Card Number Masking

Question

I'm trying to mask card number with regex. First 6 and last 2 chars can be shown other should "*".

let cardNumber: String = "5890040000000016"

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

How can I do that in swift?

There is no `mask` method and `"XXXXXXX"` is not a "regex pattern" so your code is meaningless. If there is something we are not seeing, show it, please. — matt, Nov 17 '21 at 13:54
Let me explain @matt. I want mask a card number with regex. I need a regex that can detect between first 6 and last 2 chars. I don't know if there is a regex like that. That's why it's XXXXXX. I'm just wondering. — Kaan Ozdemir, Nov 17 '21 at 14:02
Yes, you can make a regex that does that. But since this can be done easily without a regex, why do you "need" a regex? Is this just a matter of idle curiosity? And again I ask, what is the meaningless code for? If it has a meaning because you have a defined a `mask(regexPattern:)` method, please show that method. — matt, Nov 17 '21 at 14:04

Mojtaba Hosseini · Accepted Answer · 2021-11-17T14:32:06.970

2

Your question contains 2 important parts:

A regex that matches all characters but first 6 and last 2

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

and

An extension on the `String` that takes the regex and masks it:

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)
    }
}

Usage:

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

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

Output: 589004********16

edited Nov 17 '21 at 14:32

answered Nov 17 '21 at 14:21

Mojtaba Hosseini

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May look nicer to use `#"..."#` rather than `"..."` because of the annoying character escaping for regex – George Nov 17 '21 at 14:28
I have got ridden of that ugly escaping at all @George. Now it is clean as heaven – Mojtaba Hosseini Nov 17 '21 at 14:33

score 1 · Answer 2 · answered Nov 17 '21 at 13:46

1

You can concatenate substrings together:

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

answered Nov 17 '21 at 13:46

George

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where is the regex? :/ – Mojtaba Hosseini Nov 17 '21 at 13:46
@MojtabaHosseini Not a regex but it achieves the result (and will be faster too) – George Nov 17 '21 at 13:48
Good point but what about a dynamic regex? – Mojtaba Hosseini Nov 17 '21 at 13:49
Yeah thank you @George, that's cool it works but I'm wondering if I can do it with regex. – Kaan Ozdemir Nov 17 '21 at 14:03
@KaanOzdemir You're going to need to elaborate on your requirements, because the problem you've presented makes no sense to solve with a regex. – Alexander Nov 17 '21 at 15:23

score 0 · Answer 3 · answered Nov 17 '21 at 14:21

Perhaps you're looking for something like this:

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
}

Card Number Masking

3 Answers3

A regex that matches all characters but first 6 and last 2

An extension on the String that takes the regex and masks it:

An extension on the `String` that takes the regex and masks it: