在Cocoa中生成一个随机的字母数字字符串

Question

我想调用一个方法,传递它的长度并让它生成一个随机的字母数字字符串.

是否有任何实用程序库可能有一堆这些类型的函数？

Answer 1

这是一个快速而肮脏的实现.尚未经过测试.

NSString *letters = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";

-(NSString *) randomStringWithLength: (int) len {

    NSMutableString *randomString = [NSMutableString stringWithCapacity: len];

    for (int i=0; i<len; i++) {
         [randomString appendFormat: @"%C", [letters characterAtIndex: arc4random_uniform([letters length])]];
    }

    return randomString;
}

Answer 2

不完全是你问的,但仍然有用:

[[NSProcessInfo processInfo] globallyUniqueString]

样本输出:

450FEA63-2286-4B49-8ACC-9822C7D4356B-1376-00000239A4AC4FD5

Answer 3

NSString *alphabet  = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXZY0123456789";
NSMutableString *s = [NSMutableString stringWithCapacity:20];
for (NSUInteger i = 0U; i < 20; i++) {
    u_int32_t r = arc4random() % [alphabet length];
    unichar c = [alphabet characterAtIndex:r];
    [s appendFormat:@"%C", c];
}

Answer 4

当然你可以缩短它:

+(NSString*)generateRandomString:(int)num {
    NSMutableString* string = [NSMutableString stringWithCapacity:num];
    for (int i = 0; i < num; i++) {
        [string appendFormat:@"%C", (unichar)('a' + arc4random_uniform(26))];
    }
    return string;
}

Answer 5

如果您愿意仅限于十六进制字符,那么最简单的选择是生成一个UUID:

NSString *uuid = [NSUUID UUID].UUIDString;

示例输出:16E3DF0B-87B3-4162-A1A1-E03DB2F59654.

如果你想要一个较小的随机字符串,那么你只需要获取前8个字符.

这是一个版本4 UUID这意味着第三和第四组中的第一个字符不是随机的(它们将始终是4和之一8,9,A或B).

字符串中的每个其他字符都是完全随机的,您可以在数百年内每秒生成数百万个UUID,而不会产生两次生成相同UUID的风险.

Answer 6

Jeff B答案的类别版本.

的NSString + Random.h

#import <Foundation/Foundation.h>

@interface NSString (Random)

+ (NSString *)randomAlphanumericStringWithLength:(NSInteger)length;

@end

的NSString + Random.m

#import "NSString+Random.h"

 @implementation NSString (Random)

+ (NSString *)randomAlphanumericStringWithLength:(NSInteger)length
{ 
    NSString *letters = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
    NSMutableString *randomString = [NSMutableString stringWithCapacity:length];

    for (int i = 0; i < length; i++) {
        [randomString appendFormat:@"%C", [letters characterAtIndex:arc4random() % [letters length]]];
    }

    return randomString;
}

@end

Answer 7

您也可以生成一个UUID.虽然不是真正随机的,但它们复杂且独特,这使得它们在大多数用途中看起来是随机的 生成一个字符串,然后取一个等于传递长度的字符范围.

Answer 8

迅速

func randomStringWithLength(length: Int) -> String {
    let alphabet = "-_1234567890abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
    let upperBound = UInt32(count(alphabet))
    return String((0..<length).map { _ -> Character in
        return alphabet[advance(alphabet.startIndex, Int(arc4random_uniform(upperBound)))]
    })
}