if-else part is not working

Question

Why the if-else part isn't working in the code below. I checked the value of success just before if and it is returning true but the code inside if isn't working.

protected void onPostExecute(Void aVoid) {
              super.onPostExecute(aVoid);    
              dialog.dismiss();
              System.out.println(success);
              if(success == "true")
              {
                  Toast t = Toast.makeText(getApplicationContext(), "Successfully logged in", Toast.LENGTH_SHORT);
                    t.show();
                    Intent intent = new Intent(MainActivity.this, Quote_Details.class);         
                    intent.putExtra("sessionId", sessionId);
                    startActivity(intent);
              }
              else if(success == "fals")
              {
               Toast f = Toast.makeText(getApplicationContext(), "LOGIN FAILED", Toast.LENGTH_LONG);
               f.show();
              }
             }

actually `success` is `String`. Thanks for pointing out my mistake. SOLVED :) — Amit Anand, Aug 14 '14 at 08:37
@help Alright, wasn't sure as to what success was hence why I gave my answer with both boolean or String datatypes! — Juxhin, Aug 14 '14 at 08:40

score 5 · Answer 1 · answered Aug 14 '14 at 08:31

5

To compare strings, use ".equals".

E.g.:

if (success.equals("true"))

EDIT: oops, someone was faster... well, at least you have an example too now :)

answered Aug 14 '14 at 08:31

Wildcopper

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Juxhin · Answer 2 · 2014-08-18T17:40:37.580

3

Since 'success' is a boolean variable(Or so it seems), you have to compare like you would compare all primitive values. So if(success){ } or if(success == true){ }. I would like to clarify the following - if(success) tells the if-statement to return value of success, if success is true, continue the statement, if false it returns false/doesn't continue the statement. It's a nicer way of doing it rather than if(success == true)

If success is a String then, if(success.equalsIgnoreCase("true")

This is because String is an object and not a primitive datatype like int, boolean, char etc.

You also compare objects the same way incase you required to do so in the future

edited Aug 18 '14 at 17:40

answered Aug 14 '14 at 08:31

Juxhin

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2

Up voting this, as indeed, we don't know what success is exactly (String or boolean). – Wildcopper Aug 14 '14 at 08:35
1

@Wildcopper agree :) by seeing code we dont know data type of success #SMART – Android is everything for me Aug 14 '14 at 08:37

score 2 · Accepted Answer · answered Aug 14 '14 at 08:30

2

You should compare strings with equals method, not with == operator. Operator == just compares the object references, not the content of the objects.

answered Aug 14 '14 at 08:30

nogard

9,067
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oops... thanks.. solved... :) – Amit Anand Aug 14 '14 at 08:34

score 2 · Answer 4 · edited Aug 14 '14 at 08:38

When you are dealing with String-objects use equals() to compare them:

protected void onPostExecute(Void aVoid) {
              super.onPostExecute(aVoid);    
              dialog.dismiss();
              System.out.println(success);
              if(success.equals("true"))
              {
                  Toast t = Toast.makeText(getApplicationContext(), "Successfully logged in", Toast.LENGTH_SHORT);
                    t.show();
                    Intent intent = new Intent(MainActivity.this, Quote_Details.class);         
                    intent.putExtra("sessionId", sessionId);
                    startActivity(intent);
              }
              else if(success.equals("fals"))
              {
               Toast f = Toast.makeText(getApplicationContext(), "LOGIN FAILED", Toast.LENGTH_LONG);
               f.show();
              }
             }

if-else part is not working

4 Answers4