Question

我创建了一个简单的登录和注册UI一切正常，除了用户名editText字段，输入的文本总是在mysql数据库中显示为0。

这是使用Volley并处理JSON响应的类，并创建一个新用户将用户带到登录屏幕：

public class RegisterActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_register);

        final EditText etAge = (EditText) findViewById(R.id.etAge);
        final EditText etName = (EditText) findViewById(R.id.etName);
        final EditText etUserName = (EditText) findViewById(R.id.etUsername);
        final EditText etPassword = (EditText) findViewById(R.id.etPassword);
        final Button bRegister = (Button) findViewById(R.id.bRegister);

        bRegister.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                final String name = etName.getText().toString();
                final String username = etUserName.getText().toString();
                final String password = etPassword.getText().toString();
                final int age = Integer.parseInt(etAge.getText().toString());

                Response.Listener<String> responseListener = new Response.Listener<String>() {
                    @Override
                    public void onResponse(String response) {
                        try {
                            JSONObject jsonResponse = new JSONObject(response);
                            Boolean success = jsonResponse.getBoolean("success");

                            if (success) {
                                Intent intent = new Intent(RegisterActivity.this, LoginActivity.class);
                                RegisterActivity.this.startActivity(intent);
                            } else {
                                AlertDialog.Builder builder = new AlertDialog.Builder(RegisterActivity.this);
                                builder.setMessage("Register Failed")
                                        .setNegativeButton("Retry", null)
                                        .create()
                                        .show();
                            }
                        } catch (JSONException e) {
                            e.printStackTrace();
                        }
                    }
                };

                RegisterRequest registerRequest = new RegisterRequest(name, username, age, password, responseListener);
                RequestQueue queue = Volley.newRequestQueue(RegisterActivity.this);
                queue.add(registerRequest);
            }
        });
    }
}

这是发送请求并建立连接的类：

public class RegisterRequest extends StringRequest{
    private static final String REGISTER_REQUEST_URL = "http://192.168.100.105/register.php";
    private Map<String, String> params;

    public RegisterRequest(String name, String username, int age, String password, Response.Listener<String> listener) {
        super(Method.POST, REGISTER_REQUEST_URL, listener, null);
        params = new HashMap<>();
        params.put("name", name);
        params.put("username", username);
        params.put("password", password);
        params.put("age", age + "");
    }

    @Override
    public Map<String, String> getParams() {
        return params;
    }
}

这是用于验证，连接和查询数据库的php文件：

<?php
$con = mysqli_connect("localhost", "root", "password", "androidtest");
$name = $_POST["name"];
$age = $_POST["age"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO users (name, username, age, 
password) VALUES (?, ?, ?, ?)");

mysqli_stmt_bind_param($statement, "siss", $name, $username, $age, 
$password);
mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true; 

echo json_encode($response);
?>

因为我不允许上传附件图片，所以这里是指向数据库屏幕截图的链接：

https://i.gyazo.com/94e1bfe5f0c75f8b81fd9406f614f2f9.png

Answer 1

在PHP中更改此内容

mysqli_stmt_bind_param($statement, "siss", $name, $username, $age, 
$password);

到

mysqli_stmt_bind_param($statement, "ssis", $name, $username, $age, 
$password);

用户名不是整数。

android editText文本在mysql数据库中显示为0？

1 个答案: