How to draw border on just one side of a linear la

Question

I'm able to draw border to a linear layout, but it is getting drawn on all sides. I want to restrict it to right side only, like you do in CSS (border-right:1px solid red;).

I've tried this, but it still draws on all sides:

<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android" >
<item>
    <shape android:shape="rectangle" >
        <stroke
            android:height="2dp"
            android:width="2dp"
            android:color="#FF0000" />

        <solid android:color="#000000" />

        <padding
            android:bottom="0dp"
            android:left="0dp"
            android:right="1dp"
            android:top="0dp" />

        <corners
            android:bottomLeftRadius="0dp"
            android:bottomRightRadius="5dp"
            android:radius="1dp"
            android:topLeftRadius="5dp"
            android:topRightRadius="0dp" />
    </shape>
</item>

Any suggestions on how to accomplish this?

BTW, I do not want to use the hack of putting a view of width 1dp on the required side.

Answer 1

As an alternative (if you don't want to use background), you can easily do it by making a view as follows:

<View
    android:layout_width="2dp"
    android:layout_height="match_parent"
    android:background="#000000" />

For having a right border only, place this after the layout (where you want to have the border):

<View
    android:layout_width="2dp"
    android:layout_height="match_parent"
    android:background="#000000" />

For having a left border only, place this before the layout (where you want to have the border):

Worked for me...Hope its of some help....

Answer 2

I was able to achieve the effect with the following code

<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android" >
    <item android:left="0dp" android:right="-5dp" android:top="-5dp" android:bottom="-5dp">
        <shape
        android:shape="rectangle">
            <stroke android:width="1dp" android:color="#123456" />
        </shape>
    </item>
</layer-list>

You can adjust to your needs for border position by changing the direction of displacement

Answer 3

You can use this to get border on one side

<?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="#FF0000" />
    </shape>
</item>
<item android:left="5dp">
    <shape android:shape="rectangle">
        <solid android:color="#000000" />
    </shape>
</item>
</layer-list>

EDITED

As many including me wanted to have a one side border with transparent background, I have implemented a BorderDrawable which could give me borders with different size and color in the same way as we use css. But this could not be used via xml. For supporting XML, I have added a BorderFrameLayout in which your layout can be wrapped.

See my github for the complete source.

Answer 4

it is also possible to implement what you want using a single layer

<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android" >

    <item
        android:bottom="-5dp"
        android:right="-5dp"
        android:top="-5dp">
        <shape android:shape="rectangle" >
            <solid android:color="@color/color_of_the_background" />

            <stroke
                android:width="5dp"
                android:color="@color/color_of_the_border" />
        </shape>
    </item>

</layer-list>

this way only left border is visible but you can achieve any combination you want by playing with bottom, left, right and top attributes of the item element