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android – How to draw border on just one side of a linear layout?

Posted by: admin March 10, 2020 Leave a comment

Questions:

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.

How to&Answers:

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:

Easy as pie, allowing a transparent bg:

<?xml version="1.0" encoding="utf-8"?>
<shape xmlns:android="http://schemas.android.com/apk/res/android"
    android:shape="rectangle">
    <gradient
        android:angle="0"
        android:startColor="#f00"
        android:centerColor="@android:color/transparent"
        android:centerX="0.01" />
</shape>

Change the angle to change border location:

  • 0 = left
  • 90 = bottom
  • 180 = right
  • 270 = top

Answer:

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

Answer:

To get a border on just one side of a drawable, apply a negative inset to the other 3 sides (causing those borders to be drawn off-screen).

<?xml version="1.0" encoding="utf-8"?>
<inset xmlns:android="http://schemas.android.com/apk/res/android"
    android:insetTop="-2dp" 
    android:insetBottom="-2dp"
    android:insetLeft="-2dp">

    <shape android:shape="rectangle">
        <stroke android:width="2dp" android:color="#FF0000" />
        <solid android:color="#000000" />
    </shape>

</inset>

enter image description here

This approach is similar to naykah’s answer, but without the use of a layer-list.

Answer:

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:

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:

<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
    <item android:state_pressed="true" >
        <shape>
            <solid
                android:color="#f28b24" />
            <stroke
                android:width="1dp"
                android:color="#f28b24" />
            <corners
                android:radius="0dp"/>
            <padding
                android:left="0dp"
                android:top="0dp"
                android:right="0dp"
                android:bottom="0dp" />
        </shape>
    </item>
    <item>
        <shape>
            <gradient
                android:startColor="#f28b24"
                android:endColor="#f28b24"
                android:angle="270" />
            <stroke
                android:width="0dp"
                android:color="#f28b24" />
            <corners
                android:bottomLeftRadius="8dp"
                android:bottomRightRadius="0dp"
                android:topLeftRadius="0dp"
                android:topRightRadius="0dp"/>
            <padding
                android:left="10dp"
                android:top="10dp"
                android:right="10dp"
                android:bottom="10dp" />
        </shape>
    </item>
</selector>

Answer:

There is no mention about nine-patch files here. Yes, you have to create the file, however it’s quite easy job and it’s really “cross-version and transparency supporting” solution. If the file is placed to the drawable-nodpi directory, it works px based, and in the drawable-mdpi works approximately as dp base (thanks to resample).

Example file for the original question (border-right:1px solid red;) is here:

http://ge.tt/517ZIFC2/v/3?c

Just place it to the drawable-nodpi directory.

Answer:

Borders of different colors. I used 3 items.

<?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/colorAccent" />
        </shape>
    </item>
    <item android:top="3dp">
        <shape android:shape="rectangle">
            <solid android:color="@color/light_grey" />
        </shape>
    </item>
    <item
        android:bottom="1dp"
        android:left="1dp"
        android:right="1dp"
        android:top="3dp">
        <shape android:shape="rectangle">
            <solid android:color="@color/colorPrimary" />
        </shape>
    </item>
</layer-list>

Answer:

An other great example

<?xml version="1.0" encoding="UTF-8" standalone="no"?> 
<inset xmlns:android="http://schemas.android.com/apk/res/android"
     android:insetRight="-2dp">

     <shape android:shape="rectangle">
         <corners
             android:bottomLeftRadius="4dp"
             android:bottomRightRadius="0dp"
             android:topLeftRadius="4dp"
             android:topRightRadius="0dp" />
         <stroke
             android:width="1dp"
             android:color="@color/nasty_green" />
         <solid android:color="@android:color/transparent" />
     </shape>

</inset>