卡号掩蔽[关闭]

Question

我正在尝试使用正则表达式来掩盖卡号。 前 6 个和后 2 个字符可以显示为“*”。

let cardNumber: String = "5890040000000016"

print("Output: ", cardNumber.mask(regexPattern:"XXXXXXX")) // Output: 589004********16

我怎样才能快速做到这一点？

Answer 1

您可以将子字符串连接在一起：

let cardNumber = "5890040000000016"

let start = cardNumber.startIndex ..< cardNumber.index(cardNumber.startIndex, offsetBy: 6)
let end = cardNumber.index(cardNumber.endIndex, offsetBy: -2) ..< cardNumber.endIndex
let result = cardNumber[start] + Array(repeating: "*", count: cardNumber.count - 8) + cardNumber[end]
print(result)

// Prints: 589004********16

Answer 2

也许你正在寻找这样的东西：

let s = "5890040000000016"
let patt = ##"^(\d{6})(.*)(\d{2})$"## // this is your pattern
let reg = try! NSRegularExpression(pattern: patt, options: [])
if let m = reg.firstMatch(in: s, options: [], range: NSRange(s.startIndex..<s.endIndex, in: s)) {
    let r = m.range(at:2)
    let result = s.replacingCharacters(in: Range(r, in:s)!, with: String(repeating: "X", count: r.length))
    print(result) // 589004XXXXXXXX16
}

Answer 3

您的问题包含两个重要部分：

匹配除前 6 个和后 2 个以外的所有字符的正则表达式

let regexPattern = "(?<=.{6}).(?=.*.{2}$)"

和

String 上的扩展，它采用正则表达式并将其屏蔽：

extension String {

    func masked(matching regexPattern: String, with template: String = "*") throws -> String {
        let regex = try NSRegularExpression(pattern: regexPattern, options: NSRegularExpression.Options.caseInsensitive)
        let range = NSMakeRange(0, count)
        return regex.stringByReplacingMatches(in: self, options: [], range: range, withTemplate: template)
    }
}

用法：

let cardNumber = "5890040000000016"
let regexPattern = "(?<=.{6}).(?=.*.{2}$)"

print("Output:", try! cardNumber.masked(matching: regexPattern))

输出：589004********16