快速Web服务中的POST方法

Question

早安朋友

我正在尝试将数据发送到BBDD，但出现错误：

  

Error Domain = NSCocoaErrorDomain代码= 3840“无值。” UserInfo = {NSDebugDescription =无值。}。搜索信息，发现您应该添加request.addValue（“ application / json”，用于HTTPHeaderField：“ Content-Type”）

，但仍然无法正常工作。如果我尝试使用https://jsonplaceholder.typicode.com/posts的网址，它会起作用。

@IBAction func postTapped(_ sender: Any) {

    let paramater = ["Name":"Bob", "Room":"Telomer", "Teacher_Name":"Tomson"]

    /*
    guard let url = URL(string: "https://jsonplaceholder.typicode.com/posts") else {
        return
    }
     */
    guard let url = URL(string: "http://xxxxxxx/postClasses.php") else {
        return


    var request = URLRequest(url: url)

    request.httpMethod = "POST"

    request.addValue("application/json", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept")  

    guard let httpBody1 = try? JSONSerialization.data(withJSONObject: paramater, options: []) else {
        return
    }

    request.httpBody = httpBody1

    let session = URLSession.shared

    session.dataTask(with: request) { (data, response, error) in

        if let response = response {
            print(response)
        }

        if let data = data {

            do{
                let json = try JSONSerialization.jsonObject(with: data, options: [])
                print(json)

            }
            catch{
                print(error)
            }
        }

    }.resume()

}

和php

$name=         ( empty($_POST['Name']) )         ?  NULL  : $_POST['Name'];
$room=         ( empty($_POST['Room']) )         ?  ""    : $_POST['Room'];
$Teacher_Name= ( empty($_POST['Teacher_Name']) ) ?  ""    : $_POST['Teacher_Name'];

if ($name){

    if( $conexion=mysqli_connect($hostname,$username,$password,$database) ){

        $sql= "INSERT INTO classes(Name,Room,Teacher_Name) VALUES (?,?,?)";

        if ($stmt = mysqli_prepare($conexion, $sql)) {

            mysqli_stmt_bind_param($stmt, "sss", $name,$room,$Teacher_Name);

            if( mysqli_stmt_execute($stmt) ) {
                //$msg=printf("Insert", mysqli_stmt_affected_rows($stmt));
                $msg=printf("");
            }else{
                //$msg=printf("Error: %s ok", mysqli_stmt_error($stmt));
                $msg=printf("Account already exists");
            }
            mysqli_stmt_close($stmt);
        }else{
                $msg=printf("Error: %s ", mysqli_error($conexion));
        }
        mysqli_close($conexion);
    }else{
        $msg=printf("Error de conetion: %s", mysqli_connect_error());
    }
}else{
    $msg="";  
}

//echo $msg;

非常感谢您