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php - 如何在 php guzzle 上获取响应正文?

转载 作者:行者123 更新时间:2023-12-01 11:30:51 25 4
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use GuzzleHttp\Client;
$client = new Client();
$response = $client->post('http://httpbin.org/post', array());

我怎样才能得到 body react ?

getBody 未返回 响应体
echo '<pre>' . print_r($response->getBody(), true) . '</pre>';

GuzzleHttp\Psr7\Stream 对象
(
[流:GuzzleHttp\Psr7\Stream:private] => 资源 ID #80
[大小:GuzzleHttp\Psr7\Stream:private] =>
[可搜索:GuzzleHttp\Psr7\Stream:private] => 1
[可读:GuzzleHttp\Psr7\Stream:private] => 1
[可写:GuzzleHttp\Psr7\Stream:private] => 1
[uri:GuzzleHttp\Psr7\Stream:private] => php://temp
[customMetadata:GuzzleHttp\Psr7\Stream:private] => 数组
(
)

)

打印体 react 如何?

最佳答案

您可以使用 getContents拉响应正文的方法。

$response = $this->client->get("url_path", [
'headers' => ['Authorization' => 'Bearer ' . $my_token]
]);
$response_body = $response->getBody()->getContents();
print_r($response_body);

当您发出 make guzzle 请求时,您通常会将其放入 try catch 块中。此外,您还需要将该响应解码为 JSON,以便您可以像使用对象一样使用它。这是如何做到这一点的示例。在这种情况下,我正在使用服务器进行身份验证:
    try {

$response = $this->client->post($my_authentication_path, [
'headers' => ['Authorization' => 'Basic ' . $this->base_64_key,
'Content-Type' => 'application/x-www-form-urlencoded;charset=UTF-8'
],
'form_params' => ['grant_type' => 'password',
'username' => 'my_user_name',
'password' => 'my_password']
]);

$response_body = $response->getBody()->getContents();

} catch (GuzzleHttp\Exception\RequestException $e){
$response_object = $e->getResponse();
//log or print the error here.
return false;
} //end catch

$authentication_response = json_decode($response_body);
print_r($authentication_response);

关于php - 如何在 php guzzle 上获取响应正文?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32119856/

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