再会!在此程序中,我可以检索我请求的所有详细信息,但用户名返回0除外。
请参阅随附的代码
Register.php
<?php
$con = mysqli_connect("?", "?", "?", "?");
$name = $_POST["name"];
$username = $_POST["username"];
$age = $_POST["age"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO user (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); ?>
RegisterRequest.class
public class RegisterRequest extends StringRequest {
private static final String REGISTER_REQUEST_URL = "https://simonewalter.000webhostapp.com/PharmTechPH/Register.php";
private Map<String, String> params;
//ADD ORDER HERE BITCHEZ
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("age", age + "");
params.put("password", password);
}
@Override
public Map<String, String> getParams() {
return params;
}}
RegisterActivty.class
public class RegisterActivity extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_register);
final EditText etName = (EditText) findViewById(R.id.etName);
final EditText etUsername = (EditText) findViewById(R.id.etUsername);
final EditText etAge = (EditText) findViewById(R.id.etAge);
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 v) {
final String name = etName.getText().toString();
final String username = etUsername.getText().toString();
final int age = Integer.parseInt(etAge.getText().toString());
final String password = etPassword.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);
}
});
}}
您要发送的值username
是String
,并且在PHP
代码中已声明int
。
更改此:
mysqli_stmt_bind_param($statement, "siss", $name, $username, $age, $password);
........^//error here
有了这个:
mysqli_stmt_bind_param($statement, "ssss", $name, $username, $age, $password);
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