我的php脚本已连接，但android工作室不会使用它。无可见错误输出

Question

我有一个php脚本用于处理非常基本的注册。它在我本地主机的一个文件夹中；http://(ipaddresslocalhost(:3306/testing/Register.php。Android工作室指向此URL，但在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。

这是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；m

$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(!$statement) { printf("Prepare failed: %sn", mysqli_error($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);
?>

我使用的是xampp(打开了apache和mysql(，我的Register位于Htdocs中的一个文件夹中。提前感谢您的帮助！

Answer 1

您可以修改RegisterRequest:的构造函数

public RegisterRequest(String username, String password,String isAdmin, 
Response.Listener<String> listener,
Response.ErrorListener() errListener){ //add error listener
super(Method.POST, REGISTER_REQUEST_URL,listener,errListener);
......
}

在CreateUser类文件中，添加以下内容：

Response.ErrorListener errorListener = new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(context, String.valueOf(error), Toast.LENGTH_SHORT).show();
}
}

然后：

RegisterRequest registerRequest = new RegisterRequest(username,password,isadmin,
responseListener,errorListener);

现在，再次运行，当出现错误时，您可以在错误侦听器中看到它。