White background with a black border and transparency on an Android device

I'm trying to create a shape that has a white background with a black border.

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

    <item>
        <shape android:shape="rectangle" >
                <solid android:color="#000000" />

            <padding
                android:bottom="1dp"
                android:left="1dp"
                android:right="1dp"
                android:top="1dp" />
        </shape>
    </item>
    <item>
        <shape android:shape="rectangle" >
            <solid android:color="#80FFFFFF" />
        </shape>
    </item>

</layer-list>

However, changing the color code from #FFFFFF to #80FFFFFF results in some of the black showing through. How can I achieve the desired see-through effect while maintaining the black border?

androidcssshapeslayer-list

Answer №1

Here is the solution

<shape xmlns:android="http://schemas.android.com/apk/res/android" android:shape="rectangle" >
   <solid android:color="#80ffffff" />
   <stroke android:width="1dip" android:color="#000000"/>
</shape>

Answer №2

Give this a shot 

<object>
    <style android:theme="square" >
            <background android:color="#FFFFFF" />

        <border android:size="2dp" 
android:shade="#333333" />

    </style>
</object>

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