1. 使用GuzzleHttp 发送post请求,返回数据为404,造成链接状态也是404;
2. 然后laravel会以报错的形式反馈出来,如何让他不报错?
3. 代码:

  $response = $http->post($url, $http_param);

报错:

Client error: `POST http://******` resulted in a `404 Not Found` response: {"code":0,"desc":"失败","timestamp":"20180426114652","data":"eyJlcnJvckNvZGUiOjQwNCwiZXJyb3JNc2ciOiJOb3QgRm91bmQifQ==" (truncated...)

1. 注意是laravel的报错,我用try(){}catch(){}也没用,还是会以laravel报错的形式反馈出来;
2. 我不想让他报错,只想得到返回的404数据;
3. 如何解决?