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
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))
其他回答
extension UIColor {
convenience init(hex: Int, alpha: Double = 1.0) {
self.init(red: CGFloat((hex>>16)&0xFF)/255.0, green:CGFloat((hex>>8)&0xFF)/255.0, blue: CGFloat((hex)&0xFF)/255.0, alpha: CGFloat(255 * alpha) / 255)
}
}
使用这个扩展像:
let selectedColor = UIColor(hex: 0xFFFFFF)
let selectedColor = UIColor(hex: 0xFFFFFF, alpha: 0.5)
最新swift3版本
extension UIColor {
convenience init(hexString: String) {
let hex = hexString.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)
var int = UInt32()
Scanner(string: hex).scanHexInt32(&int)
let a, r, g, b: UInt32
switch hex.characters.count {
case 3: // RGB (12-bit)
(a, r, g, b) = (255, (int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6: // RGB (24-bit)
(a, r, g, b) = (255, int >> 16, int >> 8 & 0xFF, int & 0xFF)
case 8: // ARGB (32-bit)
(a, r, g, b) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
default:
(a, r, g, b) = (255, 0, 0, 0)
}
self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}
在你的类或任何你把hexcolor转换为uicolor的地方使用这种方法
let color1 = UIColor(hexString: "#FF323232")
斯威夫特2.0:
在viewDidLoad ()
var viewColor:UIColor
viewColor = UIColor()
let colorInt:UInt
colorInt = 0x000000
viewColor = UIColorFromRGB(colorInt)
self.View.backgroundColor=viewColor
func UIColorFromRGB(rgbValue: UInt) -> UIColor {
return UIColor(
red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
alpha: CGFloat(1.0)
)
}
Swift 4:结合Sulthan和Luca Torella的回答:
extension UIColor {
convenience init(hexFromString:String, alpha:CGFloat = 1.0) {
var cString:String = hexFromString.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
var rgbValue:UInt32 = 10066329 //color #999999 if string has wrong format
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.count) == 6) {
Scanner(string: cString).scanHexInt32(&rgbValue)
}
self.init(
red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
alpha: alpha
)
}
}
使用例子:
let myColor = UIColor(hexFromString: "4F9BF5")
let myColor = UIColor(hexFromString: "#4F9BF5")
let myColor = UIColor(hexFromString: "#4F9BF5", alpha: 0.5)
Swift 5:你可以在Xcode中创建颜色,如下图所示:
您应该命名颜色,因为您通过名称引用了颜色。如图2所示: