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
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))
}
其他回答
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
)
}
}
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)
以编程方式添加颜色的最简单方法是使用ColorLiteral。
只要像例子中那样添加ColorLiteral属性,Xcode就会提示你一个你可以选择的颜色列表。这样做的好处是代码更少,添加HEX值或RGB。您还将从故事板中获得最近使用的颜色。
例子: self.view.backgroundColor = ColorLiteral
这是我用的。适用于6和8字符的颜色字符串,带或不带#符号。在发布时默认为黑色,在调试时用无效字符串初始化时崩溃。
extension UIColor {
public convenience init(hex: String) {
var r: CGFloat = 0
var g: CGFloat = 0
var b: CGFloat = 0
var a: CGFloat = 1
let hexColor = hex.replacingOccurrences(of: "#", with: "")
let scanner = Scanner(string: hexColor)
var hexNumber: UInt64 = 0
var valid = false
if scanner.scanHexInt64(&hexNumber) {
if hexColor.count == 8 {
r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
a = CGFloat(hexNumber & 0x000000ff) / 255
valid = true
}
else if hexColor.count == 6 {
r = CGFloat((hexNumber & 0xff0000) >> 16) / 255
g = CGFloat((hexNumber & 0x00ff00) >> 8) / 255
b = CGFloat(hexNumber & 0x0000ff) / 255
valid = true
}
}
#if DEBUG
assert(valid, "UIColor initialized with invalid hex string")
#endif
self.init(red: r, green: g, blue: b, alpha: a)
}
}
用法:
UIColor(hex: "#75CC83FF")
UIColor(hex: "75CC83FF")
UIColor(hex: "#75CC83")
UIColor(hex: "75CC83")
我从这个回答中总结了一些想法,并针对iOS 13和Swift 5进行了更新。
extension UIColor {
convenience init(_ hex: String, alpha: CGFloat = 1.0) {
var cString = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if cString.hasPrefix("#") { cString.removeFirst() }
if cString.count != 6 {
self.init("ff0000") // return red color for wrong hex input
return
}
var rgbValue: UInt64 = 0
Scanner(string: cString).scanHexInt64(&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)
}
}
然后你可以这样使用它:
UIColor("#ff0000") // with #
UIColor("ff0000") // without #
UIColor("ff0000", alpha: 0.5) // using optional alpha value