How do I use hexadecimal color strings in Flutter?

Question

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.

Solution 1

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.

Solution 2

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

Solution 3

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

Solution 4

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

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

Hexadecimal colors example with all opacity types in Dartpad

Solution 5

Easy way:

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

Usage:

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

View more solutions

Comments

[creativecreatorormaybenot]

I needed the "leading FF" for Flutter's ThemeData.

[KevinM]

I like this solution for its simplicity, but as @creativecreatorormaybenot mentioned, the leading FF is still required.

[xhinoda]

this is really great! also works with LinearGradient.

[Alvin Konda]

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')))

[leeCoder]

the HexColor class did not work for MaterialColor for me, it keeps complaining of the second parameter. Please help here

[Łukasz Wiśniewski]

You can't have static method extension in dart yet github.com/dart-lang/language/issues/723

[Gauris Javier]

It's a good idea to save this reference, although the .withOpacity(0.2) is useful enough i most of the cases.

[HJo]

I'm just surprised that you can pass 0xFF as part of an int

[creativecreatorormaybenot]

@HamishJohnson 0x just indicates that the following digits will be parsed as a hexadecimal 🙃

[madhu131313]

your solution is working for me. Are there any downsides, anyone?

[Nipun Ravisara]

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.

[Zakria Khan]

@nipunravisara get the hex of the color and use 100% opacity

[Nipun Ravisara]

@ZakriaKhan Thanks actually I found a way. stackoverflow.com/questions/65856982/…

[Peter Mortensen]

Lots of magic numbers. Aren't named constants supported in this language / environment?

[Peter Mortensen]

Lots of magic numbers. Aren't named constants supported in this language / environment?

[Peter Mortensen]

And (1 << (4 * (len - 1 - i))) is repeated three times.

[Pratik Butani]

.toUpperCase() giving error. Removed it and working now

[Ankit]

Is there any way to use 50% or any % of these colors? Like something Color(0xFF183451)[50]?

[creativecreatorormaybenot]

@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).

[Balaji]

How do I convert #0ff this string?, It has only three digits

[creativecreatorormaybenot]

@BalajiRamadoss I think the standard convention would make this 0xff00ffff and you could write a function for it.

[Balaji]

Yes thanks, I figured it out, I have to double the characters "0ff" to "00ffff"

[TechAurelian]

Or Color(int.parse(hexString.replaceFirst('#', '0xff'))) if I know for sure that the input hex string is in #RRGGBB format (such as #45a3df).

[MNBWorld]

Now a days the best way to do this is to use the flutter hex color plugin from pub.dev and the import the package.

[creativecreatorormaybenot]

@MNBWorld I highly disagree with that sentiment. The package only implements a version of the extension class from my answer with less features. This leads to poorer customization of how you want to handle HEX codes (which might be specific to your app). And in general, it is preferrable to use the const constructors with Color.

[Tremmillicious]

What this is a bad answer and bad coding honestly, so many magic numbers, too much code for a one liner