PHP Wamp服务器未从AlamoFire请求接收参数

Question

When I receive a request from AlamoFire the script in PHP not recognize the parameter neither in POST, GET nor REQUEST variable.

The code in AlamoFire is

let parameters: [String: AnyObject] =
        ["email" : email,
         "password": password]

    Alamofire.request(.POST, url, parameters: parameters, encoding: .JSON)
        .responseJSON { response in

            print(response)

            switch response.result {

            case .Success: // Connection stablished



            case .Failure(let error): //  Connection not stablished
               print(error)
               self.showLoginError(Constants.connectionError)
            }
    }

And the code in php is:

$response = array();
header('Content-Type: application/json');
if (isset($_POST['email']) && isset($_POST['password'])) {
  ... stuff
}else{
    $response["Failure"] = 1;
    $response["message"] = "Required field(s) is missing";
    echo json_encode($response);
}

Then response in AlamoFire always is the JSON with message of required field

At one point PHP gave me this error: PHP Deprecated: Automatically populating $HTTP_RAW_POST_DATA is deprecated and will be removed in a future version. To avoid this warning set 'always_populate_raw_post_data' to '-1' in php.ini and use the php://input stream instead. in Unknown on line 0

I had change the php.ini with the option of always_populate_raw_post_data to -1 and reach me to the state what I am now.

But if the request URL is like this http://192.168.1.108:8080/labor/login.php?email=slicingit@yahoo.com&password=brown1234 the server receive the parameter. Also if I create the request using NSURLSession with the parameters as part of the HTTP Body the server does get them.

Answer 1

How do you know its not receiving it? How do you know if alomfire is sending it? use 'echo' see what alomfire brings back in your php script. if it is indeed json values remember you have to decode the json values before you can use or assign them. How to decode json

 $mymail= ($_POST['email']);
 $mypass= ($_POST['password']);
 echo $mymail;
 echo $mypass;

 //Decode your json first here

 if ($mymail && $mypass) {
     ... stuff
 }else{
 $response["Failure"] = 1;
$response["message"] = "Required field(s) is missing";
echo json_encode($response);

}