CODEWITHSUNDEEP
androidactionbarsherlockandroid-actionbarshapedrawable
user460847
user460847

Reputation: 1618

Dynamically change action bar divider color (android:bottom for programmatically generated ShapeDrawable)?

I'm trying to change the color of the divider bar at the bottom of the action bar programmatically. My strategy is to set the action bar background to a programmatically generated LayerDrawable containing ShapeDrawable rectangles, based on this XML:

<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android">
    <!-- Bottom Line -->
    <item>
        <shape android:shape="rectangle">
            <solid android:color="@color/action_bar_line_color" />
        </shape>
    </item>

    <!-- Color of your action bar -->
    <item android:bottom="2dip">
        <shape android:shape="rectangle">
            <solid android:color="@color/action_bar_color" />
        </shape>
    </item>
</layer-list>

But I've hit a roadblock: I can't figure out how to apply an android:bottom property (as in <item android:bottom="2dip">) programmatically. Obviously android:bottom is a property of the item tag, to which (I think) there's no programmatic equivalent, and I haven't been able to find any methods/properties of ShapeDrawable that look appropriate.

Code so far:

public LayerDrawable createABBackground(String color) {
    ShapeDrawable rect = new ShapeDrawable(new RectShape());
    rect.getPaint().setColor(Color.parseColor("#000000"));
    ShapeDrawable rect2 = new ShapeDrawable(new RectShape());
    rect2.getPaint().setColor(Color.parseColor(color));
    ShapeDrawable[] layers = {rect, rect2};
    LayerDrawable background = new LayerDrawable(layers);
    return background;
}

Ideas? If it matters for alternative solutions, I'm using ActionBarSherlock.

EDIT:

setLayerInset, as suggested by MH, did what I wanted. Here's a modified version of the function using it:

public LayerDrawable createABBackground(String color) {
    ShapeDrawable rect = new ShapeDrawable(new RectShape());
    rect.getPaint().setColor(Color.parseColor(color));
    ShapeDrawable rect2 = new ShapeDrawable(new RectShape());
    rect2.getPaint().setColor(Color.parseColor("#000000"));
    ShapeDrawable[] layers = {rect, rect2};
    LayerDrawable background = new LayerDrawable(layers);
    background.setLayerInset(0, 0, 3, 0, 0);
    background.setLayerInset(1, 0, 0, 0, 3);
    return background;
}

Upvotes: 4

Views: 1268

Answers (4)

brandall
brandall

Reputation: 6144

This worked for me:

<layer-list xmlns:android="http://schemas.android.com/apk/res/android" >

    <item android:id="@+id/actionBarLineParent">
        <shape
            android:id="@+id/actionBarLine"
            android:shape="rectangle" >
            <solid android:color="@color/red_all_files" />
        </shape>
    </item>

    <item
        android:id="@+id/actionBarColourParent"
        android:bottom="2dip">
        <shape
            android:id="@+id/actionBarColour"
            android:shape="rectangle" >
            <solid android:color="@android:color/white" />
        </shape>
    </item>

</layer-list>

Then calling:

final LayerDrawable ld = (LayerDrawable) getResources().getDrawable(R.drawable.actionbar_background);
ld.setDrawableByLayerId(R.id.actionBarLineParent, new ColorDrawable(Color.BLUE));
getActionBar().setBackgroundDrawable(ld);

You need to make sure you set the android:id

Upvotes: 0

MH.
MH.

Reputation: 45503

As per earlier comment:

Did you give setLayerInset(int index, int l, int t, int r, int b) a try? The docs say:

Specify modifiers to the bounds for the drawable[index]. left += l top += t; right -= r; bottom -= b;

Upvotes: 3

Tavian Barnes
Tavian Barnes

Reputation: 12932

Judging by the source code for LayerDrawable, the method which would allow you to do that is private. But you might be able to accomplish this with a LayerDrawable holding an InsetDrawable

Upvotes: 1

Simon
Simon

Reputation: 11190

If you can elaborate as to what reason you are changing the color programmatically? It seems a little pointless to do this if you are not also controlling the background color. If you are controlling the background color then you are probably using styles in which case I can think of several methods of doing this.

Simon

Upvotes: 0

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