制器调用图像生成器< /a> 使用 forward()

public function indexActionInAnotherController($name)
{
    $response = $this->forward('AcmeMyBundle:ImageGenerator:generate', array(
        'name'  => $name,
        'color' => 'green'
    ));

    // further modify the response or return it directly

    return $response;
}

If you wish to implement the image generator as a controller, you would just return the response as normal: return new Response($generated_image);.

So this could be your controller:

namespace Acme\MyBundle\Controller;
use Symfony\Component\HttpFoundation\Response;

class ImageGeneratorController
{
    public function generateAction($parameters)
    {
      //Generate an image using parameters and store it in $image
      $image = ....

      return new Response($image);
    }
}

Then, you can call your image generator from any controller using forward()

public function indexActionInAnotherController($name)
{
    $response = $this->forward('AcmeMyBundle:ImageGenerator:generate', array(
        'name'  => $name,
        'color' => 'green'
    ));

    // further modify the response or return it directly

    return $response;
}

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