Question

我有一个获取文件并将其返回给用户的Web服务（基于Symfony）。自从我用curl做这个。

我刚发现guzzlehttp，看起来很棒。但是，我不知道如何使用guzzle执行此操作而不将下载的文件（xml或txt）保存到本地文件，从文件系统读取它并将其返回给用户。我想这样做而不保存到文件系统。

Answer 1

public function streamAction()
{
     $response = $client->request(
        'GET', 'http://httpbin.org/stream-bytes/1024', ['stream' => true]
     );

     $body = $response->getBody();

     $response = new StreamedResponse(function() use ($body) {
         while (!$body->eof()) {
             echo $body->read(1024);
         }
     });

     $response->headers->set('Content-Type', 'text/xml');

     return $response;
}

Answer 2

$response = $client->request('GET', 'http://example.com/file', ['stream' => true]);
        $stream = $response->getBody();
        $content = new StreamedResponse(function () use ($stream) {
            /** @var StreamInterface $stream */
            while ($binary = $stream->read(1024)) {
                echo $binary;
                ob_flush();
                flush();
            }
        }, $response->getStatusCode(), [
            'Content-Type' => $response->getHeaderLine('Content-Type'),
            'Content-Length' => $response->getHeaderLine('Content-Length'),
            'Content-Disposition' => $response->getHeaderLine('Content-Disposition')
        ]);

        return $this->renderResponse($content->send());

Guzzle获取文件并转发它

2 个答案: