我正在开发一个用于求职的android应用程序。在这里我正在使用PHP作为后端

Question

在php代码中，当我们注册时会自动创建一个随机代码作为用户ID。 在android应用中，当我们保存个人资料信息时，我无法确保用户ID会有所不同。 因此如何在所有活动中管理用户ID，并在退出前帮助片段保持相同的用户ID。 ................ **

注册代码：

<?php
 session_start();

   if($_SERVER['REQUEST_METHOD']=='POST'){

       include_once("db_connect.php");


    $username = $_POST['user_name'];
    $useremail = $_POST['user_email'];
    $usermobile = $_POST['user_mobile'];
    $password = $_POST['password'];



 function randomstring($len) {
    $string = "";
    $chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
    for($i=0;$i<$len;$i++)
        $string.=substr($chars,rand(0,strlen($chars)),1);
    return $string;
    }
    $rndm_code=randomstring(5);




$CheckSQL = "SELECT * FROM user WHERE user_email='$useremail'";
    $check = mysqli_fetch_array(mysqli_query($con,$CheckSQL));

          if(isset($check)){

         echo 'Email Already Exist';

          }

 else{


  $Sql_Query = "insert into user (user_id,user_name,user_email,user_mobile,password) values ('$rndm_code','$username','$useremail','$usermobile','$password')";


   if (mysqli_query($con,$Sql_Query)){



  echo 'User Registration Successfully';

 }
 else
 {

 echo 'Try Again';
 }

 }
 }
 mysqli_close($con);

 ?>

Android代码：

 private void registerUser() {
             isConnectingToInternet();
            final String username = signupname.getText().toString().trim();
            final String email = signupemail.getText().toString().trim();
            final String phone = signupphone.getText().toString().trim();
            final String password = signuppassword.getText().toString().trim();


            pDialog = new ProgressDialog(SignupActivity.this);
            pDialog.setMessage("Signing Up.. Please wait...");
            pDialog.setCancelable(false);
            pDialog.show();


            StringRequest stringRequest = new StringRequest(Request.Method.POST, Config.URL_REGISTER,
                    new Response.Listener<String>() {
                        @Override
                        public void onResponse(String ServerResponse) {

                            // Hiding the progress dialog after all task complete.
                            pDialog.dismiss();

                            // Showing response message coming from server.
                            Toast.makeText(SignupActivity.this, ServerResponse, Toast.LENGTH_LONG).show();


                        }
                    },
                    new Response.ErrorListener() {
                        @Override
                        public void onErrorResponse(VolleyError volleyError) {

                            // Hiding the progress dialog after all task complete.
                            pDialog.dismiss();

                            // Showing error message if something goes wrong.
                            Toast.makeText(SignupActivity.this,"Check Internet connection", Toast.LENGTH_LONG).show();
                        }
                    }) {
                @Override
                protected Map<String, String> getParams() {

                    // Creating Map String Params.
                    Map<String, String> params = new HashMap<String, String>();

                    // Adding All values to Params.
                    params.put("user_name",username);
                    params.put("user_email",email);
                    params.put("user_mobile", phone);
                    params.put("password",password);

                    return params;
                }

            };


            RequestQueue requestQueue = Volley.newRequestQueue(SignupActivity.this);


            requestQueue.add(stringRequest);




        }**

Answer 1

您的服务器响应除了响应文本外还需要包含ID。 通常，这是通过将响应格式设置为JSON来完成的。

不仅仅是“用户注册成功”，PHP脚本的响应可能是

{
  "message":"User Registration Successfully",
  "userId":<your_user_id_goes_here>
}

您可以在PHP中使用json_encode从数组中为您创建此格式。 该代码看起来像这样

echo json_encode([
   "message" => "User Registration Successfully",
   "userId" => $rndm_code
]);
exit();

最好在发送响应后退出，这样您的JSON deos不会因为以后意外在其他地方回显某些内容而无效。

在android应用中，您将随后从JSON获取数据-我在android中没有经验，但是我认为您可以通过将其设置为JsonObjectRequest而不是StringRequest，然后显示serverResponse.getString（“ message”）来实现。并将serverResponse.getString（“ userId”）作为ID保存在某个地方的变量中。

Answer 2

<?php
 session_start();

   if($_SERVER['REQUEST_METHOD']=='POST'){

       include_once("db_connect.php");


    $username = $_POST['user_name'];
    $useremail = $_POST['user_email'];
    $usermobile = $_POST['user_mobile'];
    $password = $_POST['password'];



 function randomstring($len) {
    $string = "";
    $chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
    for($i=0;$i<$len;$i++)
        $string.=substr($chars,rand(0,strlen($chars)),1);
    return $string;
    }
    $rndm_code=randomstring(5);




$checkUser = "SELECT * FROM user WHERE user_email='$useremail' OR user_mobile='$usermobile'";

 $userStatus = mysqli_fetch_array(mysqli_query($con,$checkUser));

 if(isset($userStatus)){
 die(json_encode(array("success"=>0,"message"=>"Email or Mobile no already exist!!")));
 }
 else
 {

$sql =$con->prepare( "INSERT INTO user(user_id,user_name,user_email,user_mobile,password) VALUES ('$rndm_code','$username', '$useremail', '$usermobile','$password')");
$sql->bind_param("ssss",$rndm_code, $username, $useremail, $usermobile,$password);

if($sql->execute())
{
echo(json_encode(array("success"=>1,"message"=>"Account created successfully!! Please Login")));
}
else
{
die("Something Error".$sql."<br>".mysqli_error($con));
}

 }

}else{

  die(json_encode(array("success"=>0,"message"=>"Empty Request Parameters..!!")));

}
mysqli_close($con);

 ?>