【发布时间】：2012-05-05 11:00:39
【问题描述】：
我保存了两部分数据，用户密码和用户名以供以后使用的数据和字母的随机组合，然后检查应用程序何时再次执行以避免再次出现登录屏幕。
我昨天有这个工作，但由于某种原因，我预计我在添加功能时放错了代码，现在无法让它再次工作，我已经尝试了一整天来修复它，但我不能。
EDIT1：出于某种原因，它转到“CheckPrefs”方法的 else 语句，我无法理解。
EDIT2：它可以很好地保存首选项并达到意图，但是它无法读取它。
代码：
public static final String PREFS_NAME = "MyPregs";
public static final String PREFS_CHECK = "CheckSignedIn";
public static final String UID ="";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
CheckPrefs();
// Login button clicked
ok = (Button) findViewById(R.id.btn_login);
ok.setOnClickListener(this);
result = (TextView) findViewById(R.id.lbl_result);
}
private void CheckPrefs() {
// TODO Auto-generated method stub
//checks to see if the user has signed in before if they have it gets there data from SharedPreferences and checks against our database via HttpPost.
SharedPreferences settings = getSharedPreferences(PREFS_CHECK, 0);
String SCheck1 = settings.getString("key4", null);
if (SCheck1 != null && equals(UID)) {
Intent c = new Intent(this, CheckUserInfo.class);
startActivity(c);
} else {
}
}
//create bracket.
public void postLoginData() {
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
/* login.php returns true if username and password is equal to saranga */
HttpPost httppost = new HttpPost("http://gta5news.com/login.php");
try {
// Add user name and password
uname = (EditText) findViewById(R.id.txt_username);
String username = uname.getText().toString();
pword = (EditText) findViewById(R.id.txt_password);
String password = pword.getText().toString();
SharedPreferences signedin = getSharedPreferences(PREFS_CHECK, 0);
SharedPreferences.Editor editor1 = signedin.edit();
editor1.putString("key4", UID);
editor1.commit();
SharedPreferences settings = getSharedPreferences(PREFS_NAME, 0);
SharedPreferences.Editor editor = settings.edit();
editor.putString("key1", username);
editor.putString("key2", password);
editor.commit();
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("username", username));
nameValuePairs.add(new BasicNameValuePair("password", password));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
Log.w("HttpPost", "Execute HTTP Post Request");
HttpResponse response = httpclient.execute(httppost);
String str = inputStreamToString(response.getEntity().getContent())
.toString();
Log.w("HttpPost", str);
if (str.toString().equalsIgnoreCase("true")) {
Log.w("HttpPost", "TRUE");
result.setText("Login successful");
try {Thread.sleep(250);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Intent login = new Intent(LogIn.this, ChatService.class);
startActivity(login);
} else {
Log.w("HttpPost", "FALSE");
result.setText(str);
}
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
private StringBuilder inputStreamToString(InputStream is) {
String line = "";
StringBuilder total = new StringBuilder();
// Wrap a BufferedReader around the InputStream
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
// Read response until the end
try {
while ((line = rd.readLine()) != null) {
total.append(line);
}
} catch (IOException e) {
e.printStackTrace();
}
// Return full string
return total;
}
public void onClick(View view) {
if (view == ok) {
postLoginData();
InputMethodManager imm = (InputMethodManager) getSystemService(Context.INPUT_METHOD_SERVICE);
imm.hideSoftInputFromWindow(pword.getWindowToken(), 0);
}
// Click end
}
// if statement
}
// class ends here
【问题讨论】：
-
if (SCheck1 != null && equals(UID))？总是假的
-
怎么样？对我来说，这似乎是一个正确的说法。
-
试图得到澄清。如果我错了，请纠正我：你想从共享首选项中提取 key4 值，如果 key4 的值与常量 UID 匹配，你想进入你的 IF 语句吗？对吗？
标签： java android sharedpreferences preferences