Flutter – How to change status bar color in Flutter

dartflutterflutter-dependenciesflutter-layout

I am trying to change the status bar color to white. I found this pub on flutter. I tried to use the example code on my dart files.

Best Answer

Update Flutter 2.0 (Recommended):

On latest Flutter version, you should use:

AppBar(
  backwardsCompatibility: false,
  systemOverlayStyle: SystemUiOverlayStyle(statusBarColor: Colors.orange),
)

Only Android (more flexibility):

import 'package:flutter/services.dart';

void main() {
  SystemChrome.setSystemUIOverlayStyle(SystemUiOverlayStyle(
    systemNavigationBarColor: Colors.blue, // navigation bar color
    statusBarColor: Colors.pink, // status bar color
  ));
}

Both iOS and Android:

appBar: AppBar(
  backgroundColor: Colors.red, // status bar color
  brightness: Brightness.light, // status bar brightness
)

Related Solutions

Flutter – How to change textField underline color

Just used -:

decoration: InputDecoration(        
              enabledBorder: UnderlineInputBorder(      
                      borderSide: BorderSide(color: Colors.cyan),   
                      ),  
              focusedBorder: UnderlineInputBorder(
                      borderSide: BorderSide(color: Colors.cyan),
                   ),  
             ),

it works for me :)

Flutter – How to use hexadecimal color strings in Flutter

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.