如何使用Guzzle以JSON格式发送POST请求？

Question

有没有人知道post使用JSON 的正确方法Guzzle？

$request = $this->client->post(self::URL_REGISTER,array(
                'content-type' => 'application/json'
        ),array(json_encode($_POST)));

我收到internal server error服务器的回复.它适用于Chrome Postman.

Answer 1

对于Guzzle 5和6,您可以这样做:

use GuzzleHttp\Client;

$client = new Client();

$response = $client->post('url', [
    GuzzleHttp\RequestOptions::JSON => ['foo' => 'bar']
]);

文件

Answer 2

对于Guzzle <= 4:

这是一个原始的帖子请求,因此将JSON放入正文解决了问题

$request = $this->client->post($url,array(
                'content-type' => 'application/json'
        ),array());
$request->setBody($data); #set body!
$response = $request->send();

return $response;

Answer 3

简单而基本的方式(guzzle6):

$client = new Client([
    'headers' => [ 'Content-Type' => 'application/json' ]
]);

$response = $client->post('http://api.com/CheckItOutNow',
    ['body' => json_encode(
        [
            'hello' => 'World'
        ]
    )]
);

要获取响应状态代码和正文的内容,我执行了以下操作:

echo '<pre>' . var_export($response->getStatusCode(), true) . '</pre>';
echo '<pre>' . var_export($response->getBody()->getContents(), true) . '</pre>';

Answer 4

这对我有用(使用Guzzle 6)

$client = new Client(); 
$result = $client->post('http://api.example.com', [
            'json' => [
                'value_1' => 'number1',
                'Value_group' =>  
                             array("value_2" => "number2",
                                    "value_3" => "number3")
                    ]
                ]);

echo($result->getBody()->getContents());

Answer 5

$client = new \GuzzleHttp\Client();

$body['grant_type'] = "client_credentials";
$body['client_id'] = $this->client_id;
$body['client_secret'] = $this->client_secret;

$res = $client->post($url, [ 'body' => json_encode($body) ]);

$code = $res->getStatusCode();
$result = $res->json();

Answer 6

您可以使用硬编码json属性作为键，也可以方便地使用GuzzleHttp\RequestOptions::JSON常量。

这是使用硬编码json字符串的示例。

use GuzzleHttp\Client;

$client = new Client();

$response = $client->post('url', [
    'json' => ['foo' => 'bar']
]);

请参阅文档。

Answer 7

Guzzle 6.2对我有用:

$gClient =  new \GuzzleHttp\Client(['base_uri' => 'www.foo.bar']);
$res = $gClient->post('ws/endpoint',
                            array(
                                'headers'=>array('Content-Type'=>'application/json'),
                                'json'=>array('someData'=>'xxxxx','moreData'=>'zzzzzzz')
                                )
                    );

根据文档guzzle做json_encode

Answer 8

$client = new \GuzzleHttp\Client(['base_uri' => 'http://example.com/api']);

$response = $client->post('/save', [
    'json' => [
        'name' => 'John Doe'
    ]
]);

return $response->getBody();

Answer 9

$client->request('POST',... 的解决方案

对于那些正在使用的人来说，$client->request这是创建 JSON 请求的方法：

$client = new Client();
$res = $client->request('POST', "https://some-url.com/api", [
    'json' => [
        'paramaterName' => "parameterValue",
        'paramaterName2' => "parameterValue2",
    ]
    'headers' => [
    'Content-Type' => 'application/json',
    ]
]);

Guzzle JSON 请求参考