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
你可以在UIColor上使用这个扩展,它将你的字符串(十六进制,RGBA)转换为UIColor,反之亦然。
extension UIColor {
//Convert RGBA String to UIColor object
//"rgbaString" must be separated by space "0.5 0.6 0.7 1.0" 50% of Red 60% of Green 70% of Blue Alpha 100%
public convenience init?(rgbaString : String){
self.init(ciColor: CIColor(string: rgbaString))
}
//Convert UIColor to RGBA String
func toRGBAString()-> String {
var r: CGFloat = 0
var g: CGFloat = 0
var b: CGFloat = 0
var a: CGFloat = 0
self.getRed(&r, green: &g, blue: &b, alpha: &a)
return "\(r) \(g) \(b) \(a)"
}
//return UIColor from Hexadecimal Color string
public convenience init?(hexString: String) {
let r, g, b, a: CGFloat
if hexString.hasPrefix("#") {
let start = hexString.index(hexString.startIndex, offsetBy: 1)
let hexColor = hexString.substring(from: start)
if hexColor.characters.count == 8 {
let scanner = Scanner(string: hexColor)
var hexNumber: UInt64 = 0
if scanner.scanHexInt64(&hexNumber) {
r = CGFloat((hexNumber & 0xff000000) >> 24) / 255
g = CGFloat((hexNumber & 0x00ff0000) >> 16) / 255
b = CGFloat((hexNumber & 0x0000ff00) >> 8) / 255
a = CGFloat(hexNumber & 0x000000ff) / 255
self.init(red: r, green: g, blue: b, alpha: a)
return
}
}
}
return nil
}
// Convert UIColor to Hexadecimal String
func toHexString() -> String {
var r: CGFloat = 0
var g: CGFloat = 0
var b: CGFloat = 0
var a: CGFloat = 0
self.getRed(&r, green: &g, blue: &b, alpha: &a)
return String(
format: "%02X%02X%02X",
Int(r * 0xff),
Int(g * 0xff),
Int(b * 0xff))
}
}
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))
}
你可以在swift 5中使用它
斯威夫特5
import UIKit
extension UIColor {
static func hexStringToUIColor (hex:String) -> UIColor {
var cString:String = hex.trimmingCharacters(in: .whitespacesAndNewlines).uppercased()
if (cString.hasPrefix("#")) {
cString.remove(at: cString.startIndex)
}
if ((cString.count) != 6) {
return UIColor.gray
}
var rgbValue:UInt32 = 0
Scanner(string: cString).scanHexInt32(&rgbValue)
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)
)
}
}
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")
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
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