if-else block not working properly in android

Question

Scenario: when the focus is lost from an EditText, I'm checking if it contains null (in the first if block).
If so, then I'll show a Toast.
In the else-if block I'm checking if the EditText doesn't contain letters.
Then I'll show a toast, but when I run the application, the Toast is shown even on a correct input.
I.e.: If I enter any letter the Toast should not be shown, it should be shown only when a null or digit/special symbol is entered.

Here is the code

et1.setOnFocusChangeListener(new View.OnFocusChangeListener() {

        @Override
        public void onFocusChange(View v, boolean hasFocus) {
            // TODO Auto-generated method stub
            if(!hasFocus)
            {
                a = et1.getText().toString();
                if (a.equals(""))
                {
                    Toast.makeText(getApplicationContext(), "Your entry is incorrect!!", Toast.LENGTH_LONG).show();
                }
                else if (!a.contains("[a-z]")||!a.contains("[A-Z]")) {
                    Toast.makeText(getApplicationContext(), "Your entry is incorrect!!", Toast.LENGTH_LONG).show();

                }
                else
                {
                }

Please help

possible duplicate of [How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) — njzk2, Jul 25 '15 at 18:41
i don't understand, didn't you read the documentation for `contains`, first? — njzk2, Jul 25 '15 at 18:43

Phantômaxx · Answer 1 · 2015-07-25T18:53:52.890

1

It's not working because:

if (a == "")

won't work in Java

Use

if (a.equals(""))

instead

Also, String.contains doesn't use regular expressions, but CharacterSequences.
So, unless your string doesn't contain the exact character sequences "[a-z]" or "[A-Z]" (and only one of these 2 strings), you'll never get a match.

See: http://developer.android.com/reference/java/lang/String.html#contains(java.lang.CharSequence)

edited Jul 25 '15 at 18:53

answered Jul 25 '15 at 18:40

Phantômaxx

37,901
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1

Just to add clarity: for objects (a == "") compares the references, in memory, between a and "". Using the String.equals(a,b) method compares the values of the strings. – lacraig2 Jul 25 '15 at 18:41
1

@lacraig2 Thank you for the enrichment. – Phantômaxx Jul 25 '15 at 18:42
1

In Android we have also _TextUtils.isEmpty(String)_ which compares against _null_ and empty string. – harism Jul 25 '15 at 18:43
that's actually not the problem – njzk2 Jul 25 '15 at 18:43
This answer is becoming richer and richer, thank to your comments! – Phantômaxx Jul 25 '15 at 18:45
thanks for th answer,,,but problem is not solved completely, still if im entering letter it is showing me toast,,, why? – user5075023 Jul 25 '15 at 18:48
Because `String.contains` doesn't work on regexes, but on CharacterSequences. So, unless your string doesn't contain the exact sequances "[a-z]" or "[A-Z]" (and only one of these 2 strings), you'll never get a match. See: http://developer.android.com/reference/java/lang/String.html#contains(java.lang.CharSequence) – Phantômaxx Jul 25 '15 at 18:51
ohhkkk,,,,, so how to compare?? – user5075023 Jul 25 '15 at 18:53
This makes for a different question. – Phantômaxx Jul 25 '15 at 18:55

score 1 · Answer 2 · edited May 23 '17 at 12:21

1

The '==' operator only compares references. To compare string values you must use the equals() method.

Instead of

if (a == "")

use

if (a.equals(""))

See: What is the difference between == vs equals() in Java?

edited May 23 '17 at 12:21

Community

1
1

answered Jul 25 '15 at 18:45

Zarwan

5,537
4
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score 0 · Answer 3 · answered Jul 25 '15 at 18:42

0

The problem is:

if (a == "")

Strings can't be compared like this. Instead, check for size equal to 0, or against a specific string with the equals() method.

answered Jul 25 '15 at 18:42

Larry Schiefer

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if-else block not working properly in android

3 Answers3