Question

I created a custom button:

<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
    <item android:state_pressed="true">
        <shape>
            <gradient android:startColor="#5a803f"
                android:endColor="#446c1a" android:angle="90" />
            <stroke android:width="5dp" android:color="#3d5e1d" />
            <corners android:radius="3dp" />
            <padding android:left="8dp" android:top="8dp"
                android:right="8dp" android:bottom="8dp" />
        </shape>
    </item>
    <item android:state_focused="true">
        <shape>
            <gradient android:startColor="#5a803f"
                android:endColor="#446c1a" android:angle="90" />
            <stroke android:width="5dp" android:color="#3d5e1d" />
            <corners android:radius="3dp" />
            <padding android:left="8dp" android:top="8dp"
                android:right="8dp" android:bottom="8dp" />
        </shape>
    </item>
    <item>

    </item>

</selector>

But I need to place this button to the black background with defined margins and with white borders at the top and bottom. And this background shouldn’t be clickable. Something like this:

How can I implement this? with the help of this selector or style using. Help please with a code sample.

Answer 1

I solved this issue using the TableLayout. In this case my TextView is looks like:

   <TableRow
    android:id="@+id/tableRow3"
    style="@style/button_row"
    >
    <Button

        android:id="@+id/btnProceed"
        style="@style/custom_button"
        android:layout_centerHorizontal="true"
        android:text="Log In"
         />
    </TableRow>

etc for all UI elements.