CODEWITHSUNDEEP
androidcssshapeslayer-list
learner
learner

Reputation: 11780

Android see-through white background with black border

I have a shape that allows me to have a white background with a 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="#FFFFFF" />
        </shape>
    </item>

</layer-list>

But I need a shape with a see-through white background and a black border. If I change #FFFFFF to #80FFFFFF, then the black of #000000 shows through. How do I get this right?

Upvotes: 0

Views: 1124

Answers (2)

Usmann
Usmann

Reputation: 31

Try this 

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

        <stroke android:width="1dp" 
android:color="#FFFFFF" />

    </shape>
</item>

Upvotes: 0

learner
learner

Reputation: 11780

I got the answer

<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>

Upvotes: 1

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