Php 拉雷维尔狂饮
有人能用GuzzleHttp解决这个问题吗?我想在controller中显示$reportid,但我现在被卡住了Php 拉雷维尔狂饮,php,laravel,guzzle,Php,Laravel,Guzzle,有人能用GuzzleHttp解决这个问题吗?我想在controller中显示$reportid,但我现在被卡住了 //Select Data For SDO Homepage Report public function SDOHomepage(){ $client = new \GuzzleHttp\Client(); $url=WEBSERVICE_URL; //Response 1 $response1=$client->request('POST',
//Select Data For SDO Homepage Report
public function SDOHomepage(){
$client = new \GuzzleHttp\Client();
$url=WEBSERVICE_URL;
//Response 1
$response1=$client->request('POST', $url,['form_params'=>['tag'=>'selectSDOHomepage']]);
//$body = $response1->getBody();
//$reportid = $body['ReportID'];
//Response 2
$response2=$client->request('POST', $url,['form_params'=>['tag'=>'sdoCountReply','ReportID'=>$reportid]]);
$data1=json_decode($response1->getBody()->getContents(),true);
$data2=json_decode($response2->getBody()->getContents(),true);
//Return Data
return view('SDOHomepage',['SDO_All'=>$data1,'SDO_Reply_Count'=>$data2]);
}
你是不是错过了这个
$body = $response1->getBody();
$reportid = $body['ReportID'];
您可能需要也可能不需要执行json_解码之类的其他操作。到底出了什么问题?你的控制器在哪里?谢谢你的回答。但我有一个错误,它说不能使用GuzzleHttp\Psr7\Stream类型的对象作为数组,也许我应该在这里使用另一种方法?