有人知道用Guzzle发布JSON的正确方法吗?
$request = $this->client->post(self::URL_REGISTER,array(
'content-type' => 'application/json'
),array(json_encode($_POST)));
我从服务器得到一个内部服务器错误响应。它使用Chrome邮差工作。
有人知道用Guzzle发布JSON的正确方法吗?
$request = $this->client->post(self::URL_REGISTER,array(
'content-type' => 'application/json'
),array(json_encode($_POST)));
我从服务器得到一个内部服务器错误响应。它使用Chrome邮差工作。
当前回答
$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();
其他回答
简单地使用它将工作
$auth = base64_encode('user:'.config('mailchimp.api_key'));
//API URL
$urll = "https://".config('mailchimp.data_center').".api.mailchimp.com/3.0/batches";
//API authentication Header
$headers = array(
'Accept' => 'application/json',
'Authorization' => 'Basic '.$auth
);
$client = new Client();
$req_Memeber = new Request('POST', $urll, $headers, $userlist);
// promise
$promise = $client->sendAsync($req_Memeber)->then(function ($res){
echo "Synched";
});
$promise->wait();
这对我来说很有效(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());
@user3379466是正确的,但在这里我重写了全文:
-package that you need:
"require": {
"php" : ">=5.3.9",
"guzzlehttp/guzzle": "^3.8"
},
-php code (Digest is a type so pick different type if you need to, i have to include api server for authentication in this paragraph, some does not need to authenticate. If you use json you will need to replace any text 'xml' with 'json' and the data below should be a json string too):
$client = new Client('https://api.yourbaseapiserver.com/incidents.xml', array('version' => 'v1.3', 'request.options' => array('headers' => array('Accept' => 'application/vnd.yourbaseapiserver.v1.1+xml', 'Content-Type' => 'text/xml'), 'auth' => array('username@gmail.com', 'password', 'Digest'),)));
$url = "https://api.yourbaseapiserver.com/incidents.xml"; $data = '<事件> <名称>事件Title2a < /名称> <优先>中> < /优先 <请求者> < >电子邮件dsss@mail.ca < /电子邮件> < /请求者> <描述> description2a > < /描述 > < /事件”;
$request = $client->post($url, array('content-type' => 'application/xml',));
$request->setBody($data); #set body! this is body of request object and not a body field in the header section so don't be confused.
$response = $request->send(); #you must do send() method!
echo $response->getBody(); #you should see the response body from the server on success
die;
解决*暴饮暴食6 * - -你需要的包:
"require": {
"php" : ">=5.5.0",
"guzzlehttp/guzzle": "~6.0"
},
$client = new Client([
// Base URI is used with relative requests
'base_uri' => 'https://api.compay.com/',
// You can set any number of default request options.
'timeout' => 3.0,
'auth' => array('you@gmail.ca', 'dsfddfdfpassword', 'Digest'),
'headers' => array('Accept' => 'application/vnd.comay.v1.1+xml',
'Content-Type' => 'text/xml'),
]);
$url = "https://api.compay.com/cases.xml";
$data string variable is defined same as above.
// Provide the body as a string.
$r = $client->request('POST', $url, [
'body' => $data
]);
echo $r->getBody();
die;
解决方案$客户端->请求('POST',…
对于那些使用$client->请求的人,这是如何创建JSON请求的:
$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
对于《Guzzle 5》,《Guzzle 6》和《Guzzle 7》,你是这样做的:
use GuzzleHttp\Client;
$client = new Client();
$response = $client->post('url', [
GuzzleHttp\RequestOptions::JSON => ['foo' => 'bar'] // or 'json' => [...]
]);
Docs