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
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
)
}
}
其他回答
在Swift 2.0和Xcode 7.0.1中,你可以创建这个函数:
// Creates a UIColor from a Hex string.
func colorWithHexString (hex:String) -> UIColor {
var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
if (cString.hasPrefix("#")) {
cString = (cString as NSString).substringFromIndex(1)
}
if (cString.characters.count != 6) {
return UIColor.grayColor()
}
let rString = (cString as NSString).substringToIndex(2)
let gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
let bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)
var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
NSScanner(string: rString).scanHexInt(&r)
NSScanner(string: gString).scanHexInt(&g)
NSScanner(string: bString).scanHexInt(&b)
return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}
然后这样使用它:
let color1 = colorWithHexString("#1F437C")
Swift 4更新
func colorWithHexString (hex:String) -> UIColor {
var cString = hex.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString = (cString as NSString).substring(from: 1)
}
if (cString.characters.count != 6) {
return UIColor.gray
}
let rString = (cString as NSString).substring(to: 2)
let gString = ((cString as NSString).substring(from: 2) as NSString).substring(to: 2)
let bString = ((cString as NSString).substring(from: 4) as NSString).substring(to: 2)
var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
Scanner(string: rString).scanHexInt32(&r)
Scanner(string: gString).scanHexInt32(&g)
Scanner(string: bString).scanHexInt32(&b)
return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}
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)
)
}
RGBA版本Swift 3/4
我喜欢卢卡的回答,因为我认为它是最优雅的。
然而,我不希望我的颜色指定在ARGB。我宁愿RGBA +,我也需要在处理字符串的情况下,为每个频道指定1个字符“#FFFA”。
这个版本还增加了错误抛出+剥离'#'字符如果它包含在字符串中。 这是我修改后的Swift表格。
public enum ColourParsingError: Error
{
case invalidInput(String)
}
extension UIColor {
public convenience init(hexString: String) throws
{
let hexString = hexString.replacingOccurrences(of: "#", with: "")
let hex = hexString.trimmingCharacters(in:NSCharacterSet.alphanumerics.inverted)
var int = UInt32()
Scanner(string: hex).scanHexInt32(&int)
let a, r, g, b: UInt32
switch hex.count
{
case 3: // RGB (12-bit)
(r, g, b,a) = ((int >> 8) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17,255)
//iCSS specification in the form of #F0FA
case 4: // RGB (24-bit)
(r, g, b,a) = ((int >> 12) * 17, (int >> 8 & 0xF) * 17, (int >> 4 & 0xF) * 17, (int & 0xF) * 17)
case 6: // RGB (24-bit)
(r, g, b, a) = (int >> 16, int >> 8 & 0xFF, int & 0xFF,255)
case 8: // ARGB (32-bit)
(r, g, b, a) = (int >> 24, int >> 16 & 0xFF, int >> 8 & 0xFF, int & 0xFF)
default:
throw ColourParsingError.invalidInput("String is not a valid hex colour string: \(hexString)")
}
self.init(red: CGFloat(r) / 255, green: CGFloat(g) / 255, blue: CGFloat(b) / 255, alpha: CGFloat(a) / 255)
}
}
Swift 5.3 & SwiftUI:通过UIColor支持十六进制和CSS颜色名称
依据代码
斯威夫特包
SwiftUI包
字符串的例子:
橙子、酸橙、番茄等。 Clear, Transparent, nil和空字符串产生[UIColor clearColor] 美国广播公司 abc7 # abc7 00飞行符 # 00飞行符 00 ffff77
操场上的输出:
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: