PHP GuzzleHttp post发送json数据
1、引用 GuzzleHttp 客户端
use GuzzleHttp\Client;
2、发送请求代码示例
$http = new Client([ 'verify' => false,//绕过HTTPS证书校验 'headers' => [ 'Content-type' => 'application/json', ] ]); $form_params = [ 'json' => $params//$params 数组 ]; $url = "请求URL";//rtrim($this->api_url,'/').$uri; $response = $http->post($url,$form_params); //HTTP 响应吗 $code = $response->getStatusCode(); if ($code == 500){ throw new \Exception("请求失败,服务异常"); } //返回值 $rsp = $response->getBody()->getContents(); if(empty($rsp)){ throw new \Exception("请求失败,没有响应"); } $rspArr = json_decode($rsp,true);
注:
若没有安装GuzzleHttp包,可以参考这篇文章:php GuzzleHttp composer