Android button selector background color

Question

I'm trying to implement button with configurable colors for default and pressed states and rounded corners. Here is a solution I've found on SO and seems it work fine on my real test devices:

    GradientDrawable gradientStateNormal = new GradientDrawable(GradientDrawable.Orientation.TOP_BOTTOM, new int[]
            {buttonBaseBgColor, buttonBaseBgColor});
    gradientStateNormal.setShape(GradientDrawable.RECTANGLE);
    gradientStateNormal.setCornerRadius(cornerRadiusPX);

    GradientDrawable gradientStatePressed = new GradientDrawable(GradientDrawable.Orientation.TOP_BOTTOM, new int[]
            {colorLight, colorLight});
    gradientStatePressed.setShape(GradientDrawable.RECTANGLE);
    gradientStatePressed.setCornerRadius(cornerRadiusPX);

    StateListDrawable states = new StateListDrawable();
    states.addState(new int[] {android.R.attr.state_pressed},gradientStatePressed);
    states.addState(StateSet.WILD_CARD, gradientStateNormal);
    signUpButton.setBackground(states);

But both Android 4.3 and 4.4 Emulators draw such black corners after press on button:

Could you please explain what's wrong here and could such case appear on the real device, not only emulator?

Answer 1

You can do that by using the xml itself. Make an xml file, for eg:shape.xml as given below:

 <?xml version="1.0" encoding="utf-8"?>
 <shape xmlns:android="http://schemas.android.com/apk/res/android" >

<gradient android:angle="180"
    android:startColor="#FFFF00"
    android:centerColor="#FFFF00"
    android:endColor="#FFFF00"/>

<corners 
    android:radius="20dp"/>

<padding
    android:bottom="10dp"
    android:left="10dp"
    android:right="10dp"
    android:top="10dp" />

<stroke
    android:width="2dp"
    android:color="#000000" />

</shape>

Now use this xml as the background for the button in your main.xml like this:

 android:background="@drawable/shape