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Question

我有一个基本的注册脚本。我花了很长时间才达到这一点，但我遇到了一个“错误”。该应用程序应该向 localhost phpmyadmin 发送 3 个登录注册字符串。该应用程序运行良好，直到它会生成错误弹出消息，我对其进行了编程。但它并没有告诉我为什么注册失败。

这里是创建用户脚本。它应该从文本框中获取变量并将它们发送到 RegisterRequest 脚本。它等待响应，然后将用户踢回主菜单或创建重试消息。它会创建一条重试消息，但不会记录错误。

public class CreateUser extends AppCompatActivity {
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_create_user);
        this.setTitle("Create User");
        final EditText username1 = findViewById(R.id.Createusername);
        final EditText password1 = findViewById(R.id.CreatePassword);
        final Switch isAdmin = findViewById(R.id.isadmin);
        final Button createuser = findViewById(R.id.createuserbtn);
        if (getIntent().hasExtra("com.example.northlandcaps.crisis_response")){
            isAdmin.setVisibility(View.GONE);
        }
        createuser.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                final String username = username1.getText().toString();
                final String password = password1.getText().toString();
                final String isadmin = isAdmin.getText().toString();
                Response.Listener<String> responseListener = new Response.Listener<String>() {
                    @Override
                    public void onResponse(String response) {
                        Log.d("Response Value: ", response);
                            if (response.equals("success")){
                                Intent intent = new Intent(CreateUser.this, MainActivity.class);
                                CreateUser.this.startActivity(intent);
                            }else{
                                AlertDialog.Builder builder = new AlertDialog.Builder(CreateUser.this);
                                builder.setMessage("Register Failed")
                                        .setNegativeButton("Retry",null)
                                        .create()
                                        .show();
                        }
                    }
                };
                RegisterRequest registerRequest = new RegisterRequest(username,password,isadmin,responseListener);
                RequestQueue queue = Volley.newRequestQueue(CreateUser.this);
                queue.add(registerRequest);
            }
        });
    }

这里是注册请求。它从 CreateUser 接收字符串并将它们发布到 php 脚本。我没有从这个脚本中得到任何错误，但可能会有一些错误

public class RegisterRequest extends StringRequest {

    private static final String REGISTER_REQUEST_URL = "http://192.168.*.*:80/phptesting/Register.php";
    private Map<String, String> params;
    public RegisterRequest(String username, String password,String isAdmin, Response.Listener<String> listener){
        super(Method.POST, REGISTER_REQUEST_URL,listener,null);
        params = new HashMap<>();
        params.put("username",username);
        params.put("password",password);
        params.put("isAdmin",isAdmin+"");
    }

    public Map<String, String> getparams() {
        return params;
    }
}

这里是注册 php 脚本。

<?php
    $db_host = '192.168.*.*:3306';
    $db_user = 'root';
    $db_pass = '';
    $db_name = 'test';

    $con = mysqli_connect($db_host,'root',$db_pass,$db_name);
    if($con){
        echo "connection successful";
    }
    if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
    }
    $isAdmin ="no";
     $username="fakename";
     $password="fakepassword";
    $isAdmin = isset($_POST["isAdmin"]);
    $username = isset($_POST["username"]);
    $password = isset($_POST["password"]);
    echo $isAdmin;
    echo $username;
    echo $password;
    $statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
    mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
    mysqli_stmt_execute($statement);
    if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
    echo "success";
?>

感谢所有帮助，谢谢

（有人告诉我在 php 脚本中添加 Var_dump，然后向后工作。（感谢 riggs），所以我得到的结果是：connection successfularray(0) { } 成功

Answer 1

改变

    $isAdmin ="no";
 $username="fakename";
 $password="fakepassword";
$isAdmin = isset($_POST["isAdmin"]);
$username = isset($_POST["username"]);
$password = isset($_POST["password"]);
echo $isAdmin;
echo $username;
echo $password;
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
echo "success";

到

if(isset($_POST["isAdmin"]) &&  isset($_POST["username"]) && isset($_POST["password"]))
{
    $username = isset($_POST["username"]);
    $password =  isset($_POST["password"]);
    $isAdmin = isset($_POST["isAdmin"]);

    $statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
    mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
    mysqli_stmt_execute($statement);

    if(!$statement) 
    {
         printf("Prepare failed: %s\n", mysqli_error($con)); 
    }

    echo "success";
}
else
 echo "values not set";