我正在通过 PHP 为这个 android java 类（登录屏幕）传递值，但它给了我一个 JSONException 错误，我无法从下面的代码中识别出来。有人可以告诉我如何解决吗？先感谢您！
登录活动.java
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
// Importing all assets like buttons, text fields
inputUid = (EditText) findViewById(R.id.loginUid);
inputPassword = (EditText) findViewById(R.id.loginPassword);
btnLogin = (Button) findViewById(R.id.btnLogin);
btnLinkToRegister = (Button) findViewById(R.id.btnLinkToRegisterScreen);
loginErrorMsg = (TextView) findViewById(R.id.login_error);
// Login button Click Event
btnLogin.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
String id = inputUid.getText().toString();
String pswd = inputPassword.getText().toString();
UserFunctions userFunction = new UserFunctions();
Log.d("Button", "Login");
JSONObject json = userFunction.loginUser(id, pswd);
// check for login response
try {
if (json.getString(KEY_SUCCESS) != null) {
loginErrorMsg.setText("");
String res = json.getString(KEY_SUCCESS);
if(Integer.parseInt(res) == 1){
// user successfully logged in
// Store user details in SQLite Database
DatabaseHandler db = new DatabaseHandler(getApplicationContext());
JSONObject json_user = json.getJSONObject("user");
// Clear all previous data in database
userFunction.logoutUser(getApplicationContext());
db.addUser(json.getString(KEY_ID), json_user.getString(KEY_NAME), json_user.getString(KEY_EMAIL), json_user.getString(KEY_CREATED_AT));
// Launch Dashboard Screen
Intent dashboard = new Intent(getApplicationContext(), DashboardActivity.class);
// Close all views before launching Dashboard
dashboard.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(dashboard);
// Close Login Screen
finish();
}else{
// Error in login
loginErrorMsg.setText("Incorrect username/password");
}
}
} catch (JSONException e) {
e.printStackTrace();
}
}
});