我试图在Swift中使用十六进制颜色值,而不是UIColor允许您使用的少数标准值,但我不知道如何做到这一点。
示例:我如何使用#ffffff作为颜色?
我试图在Swift中使用十六进制颜色值,而不是UIColor允许您使用的少数标准值,但我不知道如何做到这一点。
示例:我如何使用#ffffff作为颜色?
当前回答
UIColor扩展,这将大大帮助你!(4.0版本:斯威夫特)
import UIKit
extension UIColor {
/// rgb颜色
convenience init(r: CGFloat, g: CGFloat, b: CGFloat) {
self.init(red: r/255.0 ,green: g/255.0 ,blue: b/255.0 ,alpha:1.0)
}
/// 纯色(用于灰色)
convenience init(gray: CGFloat) {
self.init(red: gray/255.0 ,green: gray/255.0 ,blue: gray/255.0 ,alpha:1.0)
}
/// 随机色
class func randomCGColor() -> UIColor {
return UIColor(r: CGFloat(arc4random_uniform(256)), g: CGFloat(arc4random_uniform(256)), b: CGFloat(arc4random_uniform(256)))
}
/// hex颜色-Int
convenience init(hex:Int, alpha:CGFloat = 1.0) {
self.init(
red: CGFloat((hex & 0xFF0000) >> 16) / 255.0,
green: CGFloat((hex & 0x00FF00) >> 8) / 255.0,
blue: CGFloat((hex & 0x0000FF) >> 0) / 255.0,
alpha: alpha
)
}
/// hex颜色-String
convenience init(hexString: String){
var red: CGFloat = 0.0
var green: CGFloat = 0.0
var blue: CGFloat = 0.0
var alpha: CGFloat = 1.0
let scanner = Scanner(string: hexString)
var hexValue: CUnsignedLongLong = 0
if scanner.scanHexInt64(&hexValue) {
switch (hexString.characters.count) {
case 3:
red = CGFloat((hexValue & 0xF00) >> 8) / 15.0
green = CGFloat((hexValue & 0x0F0) >> 4) / 15.0
blue = CGFloat(hexValue & 0x00F) / 15.0
case 4:
red = CGFloat((hexValue & 0xF000) >> 12) / 15.0
green = CGFloat((hexValue & 0x0F00) >> 8) / 15.0
blue = CGFloat((hexValue & 0x00F0) >> 4) / 15.0
alpha = CGFloat(hexValue & 0x000F) / 15.0
case 6:
red = CGFloat((hexValue & 0xFF0000) >> 16) / 255.0
green = CGFloat((hexValue & 0x00FF00) >> 8) / 255.0
blue = CGFloat(hexValue & 0x0000FF) / 255.0
case 8:
alpha = CGFloat((hexValue & 0xFF000000) >> 24) / 255.0
red = CGFloat((hexValue & 0x00FF0000) >> 16) / 255.0
green = CGFloat((hexValue & 0x0000FF00) >> 8) / 255.0
blue = CGFloat(hexValue & 0x000000FF) / 255.0
default:
log.info("Invalid RGB string, number of characters after '#' should be either 3, 4, 6 or 8")
}
} else {
log.error("Scan hex error")
}
self.init(red:red, green:green, blue:blue, alpha:alpha)
}}
其他回答
斯威夫特2.3: 用户界面颜色扩展。我认为这样更简单。
extension UIColor {
static func colorFromHex(hexString: String, alpha: CGFloat = 1) -> UIColor {
//checking if hex has 7 characters or not including '#'
if hexString.characters.count < 7 {
return UIColor.whiteColor()
}
//string by removing hash
let hexStringWithoutHash = hexString.substringFromIndex(hexString.startIndex.advancedBy(1))
//I am extracting three parts of hex color Red (first 2 characters), Green (middle 2 characters), Blue (last two characters)
let eachColor = [
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex...hexStringWithoutHash.startIndex.advancedBy(1)),
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(2)...hexStringWithoutHash.startIndex.advancedBy(3)),
hexStringWithoutHash.substringWithRange(hexStringWithoutHash.startIndex.advancedBy(4)...hexStringWithoutHash.startIndex.advancedBy(5))]
let hexForEach = eachColor.map {CGFloat(Int($0, radix: 16) ?? 0)} //radix is base of numeric system you want to convert to, Hexadecimal has base 16
//return the color by making color
return UIColor(red: hexForEach[0] / 255, green: hexForEach[1] / 255, blue: hexForEach[2] / 255, alpha: alpha)
}
}
用法:
let color = UIColor.colorFromHex("#25ac09")
斯威夫特4.0
使用这种单行方法
override func viewDidLoad() {
super.viewDidLoad()
let color = UIColor(hexColor: "FF00A0")
self.view.backgroundColor = color
}
你必须创建新的类或使用任何控制器,你需要使用十六进制颜色。这个扩展类为您提供UIColor,将十六进制转换为RGB颜色。
extension UIColor {
convenience init(hexColor: String) {
let scannHex = Scanner(string: hexColor)
var rgbValue: UInt64 = 0
scannHex.scanLocation = 0
scannHex.scanHexInt64(&rgbValue)
let r = (rgbValue & 0xff0000) >> 16
let g = (rgbValue & 0xff00) >> 8
let b = rgbValue & 0xff
self.init(
red: CGFloat(r) / 0xff,
green: CGFloat(g) / 0xff,
blue: CGFloat(b) / 0xff, alpha: 1
)
}
}
我做了一个小函数,把它放在我可以全局使用它的地方,在swift 2.1中工作得很好:
func getColorFromHex(rgbValue:UInt32)->UIColor{
let red = CGFloat((rgbValue & 0xFF0000) >> 16)/255.0
let green = CGFloat((rgbValue & 0xFF00) >> 8)/255.0
let blue = CGFloat(rgbValue & 0xFF)/255.0
return UIColor(red:red, green:green, blue:blue, alpha:1.0)
}
用法:
getColorFromHex(0xffffff)
另一种方法
斯威夫特3.0
为UIColor写一个扩展
// To change the HexaDecimal value to Corresponding Color
extension UIColor
{
class func uicolorFromHex(_ rgbValue:UInt32, alpha : CGFloat)->UIColor
{
let red = CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0
let green = CGFloat((rgbValue & 0xFF00) >> 8) / 255.0
let blue = CGFloat(rgbValue & 0xFF) / 255.0
return UIColor(red:red, green:green, blue:blue, alpha: alpha)
}
}
你可以像这样用hex直接创建UIColor
let carrot = UIColor.uicolorFromHex(0xe67e22, alpha: 1))
Xcode 13.2.1, M1, Swift 5.5
我们可以在ColorLiterals中使用Hex
输入#colorLiteral(在Xcode中,这将触发并修复与ColorLiterals相关的错误
然后点击其他
然后选择RGB滑块,你现在可以看到十六进制面板