我在一个应用程序中有 3 个 fragment ,在其中一个 fragment 中,我显示了 SQLite 数据库中的用户名。当我注册一个新用户并第一次使用它登录时,会发生什么情况,在用户名应该出现的 TextView 中,它显示 NULL 值,但是当我注销并再次使用同一用户登录时,名称会按原样显示。
注册用户后,所有数据都插入到数据库中,我已经检查过了。
有什么想法会导致这个问题吗?我会根据要求添加一些代码,因为我不知道代码、 fragment 或 Java 文件的哪一部分可能会导致这种情况。
已编辑
我添加了一些代码来帮助解决这个问题。
主屏幕内的登录功能(应用启动后):
private void checkLogin(final String email, final String password) {
// Tag used to cancel the request
String tag_string_req = "req_login";
pDialog.setMessage("Logging in ...");
showDialog();
StringRequest strReq = new StringRequest(Request.Method.POST,
AppConfig.URL_LOGIN, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.d(TAG, "Login Response: " + response.toString());
hideDialog();
try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
// Check for error node in json
if (!error) {
// user successfully logged in
// Create login session
session.setLogin(true);
// Now store the user in SQLite
String uid = jObj.getString("uid");
JSONObject user = jObj.getJSONObject("user");
String name = user.getString("name");
String email = user.getString("email");
String created_at = user.getString("created_at");
// Inserting row in users table
db.addUser(name, email, uid, created_at);
// Launch main activity
Intent intent = new Intent(Main.this,
Logged.class);
startActivity(intent);
finish();
} else {
error_msg.setVisibility(View.VISIBLE);
String msg = jObj.getString("error_msg");
error_msg.setText(msg);
}
} catch (JSONException e) {
// JSON error
e.printStackTrace();
Toast.makeText(getApplicationContext(), "Json error: " + e.getMessage(), Toast.LENGTH_LONG).show();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "Login Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_LONG).show();
hideDialog();
}
}) {
@Override
protected Map<String, String> getParams() {
// Posting parameters to login url
Map<String, String> params = new HashMap<String, String>();
params.put("email", email);
params.put("password", password);
return params;
}
};
// Adding request to request queue
AppController.getInstance().addToRequestQueue(strReq, tag_string_req);
}
onViewCreated func inside fragment where users name should be diplayed:
@Override
public void onViewCreated(View view, Bundle savedInstanceState){
profileName = (TextView) getActivity().findViewById(R.id.profileName);
// SqLite database handler
db = new SQLiteHandler(getActivity().getApplicationContext());
// session manager
session = new SessionManager(getActivity().getApplicationContext());
if (!session.isLoggedIn()) {
logoutUser();
}
// Fetching user details from SQLite
HashMap<String, String> user = db.getUserDetails();
String name = user.get("name");
// Displaying the user details on the screen
profileName.setText(name);
}
注册函数部分:
public void onResponse(String response) {
Log.d(TAG, "Register Response: " + response.toString());
hideDialog();
try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
if (!error) {
// User successfully stored in MySQL
// Now store the user in sqlite
String uid = jObj.getString("uid");
JSONObject user = jObj.getJSONObject("user");
String name = user.getString("name");
String email = user.getString("email");
String created_at = user.getString("created_at");
// Inserting row in users table
db.addUser(name, email, uid, created_at);
// Launch login activity
Intent intent = new Intent(
Register.this,
Main.class);
startActivity(intent);
finish();
} else {
// Error occurred in registration. Get the error
// message
error_msg.setVisibility(View.VISIBLE);
String msg = jObj.getString("error_msg");
error_msg.setText(msg);
}
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "Registration Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_LONG).show();
hideDialog();
}
}) {
最佳答案
我设法自己解决了这个问题。问题不在 java 文件中,而是在出于某种原因发送到 java 的 php 文件中。
在我使用的 php 函数中:
$stmt->bind_result($id, $unique_id, $name, $email, $encrypted_password, $salt, $created_at, $updated_at);
$user = $stmt->fetch();
但后来改成了
$user = $stmt->get_result()->fetch_assoc();
这就解决了问题..
关于java - 在第一个用户登录 android 后显示空值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33677786/