ShapeDrawable with black border on top

Question

I'm trying to create a ShapeDrawable for the background of my table cells. At the moment, I have a white rectangle with a 1px black border:

<?xml version="1.0" encoding="utf-8"?>
<shape
    xmlns:android="http://schemas.android.com/apk/res/android"
    android:shape= "rectangle"  >
    <solid android:color="#fff"/>
    <stroke android:width="1dp"  android:color="#000"/>
</shape>

I'd like to get this exact thing but with the border only along the top and bottom of the shape. Can someone show how to do this, or link to a page explaining how to use ShapeDrawable? Thanks!

use a nine patch drawable then – pskink Sep 02 '15 at 07:19 — pskink, Sep 02 '15 at 07:19

score 0 · Answer 1 · answered Sep 02 '15 at 02:51

You use a layer-list take a look at this

<layer-list xmlns:android="http://schemas.android.com/apk/res/android">
<item>
    <shape android:shape="rectangle">
        <solid android:color="#FF0000" />
    </shape>
</item>
<item android:top="5dp" android:bottom="5dp">
    <shape android:shape="rectangle">
        <solid android:color="#000000" />
    </shape>
</item>

Or you may just do a padding on your layout like this

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical" android:layout_width="match_parent"
android:layout_height="match_parent"
android:background="#f00">

<LinearLayout
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:background="#fff"
    android:layout_marginTop="5dp"android:layout_marginBottom="5dp"></LinearLayout>

</LinearLayout>

ShapeDrawable with black border on top

1 Answers1