如何使用 guzzle 下载文件并将其转发到响应而不在本地保存

Question

I'm using Laravel and I want to get files from api (with providing token and some headers) then forward it directly to the response without saving the file locally.

Currently I'm doing it like this:

$response = $client->get($requestUrl, [
        'query' => [
            'hash' => $hash
        ],
        'headers' => [
            '_token_' => $oauthConfig['api_token'],
            '_token_issuer_' => 1
        ],
        'stream' => true,
    ]);

    $type = $response->getHeader('Content-Type')[0];
    $body = $response->getBody();

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

    $response->headers->set('Content-Type', $type);

    return $response;

But I'm getting this exception:

LogicException
The content cannot be set on a StreamedResponse instance.

Any help / guidance is greatly appreciated. Thanks.

Answer 1

It seems the right thing to do is

http://docs.guzzlephp.org/en/stable/psr7.html

use GuzzleHttp\Stream\Stream;



$response = $client->get($requestUrl, [
        'query' => [
            'hash' => $hash
        ],
        'headers' => [
            '_token_' => $oauthConfig['api_token'],
            '_token_issuer_' => 1
        ],
        'stream' => true,
    ]);

$result = '';
while(!$response->getBody()->eof()){
   $result .= $response->getBody()->read(8192);
}

$filename = 'test.gif';
return response()->streamDownload(function () use ($result) {
    echo $result;
}, $filename);