Question

I have a ListView that sits on the left side of a tablet-size screen. My goal was to give it a solid background with a border on the right, then apply an overlapping background on the list element to break that border so that it appears to be a part of the view on the right.

---

The ListView Background

I achieved the right border using a <layer-list> drawable as suggested by Emile in another question:

rightborder.xml

<?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="@color/black" />
        </shape>
    </item>
    <item android:right="2dp">
        <shape android:shape="rectangle">
            <solid android:color="@color/white" />
        </shape>
    </item>

</layer-list>

…and here’s the ListView definition for good measure:

<ListView
    android:id="@+id/msglist"
    android:layout_width="300dp"
    android:layout_height="match_parent"
    android:divider="@color/black"
    android:dividerHeight="1dp"
    android:background="@drawable/rightborder"
    android:paddingRight="0dip">
</ListView>
<!-- I added the android:paddingRight after reading something 
about shape drawables and padding, don't think it actually
did anything. -->

---

Attempting to override it with a color

In order to achieve the desired effect, I placed the following in the getView function of my adapter:

//If it's selected, highlight the background
if(position == mSelectedIndex)
    convertView.setBackgroundColor(R.color.light_gray);

else
    convertView.setBackgroundResource(0);

However, using this method, the black border of the ListView‘s drawable remained visible, and only the white part of the background was replaced by gray. Here’s a screen capture:

---

Fixing it with a drawable

On a hunch, I replaced the color I was assigning with a shape drawable:

selectedmessage.xml:

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

getView snippet:

//If it's selected, highlight the background
if(position == mSelectedIndex)
    convertView.setBackgroundResource(R.drawable.selectedmessage);

else
    convertView.setBackgroundResource(0);

This achieves the desired result, as shown below:

---

The question:

Why does assigning a rectangle as the background for my ListView element cover the entire view, while assigning the color allows the black border to show through? I’m happy it’s working, but I’d like to know why Android is rendering the view this way so I can learn more about how Android renders Views.

Other notes:

• I’m running the project in the stock Android 3.2 emulator, if that makes any
  difference.
• One clue may be that the light_gray color background seems to render darker than the light_gray shape resource.

• I doubt it makes a difference, but light_gray is:

  <color name="light_gray">#FFCCCCCC</color>

Answer 1

You can’t do this:

 convertView.setBackgroundColor(R.color.light_gray);

setBackgroundColor does not take a resource id : http://developer.android.com/reference/android/view/View.html#setBackgroundColor(int)

So your getting some incidental behaviour that isn’t doing what your expecting.

You would have to do:

 convertView.setBackgroundColor(getResources().getColor(R.color.light_gray);

http://developer.android.com/reference/android/content/res/Resources.html#getColor(int)