laravel 使用 Guzzle 用 JSON 发送 POST 请求
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Using Guzzle to send POST request with JSON
提问by Kira
$client = new Client();
$url = 'api-url';
$request = $client->post($url, [
'headers' => ['Content-Type' => 'application/json'],
'json' => ['token' => 'foo']
]);
return $request;
And I get back 502 Bad Gatewayand Resource interpreted as Document but transferred with MIME type application/json
我回来了502 Bad Gateway，资源被解释为文档，但用 MIME 类型应用程序/json 传输
I need to make a POST request with some json. How can I do that with Guzzle in Laravel?
我需要用一些 json 发出 POST 请求。我怎样才能在 Laravel 中用 Guzzle 做到这一点？
回答by Morteza Rajabi
Give it a try
试一试
$response = $client->post('http://api.example.com', [
'json' => [
'key' => 'value'
]
]);
dd($response->getBody()->getContents());
回答by rAm
Take a look..
看一看..
$client = new Client();
$url = 'api-url';
$headers = array('Content-Type: application/json');
$data = array('json' => array('token' => 'foo'));
$request = new Request("POST", $url, $headers, json_encode($data));
$response = $client->send($request, ['timeout' => 10]);
$data = $response->getBody()->getContents();
回答by pankaj kumar
you can also try this solution. that is working on my end. I am using Laravel 5.7.
你也可以试试这个解决方案。这对我有用。我正在使用Laravel 5.7。
This is an easy solution of Make a POST Requestfrom PHPWith Guzzle
这是使用Guzzle从PHP发出POST 请求的简单解决方案
function callThirdPartyPostAPI( $url,$postField )
{
$client = new Client();
$response = $client->post($url , [
//'debug' => TRUE,
'form_params' => $postField,
'headers' => [
'Content-Type' => 'application/x-www-form-urlencoded',
]
]);
return $body = $response->getBody();
}
For Use this method
对于使用此方法
$query['schoolCode'] =$req->schoolCode;
$query['token']=rand(19999,99999);
$query['cid'] =$req->cid;
$query['examId'] =$req->examId;
$query['userId'] =$req->userId;
$tURL = "https://www.XXXXXXXXXX/tabulation/update";
$response = callThirdPartyPostAPI($tURL,$query);
if( json_decode($response,true)['status'] )
{
return success(["data"=>json_decode($response,true)['data']]);
}
