php - 创建 PHP 页面的自定义响应

Question

我正在创建一个连接到 MYSQL 数据库的应用程序，用户可以在其中登录或注册。在注册类中，我使用 OKHTTP3 来注册新用户。 OKHTTP3 向执行查询的 php 页面发出请求。页面的响应始终相同(代码:200，消息:“空”)。我想自定义 php 页面的响应，例如代码和消息，这样我就可以知道注册是否成功。我该怎么做？

请尽可能简单。

我的 OKHTTP3 请求:

String reg_name = username.getText().toString();
            String reg_pass = password.getText().toString();
            String reg_url = "https://collectcards.000webhostapp.com/register.php";

            OkHttpClient client = new OkHttpClient();

            RequestBody formBody = new FormBody.Builder()
                    .add("user_name", reg_name)
                    .add("user_pass", reg_pass)
                    .build();

            Request request = new Request.Builder()
                    .url(reg_url)
                    .post(formBody)
                    .build();

            client.newCall(request).enqueue(new Callback() {
                @Override
                public void onFailure(Call call, IOException e) {
                    // Request failed
                }

                @Override
                public void onResponse(Call call, Response response) throws IOException {
                    // Handle whatever comes back from the server
                    String res = response.message(); //MESSAGE IS ALWAYS EMPTY, COSE IS ALWAYS 200
                    if(res.equals("OK")) { //ID LIKE TO CUSTOM THE MESSAGGE TO SAY OK WHEN IT GOES WELL AND NOTOK WHEN IT FAIL
                        ok = true;
                        risposta = "Registrazione Avvenuta";
                    }else {
                        risposta = "Registrazione Fallita";
                        ok = false;
                    }
                    mHandler.post(new Runnable() {
                        @Override
                        public void run() {
                            Toast.makeText(thisContext, risposta, Toast.LENGTH_LONG).show();
                            if (ok) {
                                Intent myIntent = new Intent(thisContext, mainMenuActivity.class);
                                thisContext.startActivity(myIntent);
                            }else{
                                password.setText("");
                                username.setText("");
                            }
                        }
                    });



                }
            });

我的 php 页面:

<?php  

 $connessione = mysqli_connect("localhost", "usr", "psw", "db");

 $user_name = $_POST['user_name']; 

 $user_pass = $_POST['user_pass'];  

 $query =  "INSERT INTO `player` (`username`, `password`, `money`) VALUES ('".$user_name."', '".$user_pass."', '100');";

 $risultato = mysqli_query($connessione, $query);

if($risultato){
 //CUSTOM MESSAGGE
}else{

}

 ?>

Answer 1

您可以使用http_response_code();设置输出的代码，例如:

http_response_code(500); 

如果出现服务器错误，400 表示错误请求等。对于正文，您应该能够在正文中回显它。我建议做类似 json_encode() 'ing 数据数组的事情。

请不要使用 PHP 关闭标签。

Why would one omit the close tag?