IBAN Validator Swift

|▌冷眼眸甩不掉的悲伤 提交于 2019-12-10 14:22:54

问题


I am writing an algorithm to validate IBAN (International Bank Account Number) in Swift 3 and not able to figure one of the validation.

Example IBAN - BE68539007547034

Here are the rules to validate -

  1. Input number should be of length 16.
  2. First 2 characters are country code (not numeric).
  3. Last 14 are numeric.
  4. Last 2 characters are the modulo 97 result of the previous 12 numeric characters.

While #1 - #3 are clear I need clarity on #4. If anyone have done this before and know about it then please let me know.


回答1:


From Wikipedia

let IBAN = "GB82WEST12345698765432" // uppercase, no whitespace !!!!
var a = IBAN.utf8.map{ $0 }
while a.count < 4 {
    a.append(0)
}
let b = a[4..<a.count] + a[0..<4]
let c = b.reduce(0) { (r, u) -> Int in
    let i = Int(u)
    return i > 64 ? (100 * r + i - 55) % 97: (10 * r + i - 48) % 97
}
print( "IBAN \(IBAN) is", c == 1 ? "valid": "invalid")

prints

IBAN GB82WEST12345698765432 is valid

With IBAN from your question it prints

IBAN BE68539007547034 is valid



回答2:


The validation algorithm is rather simple if you follow the algorithm on wikipedia:

extension String {
    private func mod97() -> Int {
        let symbols: [Character] = Array(self)
        let swapped = symbols.dropFirst(4) + symbols.prefix(4)

        let mod: Int = swapped.reduce(0) { (previousMod, char) in
            let value = Int(String(char), radix: 36)! // "0" => 0, "A" => 10, "Z" => 35
            let factor = value < 10 ? 10 : 100          
            return (factor * previousMod + value) % 97
        }

        return mod
    }    

    func passesMod97Check() -> Bool {
        guard self.characters.count >= 4 else {
            return false
        }

        let uppercase = self.uppercased()

        guard uppercase.range(of: "^[0-9A-Z]*$", options: .regularExpression) != nil else {
            return false
        }

        return (uppercase.mod97() == 1)
    }
}

Usage:

let iban = "XX0000000..."
let valid = iban.passesMod97Check()

If you want to validate the format for a specific country, just modify the regular expression, e.g.

"^[A-Z]{2}[0-9]{14}$"

or directly

"^BE\\d{14}$"



回答3:


I finded a great solution that work for me in Objective-C https://gist.github.com/0xc010d/5301790 you can rewrite for Swift or use bridging header. Objective-C implementation of mod97 IBAN checking algorithm

#import <Foundation/Foundation.h>

@interface NSString (Mod97Check)

- (BOOL)passesMod97Check; // Returns result of mod 97 checking algorithm. Might be used to check IBAN.
                          // Expects string to contain digits and/or upper-/lowercase letters; space and all the rest symbols are not acceptable.

@end
#import "NSString+Mod97Check.h"

@implementation NSString (Mod97Check)

- (BOOL)passesMod97Check {
    NSString *string = [self uppercaseString];
    NSInteger mod = 0, length = [self length];
    for (NSInteger index = 4; index < length + 4; index ++) {
        unichar character = [string characterAtIndex:index % length];
        if (character >= '0' && character <= '9') {
            mod = (10 * mod + (character - '0')) % 97; // '0'=>0, '1'=>1, ..., '9'=>9
        }
        else if (character >= 'A' && character <= 'Z') {
            mod = (100 * mod + (character - 'A' + 10)) % 97; // 'A'=>10, 'B'=>11, ..., 'Z'=>35
        }
        else {
            return NO;
        }
    }
    return (mod == 1);
}

@end
-(BOOL)isValidIBAN {
    NSString *iban = self;
    static NSString* const LettersAndDecimals = @"ABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789";
    iban = [[iban stringByReplacingOccurrencesOfString:@" " withString:@""] uppercaseString];
    NSCharacterSet *invalidChars = [[NSCharacterSet characterSetWithCharactersInString:LettersAndDecimals] invertedSet];

    if ([iban rangeOfCharacterFromSet:invalidChars].location != NSNotFound)
    {
        return NO;
    }

    int checkDigit = [iban substringWithRange:NSMakeRange(2, 2)].intValue;
    iban = [NSString stringWithFormat:@"%@%@",[iban substringWithRange:NSMakeRange(4, iban.length - 4)], [iban substringWithRange:NSMakeRange(0, 4)]] ;

    for (int i = 0; i < iban.length; i++) {
        unichar c = [iban characterAtIndex:i];
        if (c >= 'A' && c <= 'Z') {
            iban = [NSString stringWithFormat:@"%@%d%@", [iban substringWithRange:NSMakeRange(0, i)], (c - 'A' + 10),[iban substringWithRange:NSMakeRange(i+1, iban.length - i - 1)]];
        }

    }
    iban = [[iban substringWithRange:NSMakeRange(0, iban.length - 2)] stringByAppendingString:@"00"];

    while(true)
    {
        int iMin = (int)MIN(iban.length, 9);
        NSString* strPart = [iban substringWithRange:NSMakeRange(0, iMin)];
        int decnumber = strPart.intValue;
        if(decnumber < 97 || iban.length < 3)
            break;
        int del = decnumber % 97;
        iban =  [NSString stringWithFormat:@"%d%@", del, [iban substringFromIndex:iMin]];
    }
    int check = 98 - iban.intValue;

    return checkDigit == check;
}



回答4:


Here you go :

func isValidIBAN(text:String) -> Bool {
        let ibanRegEx = "[a-zA-Z]{2}+[0-9]{2}+[a-zA-Z0-9]{4}+[0-9]{7}([a-zA-Z0-9]?){0,16}"
        let ibanTest = NSPredicate(format:"SELF MATCHES %@", ibanRegEx)
        return ibanTest.evaluate(with: text)
    }

It's clean, and it works.



来源:https://stackoverflow.com/questions/43803956/iban-validator-swift

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!