Change EditText password mask character to asterisk (*)

Question

Is there any way to change the password text from dot(.) to asterisk(*) .

Password is entering in edittext.

<EditText
        android:id="@+id/passWord1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:gravity="center"
        android:inputType="number"
        android:password="true"/>

Note to replies: the EditText shown *is already* `password=true`. The question is about how to display a `*` instead of a `·` character. — , Dec 27 '12 at 09:07
Example app doing just that: https://github.com/luskan/TextBoxWithPassword — marcinj, Jul 18 '23 at 11:13

Najib.Nj · Accepted Answer · 2022-01-09T07:36:19.637

Insert edittext in your xml file,

<EditText
    android:id="@+id/passWordEditText"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:gravity="center"
    android:inputType="textPassword"/>

and your class file go on and get findViewById from edittext and implement for this,

EditText edittext = (EditText)findViewById(R.id.passWordEditText);
edittext.setTransformationMethod(new AsteriskPasswordTransformationMethod());

and This class implement for that,

public class AsteriskPasswordTransformationMethod extends PasswordTransformationMethod {
    @Override
    public CharSequence getTransformation(CharSequence source, View view) {
        return new PasswordCharSequence(source);
    }
 
    private class PasswordCharSequence implements CharSequence {
        private CharSequence mSource;
        public PasswordCharSequence(CharSequence source) {
            mSource = source; // Store char sequence
        }
        public char charAt(int index) {
            return '*'; // This is the important part
        }
        public int length() {
            return mSource.length(); // Return default
        }
        public CharSequence subSequence(int start, int end) {
            return mSource.subSequence(start, end); // Return default
        }
    }
};

And If your code is Kotlin then you have to make separate java file then you have to use java with kotlin code.

Thanks, this works for me. I my case, I have to hide last 10 numbers, NOT all. So, I just have to override the method charAt(int index) to return '*' if index > 9 else return mSource.charAt(index). — toantran, Jun 05 '13 at 08:45
Nice work mate! Do you know what is the default char? In my case I have to show the first 3 char and than just the default password masking char, NOT the '*'. — narancs, May 10 '17 at 14:14
This removes the lapse where the actual text is displayed first before the 'password' character. Any idea how it will be retained? — sticky, May 27 '17 at 17:05
Has anyone tried putting a log in charAt() method, I see that its been called thousands of times. — abhishek maharajpet, Aug 05 '19 at 12:59

Ram kiran Pachigolla · Answer 2 · 2019-07-18T10:33:35.150

19

<EditText
        android:id="@+id/passWord1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:gravity="center"
        android:inputType="textPassword"//here is the change. check it once in your xml
        android:password="true"/>

In eclipse there will be hints when you click Ctrl + Space when you move cursor at android:inputType. Then you can see list of options. there you can select textPassword

If you want to see * in place of . then check this Android: Asterisk Password Field

edited Jul 18 '19 at 10:33

answered Dec 27 '12 at 09:04

Ram kiran Pachigolla

20,897
15
57
78

I think android:password="true" do the same. The password should be number . So I give android:inputType="number" – Jithin Sunny Dec 27 '12 at 09:11

score 10 · Answer 3 · answered Aug 17 '16 at 05:55

Got answer with help of link posted by Ram kiran

text.setTransformationMethod(new AsteriskPasswordTransformationMethod());


public class AsteriskPasswordTransformationMethod extends PasswordTransformationMethod {
@Override
public CharSequence getTransformation(CharSequence source, View view) {
    return new PasswordCharSequence(source);
}

private class PasswordCharSequence implements CharSequence {
    private CharSequence mSource;
    public PasswordCharSequence(CharSequence source) {
        mSource = source; // Store char sequence
    }
    public char charAt(int index) {
        return '*'; // This is the important part
    }
    public int length() {
        return mSource.length(); // Return default
    }
    public CharSequence subSequence(int start, int end) {
        return mSource.subSequence(start, end); // Return default
    }
}
};

thanks, this is exactly what i needed! what if i want to make it display the character and then asterisk-ize it after user types in another character, would i use this method alongside a textwatcher interface? — chornge, Aug 28 '17 at 22:40

score 5 · Answer 4 · answered Mar 19 '21 at 10:42

Kotlin

class AsteriskPasswordTransformationMethod : PasswordTransformationMethod() {

    override fun getTransformation(source: CharSequence, view: View): CharSequence {
        return PasswordCharSequence(source)
    }

    inner class PasswordCharSequence (private val source: CharSequence) : CharSequence {

        override val length: Int
            get() = source.length

        override fun get(index: Int): Char = '*'

        override fun subSequence(startIndex: Int, endIndex: Int): CharSequence {
            return source.subSequence(startIndex, endIndex)
        }
        
    }
    
}

score 3 · Answer 5 · answered Sep 29 '20 at 14:48

In Kotlin way:

class AsteriskPasswordTransformationMethod: PasswordTransformationMethod() {

override fun getTransformation(source: CharSequence?, view: View?): CharSequence {
    return super.getTransformation(source, view)
}


 abstract inner class PasswordCharSequence(val source: CharSequence) : CharSequence {
     override val length: Int
         get() = source.length

     override fun get(index: Int): Char =
         '*'

    override fun subSequence(start: Int, end: Int): CharSequence {
        return source.subSequence(start, end) // Return default
    }
}

}`

score 2 · Answer 6 · edited Apr 13 '18 at 13:28

2

EditText.setTransformationMethod(new PasswordTransformationMethod());

edited Apr 13 '18 at 13:28

rollstuhlfahrer

3,988
9
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38

answered Nov 28 '17 at 06:22

Suresh

701
1
6
20

score 0 · Answer 7 · answered Dec 27 '12 at 09:03

0

For android:inputType, there is a type of password.

answered Dec 27 '12 at 09:03

TieDad

9,143
5
32
58

score 0 · Answer 8 · answered Dec 27 '12 at 09:28

0

Try this

android:inputType="textPassword"

answered Dec 27 '12 at 09:28

NagarjunaReddy

8,621
10
63
98

I recommend this answer. `password` property is deprecated. – Capella Oct 06 '16 at 09:35

Change EditText password mask character to asterisk (*)

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