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
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)
其他回答
斯威夫特2.0
下面的代码是在xcode 7.2上测试的
import UIKit
extension UIColor{
public convenience init?(colorCodeInHex: String, alpha: Float = 1.0){
var filterColorCode:String = colorCodeInHex.stringByReplacingOccurrencesOfString("#", withString: "")
if filterColorCode.characters.count != 6 {
self.init(red: 0.0, green: 0.0, blue: 0.0, alpha: CGFloat(alpha))
return
}
filterColorCode = filterColorCode.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString
var range = Range(start: filterColorCode.startIndex.advancedBy(0), end: filterColorCode.startIndex.advancedBy(2))
let rString = filterColorCode.substringWithRange(range)
range = Range(start: filterColorCode.startIndex.advancedBy(2), end: filterColorCode.startIndex.advancedBy(4))
let gString = filterColorCode.substringWithRange(range)
range = Range(start: filterColorCode.startIndex.advancedBy(4), end: filterColorCode.startIndex.advancedBy(6))
let bString = filterColorCode.substringWithRange(range)
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)
self.init(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(alpha))
return
}
}
支持7十六进制颜色类型
有7种十六进制颜色格式:""#FF0000","0xFF0000", "FF0000", "F00", "red", 0x00FF00, 16711935
NSColorParser.nsColor("#FF0000",1)//red nsColor
NSColorParser.nsColor("FF0",1)//red nsColor
NSColorParser.nsColor("0xFF0000",1)//red nsColor
NSColorParser.nsColor("#FF0000",1)//red nsColor
NSColorParser.nsColor("FF0000",1)//red nsColor
NSColorParser.nsColor(0xFF0000,1)//red nsColor
NSColorParser.nsColor(16711935,1)//red nsColor
注意:这不是一个“单文件解决方案”,有一些依赖关系,但查找它们可能比从头开始研究更快。
https://github.com/eonist/swift-utils/blob/2882002682c4d2a3dc7cb3045c45f66ed59d566d/geom/color/NSColorParser.swift
永久链接: https://github.com/eonist/Element/wiki/Progress#supporting-7-hex-color-types
extension UIColor {
convenience init(r: CGFloat, g: CGFloat, b: CGFloat, a: CGFloat = 1) {
self.init(red: r/255, green: g/255, blue: b/255, alpha: a)
}
convenience init(hex: Int, alpha: CGFloat = 1) {
self.init(r: CGFloat((hex >> 16) & 0xff), g: CGFloat((hex >> 08) & 0xff), b: CGFloat((hex >> 00) & 0xff), a: alpha)
}
}
iOS 14, SwiftUI 2.0, swift 5.1, Xcode beta12
extension Color {
static func hexColour(hexValue:UInt32)->Color
{
let red = Double((hexValue & 0xFF0000) >> 16) / 255.0
let green = Double((hexValue & 0xFF00) >> 8) / 255.0
let blue = Double(hexValue & 0xFF) / 255.0
return Color(red:red, green:green, blue:blue)
}
}
用十六进制数表示
let red = Color.hexColour(hexValue: 0xFF0000)
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))