Laravel Guzzle不起作用,但Curl确实如此

时间:2015-12-01 23:54:32

标签: php laravel curl guzzle

我正在使用Guzzle来处理外部API。

我这样使用它:

$client = new Client;
$request = $client->request('GET', 'https://api.callrail.com/v1/companies.json', [
    'headers' => [
        'Authorization' => 'Token token="my_api_string"'
    ]
]);

return dd($request);

这是输出

Stream {#255 ▼
  -stream: stream resource @280 ▶}
  -size: null
  -seekable: true
  -readable: true
  -writable: true
  -uri: "php://temp"
  -customMetadata: []
}

但是当我像这样使用 curl

$api_key = 'my_api_string';

$ch = curl_init($api_url);

curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array("Authorization: Token token=\"{$api_key}\""));

$json_data = curl_exec($ch);

return dd($json_data);

输出看起来像预期的那样

{"page":1,"per_page":100,"total_pages":1,"total_records":11,
....
....

我对Guzzle做错了什么?

1 个答案:

答案 0 :(得分:7)

您已正确设置Guzzle请求,只需在检索后使用$request执行更多操作。

在您的请求后添加此行:

$result = json_decode($request->getBody());

将其添加到您的代码中,它将如下所示:

$client = new Client;
$request = $client->request('GET', 'https://api.callrail.com/v1/companies.json', [
    'headers' => [
        'Authorization' => 'Token token="my_api_string"'
    ]
]);

$result = json_decode($request->getBody());

return dd($result);