Create a new color drawable

Question

I am trying to convert a hex value to an int so I can create a new color drawable. I'm not sure if this is possible, but according to the documentation, it should. It plainly asks for

public ColorDrawable (int color)

Added in API level 1 Creates a new ColorDrawable with the specified color.

Parameters color The color to draw.

So, my code isn't working because I'm getting an Invalid int: "FF6666" error. Any ideas?

int decode = Integer.decode("FF6666");
ColorDrawable colorDrawable = new ColorDrawable(decode);

if one of the answers provided solved your answer, think of validating it as an accepted answer so the other knows your problem is solved. — HpTerm, Oct 17 '13 at 11:30
There is a nice utility class android.graphics.Color that have a method parseColor. Try using it instead and don't forget to append the sharp (#) prefix to your colors string representation, e.g. #FF6666 — Игорь Комаров, Feb 27 '18 at 16:01

Enrichman · Accepted Answer · 2013-10-16T15:49:53.147

191

Since you're talking about hex you have to start with 0x and don't forget the opacity.

So basically: 0xFFFF6666

ColorDrawable cd = new ColorDrawable(0xFFFF6666);

You can also create a new colors.xml file into /res and define the colors like:

<?xml version="1.0" encoding="utf-8"?>
<resources>
    <color name="mycolor">#FF6666</color>
</resources>

and simply get the color defined in R.color.mycolor

getResources().getColor(R.color.mycolor)

edited Oct 16 '13 at 15:49

answered Oct 16 '13 at 15:44

Enrichman

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what i needed is `new ColorDrawable(getResources().getColor(R.color.red)))` thanks – shareef Dec 22 '15 at 18:55
9

ContextCompat.getColor(getContext(),R.color.red) for compatiability – loshkin Apr 07 '16 at 14:37
can we change background color by using this **cd** which is colorDrawable type variable – Jul 26 '16 at 08:53
On a side note, you can use ColorDrawable's for objects that require a Drawable such as DividerItemDecoration.setDrawable(); instead of using an actual drawable layout you can pass a ColorDrawable to simply change the divider's color. – 6rchid Feb 05 '20 at 18:29

score 30 · Answer 2 · answered May 18 '16 at 14:54

30

For using with ContextCompat and rehuse the color you can do something like this:

ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(this, R.color.white));

answered May 18 '16 at 14:54

JpCrow

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What's meant by `this`? – Christian Oct 16 '18 at 08:01
@Christian Context – JpCrow Oct 16 '18 at 13:32

score 10 · Answer 3 · edited May 23 '17 at 12:34

10

It should be like this...

ColorDrawable cd = new ColorDrawable(0xffff6666);

Note I used 8 hex digits, not 6 hex digit . which add to transparency

edited May 23 '17 at 12:34

Community

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answered Oct 16 '13 at 15:46

CRUSADER

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But you'd like to have such values in appropriate 'res' files, not hardcoded. Enrichman answer is complete :) – aMiGo Jun 02 '14 at 12:39
How is `cd` used in relation to `Toast`? – Azurespot Mar 11 '15 at 04:17

score 7 · Answer 4 · answered Dec 05 '17 at 09:11

7

By followingthe above advice,to be a summary of this question:

ColorDrawable colorDrawable = new ColorDrawable(Color.parseColor("#ce9b2c"));`
ColorDrawable colorDrawable = new ColorDrawable(0xFFCE9B2C); Note there is 8 hex digits, not 6 hex digit,which no work. Case all
ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(mContext,R.color.default_color));

Selecting up to you!

answered Dec 05 '17 at 09:11

BertKing

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If I have an array of color then ? – Prince Dholakiya Aug 08 '18 at 07:32
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@DPrince ，You can try this, `int color = Color.HSVToColor(new float[]{...}) ` and then use above . – BertKing Dec 29 '18 at 03:19

HpTerm · Answer 5 · 2013-10-17T21:19:27.177

I think you have to use :

public static int parseColor (String colorString)

Added in API level 1 Parse the color string, and return the corresponding color-int. If the string cannot be parsed, throws an IllegalArgumentException exception. Supported formats are: #RRGGBB #AARRGGBB red, blue, green, black, white, gray, cyan, magenta, yellow, lightgray, darkgray, grey, lightgrey, darkgrey, aqua, fuschia, lime, maroon, navy, olive, purple, silver, teal

score 1 · Answer 6 · answered Nov 04 '18 at 19:28

1

This is how I converted a Hex color to int and applied to a Background of a View

Let's say that we have a color #8080000.

1) Hex to int conversion

int myColor = Color.parseColor("#808000");

2) Set background

view.setBackgroundColor(context.getColor(myColor));

answered Nov 04 '18 at 19:28

Rohit Singh

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`(context.getColor(myColor)` are you serious?! `(context.getColor(COLOR_RESOURCE_ID)`, it resource color id – user924 Feb 09 '21 at 21:57
What do you mean? @user924 – Rohit Singh Feb 10 '21 at 03:33

score 0 · Answer 7 · answered Nov 28 '22 at 15:33

0

Xamarin/Maui :

  protected override void OnCreate(Bundle savedInstanceState)
    {
        base.OnCreate(savedInstanceState);      

        Window.SetBackgroundDrawable(new ColorDrawable(ColorExtensions.ToAndroid(Colors.Black)));
    }

answered Nov 28 '22 at 15:33

Nick Kovalsky

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Create a new color drawable

7 Answers7

1) Hex to int conversion

2) Set background

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