655

How do I convert a hexadecimal color string like #b74093 to a Color in Flutter?

I want to use a HEX color code in Dart.

2
  • 2
    in google chrome devtools you can quickly get the decimal numbers for a hex-style color; just click on the tiny color rectangle in the styles sidebar; in the popup you can switch the representation between HSLA, RGBA, HEX by clicking the chevron. Then you can use Color.fromRGBO()
    – ccpizza
    Commented Dec 4, 2020 at 17:25
  • Nice, but this is a programmer website so we generally want to know how to do it in code. Commented Sep 6, 2022 at 12:20

33 Answers 33

983

In Flutter, the Color class only accepts integers as parameters, or there is the possibility to use the named constructors fromARGB and fromRGBO.

So we only need to convert the string #b74093 to an integer value. Also we need to respect that opacity always needs to be specified.
255 (full) opacity is represented by the hexadecimal value FF. This already leaves us with 0xFF. Now, we just need to append our color string like this:

const color = const Color(0xffb74093); // Second `const` is optional in assignments.

The letters can by choice be capitalized or not:

const color = const Color(0xFFB74093);

If you want to use percentage opacity values, you can replace the first FF with the values from this table (also works for the other color channels).

Extension class

Starting with Dart 2.6.0, you can create an extension for the Color class that lets you use hexadecimal color strings to create a Color object:

extension HexColor on Color {
  /// String is in the format "aabbcc" or "ffaabbcc" with an optional leading "#".
  static Color fromHex(String hexString) {
    final buffer = StringBuffer();
    if (hexString.length == 6 || hexString.length == 7) buffer.write('ff');
    buffer.write(hexString.replaceFirst('#', ''));
    return Color(int.parse(buffer.toString(), radix: 16));
  }

  /// Prefixes a hash sign if [leadingHashSign] is set to `true` (default is `true`).
  String toHex({bool leadingHashSign = true}) => '${leadingHashSign ? '#' : ''}'
      '${alpha.toRadixString(16).padLeft(2, '0')}'
      '${red.toRadixString(16).padLeft(2, '0')}'
      '${green.toRadixString(16).padLeft(2, '0')}'
      '${blue.toRadixString(16).padLeft(2, '0')}';
}

The fromHex method could also be declared in a mixin or class because the HexColor name needs to be explicitly specified in order to use it, but the extension is useful for the toHex method, which can be used implicitly. Here is an example:

void main() {
  final Color color = HexColor.fromHex('#aabbcc');

  print(color.toHex());
  print(const Color(0xffaabbcc).toHex());
}

Disadvantage of using hex strings

Many of the other answers here show how you can dynamically create a Color from a hex string, like I did above. However, doing this means that the color cannot be a const.
Ideally, you would assign your colors the way I explained in the first part of this answer, which is more efficient when instantiating colors a lot, which is usually the case for Flutter widgets.

14
  • You can't have static method extension in dart yet github.com/dart-lang/language/issues/723 Commented May 18, 2020 at 15:03
  • I'm just surprised that you can pass 0xFF as part of an int
    – HJo
    Commented Jun 18, 2020 at 14:09
  • @HamishJohnson 0x just indicates that the following digits will be parsed as a hexadecimal 🙃 Commented Jun 18, 2020 at 16:11
  • Is there any way to use 50% or any % of these colors? Like something Color(0xFF183451)[50]?
    – Ankit
    Commented Aug 12, 2021 at 12:48
  • @Ankit You can use this table and replace the FF in the leading 0xFF of your Color object with the value you need. So in your case that would be Color(0x80183451). Commented Aug 12, 2021 at 12:59
282

The Color class expects an ARGB integer. Since you try to use it with an RGB value, represent it as int and prefix it with 0xff.

Color mainColor = Color(0xffb74093);

If you get annoyed by this and still wish to use strings, you can extend Color and add a string constructor

class HexColor extends Color {
  static int _getColorFromHex(String hexColor) {
    hexColor = hexColor.toUpperCase().replaceAll("#", "");
    if (hexColor.length == 6) {
      hexColor = "FF" + hexColor;
    }
    return int.parse(hexColor, radix: 16);
  }

  HexColor(final String hexColor) : super(_getColorFromHex(hexColor));
}

Usage

Color color1 = HexColor("b74093");
Color color2 = HexColor("#b74093");
Color color3 = HexColor("#88b74093"); // If you wish to use ARGB format
2
  • 2
    the HexColor class did not work for MaterialColor for me, it keeps complaining of the second parameter. Please help here
    – CanCoder
    Commented Jan 9, 2020 at 22:51
  • .toUpperCase() giving error. Removed it and working now Commented Jun 11, 2021 at 7:58
104

If you want to use the hexadecimal code of a color which is in the format #123456, then it is very easy to do. Create a variable of type Color and assign the following values to it.

Color myHexColor = Color(0xff123456)

// Here you notice I use the 0xff and that is the opacity or transparency
// of the color and you can also change these values.

Use myHexColor and you are ready to go.

If you want to change the opacity of color direct from the hexadecimal code, then change the ff value in 0xff to the respective value from the table below. (alternatively you can use

myHexColor.withOpacity(0.2)

it is easier way to do it. 0.2 is mean 20% opacity)

Hexadecimal opacity values

100% — FF

95% — F2

90% — E6

85% — D9

80% — CC

75% — BF

70% — B3

65% — A6

60% — 99

55% — 8C

50% — 80

45% — 73

40% — 66

35% — 59

30% — 4D

25% — 40

20% — 33

15% — 26

10% — 1A

5% — 0D

0% — 00
4
  • 2
    It's a good idea to save this reference, although the .withOpacity(0.2) is useful enough i most of the cases. Commented Jun 8, 2020 at 11:19
  • This is grate but sometimes It shows whats going below the widgets for example when SliverAppBar scrolls it shows whats below that widget because of I'm reducing the opacity. Is there anyway to fade the color without reducing opacity. Commented Jan 23, 2021 at 7:58
  • @nipunravisara get the hex of the color and use 100% opacity Commented Jan 24, 2021 at 7:37
  • @ZakriaKhan Thanks actually I found a way. stackoverflow.com/questions/65856982/… Commented Jan 24, 2021 at 11:22
75

How to use a hexadecimal color code #B74093 in Flutter

Simply remove the # sign from the hexadecimal color code and add 0xFF with the color code inside the Color class:

#b74093 will become Color(0xffb74093) in Flutter

#B74093 will become Color(0xFFB74093) in Flutter

The ff or FF in Color(0xFFB74093) defines the opacity.

Hexadecimal colors example with all opacity types in Dartpad

Enter image description here

59

Easy way:

String color = yourHexColor.replaceAll('#', '0xff');

Usage:

Container(
  color: Color(int.parse(color)),
)
0
31

Thanks for asking this question, simples solution is as:

// Color to Hex String

colorToHexString(Color color) {
  return '#FF${color.value.toRadixString(16).substring(2, 8)}';
}

// Hex String to Color

hexStringToColor(String hexColor) {
  hexColor = hexColor.toUpperCase().replaceAll("#", "");
  if (hexColor.length == 6) {
    hexColor = "FF" + hexColor;
  }
  return Color(int.parse(hexColor, radix: 16));
}

// How to call function

String hexCode = colorToHexString(Colors.green);
Color bgColor = hexStringToColor(hexCode);
print("$hexCode = $bgColor");

Enjoy code and help others :)

0
28

A simple function without using a class:

Color _colorFromHex(String hexColor) {
  final hexCode = hexColor.replaceAll('#', '');
  return Color(int.parse('FF$hexCode', radix: 16));
}

You can use it like this:

Color color1 = _colorFromHex("b74093");
Color color2 = _colorFromHex("#b74093");
1
  • 6
    Thansk to @jeroen-meijer for the edit. In fact you can use this one liner as well if you feel fancy Color(int.parse('#000000'.replaceAll('#', '0xff'))) Commented Dec 20, 2019 at 20:21
27

To convert from a hexadecimal string to integer, do:

int hexToInt(String hex)
{
  int val = 0;
  int len = hex.length;
  for (int i = 0; i < len; i++) {
    int hexDigit = hex.codeUnitAt(i);
    if (hexDigit >= 48 && hexDigit <= 57) {
      val += (hexDigit - 48) * (1 << (4 * (len - 1 - i)));
    } else if (hexDigit >= 65 && hexDigit <= 70) {
      // A..F
      val += (hexDigit - 55) * (1 << (4 * (len - 1 - i)));
    } else if (hexDigit >= 97 && hexDigit <= 102) {
      // a..f
      val += (hexDigit - 87) * (1 << (4 * (len - 1 - i)));
    } else {
      throw new FormatException("Invalid hexadecimal value");
    }
  }
  return val;
}

Call example:

Color color = new Color(hexToInt("FFB74093"));
3
  • Lots of magic numbers. Aren't named constants supported in this language / environment? Commented Apr 26, 2021 at 20:56
  • And (1 << (4 * (len - 1 - i))) is repeated three times. Commented May 4, 2021 at 14:00
  • 1
    This can be done much easier: int hexToInt(String hex){ return int.parse('0x$hex'); }
    – Ivo
    Commented Aug 4, 2022 at 6:46
26

There is another solution. If you store your color as a normal hexadecimal string and don't want to add opacity to it (leading "FF"):

  1. Convert your hexadecimal string to int To convert a hexadecimal string to an integer, do one of the following:

     var myInt = int.parse(hexString, radix: 16);
    

    or

     var myInt = int.parse("0x$hexString");
    

    as a prefix of 0x (or -0x) will make int.parse default to a radix of 16.

  2. Add opacity to your color via code

     Color color = new Color(myInt).withOpacity(1.0);
    
3
  • 1
    I needed the "leading FF" for Flutter's ThemeData. Commented Jul 11, 2018 at 20:56
  • I like this solution for its simplicity, but as @creativecreatorormaybenot mentioned, the leading FF is still required.
    – KevinM
    Commented Jan 16, 2019 at 18:45
  • - withOpacity is important Commented Jan 26 at 11:16
21

This was the solution for me:

String hexString = "45a3df";
Color(int.parse("0xff${hexString}"));

It was the only way that didn't require additional steps.

3
  • 2
    your solution is working for me. Are there any downsides, anyone? Commented Aug 6, 2020 at 16:36
  • Or Color(int.parse(hexString.replaceFirst('#', '0xff'))) if I know for sure that the input hex string is in #RRGGBB format (such as #45a3df). Commented Nov 3, 2021 at 12:46
  • Do not forget that hex can contain 8 digits Commented Mar 16, 2023 at 12:18
16

Use hexcolor for bringing hexadecimal colors to the Dart hexcolorPlugin:

hexcolor: ^2.0.3

Sample usage

import 'package:hexcolor/hexcolor.dart';
Container(
    decoration: new BoxDecoration(
        color: Hexcolor('#34cc89'),
    ),
    child: Center(
        child: Text(
            'Running on: $_platformVersion\n',
            style: TextStyle(color: Hexcolor("#f2f2f2")),
        ),
    ),
),
15

In Flutter, to create a color from RGB with alpha, use:

return new Container(
  color: new Color.fromRGBO(0, 0, 0, 0.5),
);

How to use hexadecimal color:

return new Container(
  color: new Color(0xFF4286f4),
);
// 0xFF -> the opacity (FF for opaque)
// 4286f4 -> the hexadecimal color

Hexadecimal color with opacity:

return new Container(
  color: new Color(0xFF4286f4).withOpacity(0.5),
);

// Or change the "FF" value

100% — FF
 95% — F2
 90% — E6
 85% — D9
 80% — CC
 75% — BF
 70% — B3
 65% — A6
 60% — 99
 55% — 8C
 50% — 80
 45% — 73
 40% — 66
 35% — 59
 30% — 4D
 25% — 40
 20% — 33
 15% — 26
 10% — 1A
 5% — 0D
 0% — 00

For more, see the official documentation page, Color class - dart:ui library - Dart API.

12

You can use this

Color getColorFromColorCode(String code){
  return Color(int.parse(code.substring(1, 7), radix: 16) + 0xFF000000);
}
0
12

To use hex color code inside flutter you can simply use following trick:

Example :

Your color code : #F8485E

Now remove # and use following declaration:

Color(0xFFF8485E) //Add 0xFF and Replace # with your hex code

Happy Fluttering!

11

Simple extension based on @Serdar answer https://stackoverflow.com/a/57943307/4899849

extension HexString on String {
  int getHexValue() => int.parse(replaceAll('#', '0xff'));
}

Usage:

'#b74093'.getHexValue()
10

File utils.dart

///
/// Convert a color hex-string to a Color object.
///
Color getColorFromHex(String hexColor) {
  hexColor = hexColor.toUpperCase().replaceAll('#', '');

  if (hexColor.length == 6) {
    hexColor = 'FF' + hexColor;
  }

  return Color(int.parse(hexColor, radix: 16));
}

Example usage

Text(
  'Hello, World!',
  style: TextStyle(
    color: getColorFromHex('#aabbcc'),
    fontWeight: FontWeight.bold,
  )
)
10

There isn't any need to use functions.

For example, to give a color to a container using colorcode:

Container
(
    color:Color(0xff000000)
)

Here the 0xff is the format followed by color code

7

Add this function in your file -


Color parseColor(String color) {
  String hex = color.replaceAll("#", "");
  if (hex.isEmpty) hex = "ffffff";
  if (hex.length == 3) {
    hex = '${hex.substring(0, 1)}${hex.substring(0, 1)}${hex.substring(1, 2)}${hex.substring(1, 2)}${hex.substring(2, 3)}${hex.substring(2, 3)}';
  }
  Color col = Color(int.parse(hex, radix: 16)).withOpacity(1.0);
  return col;
}

And use it like -

Container(
    color: parseColor("#b74093")
)
0
7

For general reference. There is a simpler way using the library Supercharged. Although you can use extension methods with all solutions mentioned, you find practical user library toolkit.

"#ff00ff".toColor(); // Painless hex to color
"red".toColor(); // Supports all web color names

Easier, right?

Supercharged

7

The easiest way is to convert it into an integer. For example, #BCE6EB. You would add 0xFF and you would then remove the hashtag making it:

0XFFBCE6EB

Then let’s say you were to implement it by doing:

backgroundColor: Color(0xffbce6eb)

If you can only use a hexadecimal then I suggest using the Hexcolor package.

0
5

"#b74093"? OK...

To HEX Recipe

int getColorHexFromStr(String colorStr)
{
  colorStr = "FF" + colorStr;
  colorStr = colorStr.replaceAll("#", "");
  int val = 0;
  int len = colorStr.length;
  for (int i = 0; i < len; i++) {
    int hexDigit = colorStr.codeUnitAt(i);
    if (hexDigit >= 48 && hexDigit <= 57) {
      val += (hexDigit - 48) * (1 << (4 * (len - 1 - i)));
    } else if (hexDigit >= 65 && hexDigit <= 70) {
      // A..F
      val += (hexDigit - 55) * (1 << (4 * (len - 1 - i)));
    } else if (hexDigit >= 97 && hexDigit <= 102) {
      // a..f
      val += (hexDigit - 87) * (1 << (4 * (len - 1 - i)));
    } else {
      throw new FormatException("An error occurred when converting a color");
    }
  }
  return val;
}
2
  • Lots of magic numbers. Aren't named constants supported in this language / environment? Commented Apr 26, 2021 at 20:56
  • What this is a bad answer and bad coding honestly, so many magic numbers, too much code for a one liner Commented Mar 2, 2022 at 11:01
5

As the Color constructor does not support hexadecimal string, so we should find other alternatives.

There are several possibilities:

1- The first one is to create a small function that will allow you to convert a color hex-string to a Color object.

Code:

   Color colorFromHex(String hexColor) {
   final hexCode = hexColor.replaceAll('#', '');
   if (hexColor.length == 6) {
    hexColor = 'FF' + hexColor; // FF as the opacity value if you don't add it.
   }
  return Color(int.parse('FF$hexCode', radix: 16));
}

Usage:

 Container(
          color: colorFromHex('abcdff'),
          child: Text(
            'Never stop learning',
            style: TextStyle(color: colorFromHex('bbffffcc')),
          ),
        )

2- The second possibility is to use the supercharged package. Supercharged brings all the comfort features from languages like Kotlin to all Flutter developers.

Add the dependency supercharged: ^1.X.X (find recent version) to your project and start using Supercharged everywhere:

import 'package:supercharged/supercharged.dart';

Now ,transform any String to colors

Code :

"#ff00ff".toColor(); // Painless hex to color
"red".toColor(); // Supports all web color names

You can also use the hexcolor package which is also great.

5
import 'package:flutter/material.dart';
class HexToColor extends Color{
  static _hexToColor(String code) {
    return int.parse(code.substring(1, 7), radix: 16) + 0xFF000000;
  }
  HexToColor(final String code) : super(_hexToColor(code));
}

Import the new class and use it like this:

HexToColor('#F2A03D')
5

Use hexadecimal numbers with the fromRGB constructor:

Color.fromRGBO(0xb7, 0x40, 0x93, 1),
5

While the existing answer is correct, the solution is overcomplicated. Dart has simpler methods to convert hexadecimal strings to Colors.

This function converts a hex string to an integer

int hexToInteger(String hex) => int.parse(hex, radix: 16);

Usage:

final red = Color(hexToInteger('FFFF0000'));

You can also extend String to convert to a Color

extension StringColorExtensions on String {
  Color toColor() => Color(hexToInteger(this));
}

Usage:

final red = 'FFFF0000'.toColor();

You can read more about these functions in this blog post.

0
3

You can click on Color Widget and it tells you with much deeper information what those letters stand for.

You can also use the Color.fromARGB() method to create custom colors which is much easier to me. Use the Flutter Doctor Color Picker website to pick any color you want for your Flutter application.

3

I have created this Flutter extention function of String class.. kinda useful if you also hate 0xFFF 😎

extension on String {
  Color parse() {
    var hexColor = this.replaceAll("#", "");
    if (hexColor.length == 6) {
      hexColor = "FF" + hexColor;
    }
    if (hexColor.length == 8) {
      return Color(int.parse("0x$hexColor"));
    }
  }
}

you can use to any hexadecimal color code string as follows...

'#bdbdbd'.parse() // this will return Color class object which you use in widget... 
2

You can use the package from_css_color to get Color out of a hexadecimal string. It supports both three-digit and six-digit RGB hexadecimal notation.

Color color = fromCSSColor('#ff00aa')

For optimisation sake, create a Color instance once for each color and store it somewhere for later usage.

2

I'm using this material_color_gen package it works like a charm

material_color_gen: ^2.0.0

Using :

import 'package:material_color_gen/material_color_gen.dart';
primarySwatch: Color(0xFFFF0000).toMaterialColor()

This is a HexColor example: #ff0000 Change # with 0xFF result is: 0xFFFF0000

Official link : https://pub.dev/packages/material_color_gen

2

The best way is to use the flutter hex color plugin from pub.dev and the import the package.

import 'package:hexcolor/hexcolor.dart';

and then use that in this particular way.

Text( 
     'Running on: $_platformVersion\n',
      style: TextStyle(color: HexColor("#f2f2f2")),
    ),
Text(
      "Hex From Material  $textColor",
       style: TextStyle(color: ColorToHex(Colors.teal)),
    ),

particular plugin here.

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