我有一个php脚本来处理非常基本的注册。它在我的本地主机的一个文件夹中;http://(ipaddress/localhost):3306/testing/register.php。androidstudio指向这个url,但是当我运行模拟器时没有得到错误输出。当我在浏览器中打开php脚本时,一切看起来都很好。
下面是registerrequest java脚本:
public class RegisterRequest extends StringRequest {
private static final String REGISTER_REQUEST_URL = "http://(ipaddress/localhost):3306/testing/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;
}
}
这是我的createuser脚本
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) {
try {
JSONObject jsonResponse = new JSONObject(response);
boolean success = jsonResponse.getBoolean("success");
if (success){
Intent intent = new Intent(CreateUser.this, MainActivity.class);
startActivity(intent);
}else{
AlertDialog.Builder builder = new AlertDialog.Builder(CreateUser.this);
builder.setMessage("Register Failed")
.setNegativeButton("Retry",null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
};
RegisterRequest registerRequest = new RegisterRequest(username,password,isadmin,responseListener);
RequestQueue queue = Volley.newRequestQueue(CreateUser.this);
queue.add(registerRequest);
}
});
}
最后,这里是我的php脚本,register;米
$db_host = 'localhost:3306';
$db_user = 'root';
$db_pass = '';
$db_name = 'test';
$con = mysqli_connect($db_host,'user',$db_pass,$db_name);
if($con){
echo "connection successful";
}else{
echo "connection failed";
}
$age = $_POST["isAdmin"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
if(!$statement) { return json_encode(['status'=>'failed','message'=>mysqli_error($con)]); }
mysqli_stmt_bind_param($statement, "ssi",$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(mysqli_error($statement)) { return json_encode(['status'=>'failed','message'=>mysqli_error($con)]); }
$response = array();
$response["success"] = true;
echo json_encode($response);
?>
我使用xampp(打开了apache和mysql),我的寄存器在htdocs的一个文件夹中。提前感谢您的帮助!
1条答案
按热度按时间q8l4jmvw1#
你可以修改
RegisterRequest
:而且在
CreateUser
类文件,添加以下内容:然后:
现在,再次运行,当错误发生时,您可以在错误侦听器中看到它。