Can't Compare EditText Value with SharedPreference Value

Question

When I enter a value in the EditText, i can't to compare it with a Value in My SharedPreference .

This is my EditText Preference

<EditTextPreference

        android:defaultValue="1234"
        android:inputType="number"
        android:key="userid"
        android:maxLength="4"
        android:singleLine="true"
        android:summary="To Access App setting set a PIN"
        android:title="Set PIN" />

And This is My Methode code:

private void GoToSettings(){
    SharedPreferences prefs = PreferenceManager.getDefaultSharedPreferences(getApplication());
    String userpin = prefs.getString("userid", "");
    String userpin1 = txtpin.getText().toString();
    if(userpin.equals(userpin1)){
                        /*Intent intent = new Intent(FingerprintScannerActivity.this, PreferencesActivity.class);
                        startActivity(intent);*/
        Toast.makeText(getApplicationContext(),
                "Good",
                Toast.LENGTH_SHORT).show();
    }else Toast.makeText(getApplicationContext(),
            "Error PIN",
            Toast.LENGTH_SHORT).show();

}

Solved : I forget to load PreferenceManager.setDefaultValues(this, R.xml.preferences_xml, false); — Ysen Eycen, Nov 04 '16 at 21:19

score 0 · Answer 1 · edited May 23 '17 at 12:00

0

Whenever comparing Strings use the following approach

string1.equals(string2)

Don't use '==' operator for strings
See this for more detailed answer Java String.equals versus ==

edited May 23 '17 at 12:00

Community

1
1

answered Nov 04 '16 at 20:31

Harjot Singh Oberai

282
2
14

The same problem , When I click the button message appears in the instruction else – Ysen Eycen Nov 04 '16 at 20:40

score 0 · Accepted Answer · answered Nov 04 '16 at 20:43

0

SharedPreferences prefs = PreferenceManager.getDefaultSharedPreferences(getApplication());
String userpin = prefs.getString("userid", "");

change it to : String userpin = prefs.getString("userid", null); its more efficient to return null instead of empty string and then you can use:

if((userpin == null) ? (userpin1 == null) : userpin.equalsIgnoreCase(userpin1))

answered Nov 04 '16 at 20:43

Lior

832
6
6

Error PIN Appear Again :s :s – Ysen Eycen Nov 04 '16 at 20:49
@YsenEycen it doesnt mean that the comparison was worng. print here the value of the 'userid' value in the sharedPref `Log.d(TAG, userpin + " " userpin1);` – Lior Nov 04 '16 at 20:58
Code work only when i open the setting activity and i set the value in the editText , otherwise without opening the setting activity code dosn't work – Ysen Eycen Nov 04 '16 at 21:08
Solved : I forget to load PreferenceManager.setDefaultValues(this, R.xml.preferences_xml, false); – Ysen Eycen Nov 04 '16 at 21:17

Can't Compare EditText Value with SharedPreference Value

2 Answers2