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具有这些类型的功能？
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94784 次
20 回答
316
这是一个快速而肮脏的实现。没有经过测试。
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;
}
于 2010-04-13T23:42:57.817 回答
107
不完全符合您的要求，但仍然有用：
[[NSProcessInfo processInfo] globallyUniqueString]
样本输出：
450FEA63-2286-4B49-8ACC-9822C7D4356B-1376-00000239A4AC4FD5
于 2014-01-18T18:37:53.263 回答
67
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];
}
于 2012-01-03T12:47:41.587 回答
46
当然，您可以缩短此时间：
+(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;
}
于 2013-05-10T19:51:03.240 回答
28
如果您愿意将自己限制为仅使用十六进制字符，那么最简单的选择是生成 UUID：
NSString *uuid = [NSUUID UUID].UUIDString;
示例输出：16E3DF0B-87B3-4162-A1A1-E03DB2F59654.
如果你想要一个更小的随机字符串，那么你可以只抓取前 8 个字符。
这是第 4 版 UUID，这意味着第 3 组和第 4 组中的第一个字符不是随机的（它们始终是、或中4的一个）。89AB
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，而不会产生相同 UUID 被生成两次的风险。
于 2014-02-19T01:22:14.803 回答
24
Jeff B 答案的类别版本。
NSString+随机.h
#import <Foundation/Foundation.h>
@interface NSString (Random)
+ (NSString *)randomAlphanumericStringWithLength:(NSInteger)length;
@end
NSString+随机.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
于 2013-03-07T15:24:56.017 回答
7
您也可以只生成一个 UUID。虽然不是真正随机的，但它们复杂而独特，这使得它们在大多数用途中显得随机。生成一个字符串，然后取一个等于传递长度的字符范围。
于 2010-04-14T00:57:04.477 回答
5
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)))]
})
}
于 2015-06-20T01:24:20.363 回答
4
这是解决它的另一种方法。您可以在整数和字符之间进行转换，而不是使用准备好的字符串，并生成一个动态的字符列表以供选择。它非常精简和快速，但有更多的代码。
int charNumStart = (int) '0';
int charNumEnd = (int) '9';
int charCapitalStart = (int) 'A';
int charCapitalEnd = (int) 'Z';
int charLowerStart = (int) 'a';
int charLowerEnd = (int) 'z';
int amountOfChars = (charNumEnd - charNumStart) + (charCapitalEnd - charCapitalStart) + (charLowerEnd - charLowerStart); // amount of the characters we want.
int firstGap = charCapitalStart - charNumEnd; // there are gaps of random characters between numbers and uppercase letters, so this allows us to skip those.
int secondGap = charLowerStart - charCapitalEnd; // similar to above, but between uppercase and lowercase letters.
// START generates a log to show us which characters we are considering for our UID.
NSMutableString *chars = [NSMutableString stringWithCapacity:amountOfChars];
for (int i = charNumStart; i <= charLowerEnd; i++) {
if ((i >= charNumStart && i <= charNumEnd) || (i >= charCapitalStart && i <= charCapitalEnd) || (i >= charLowerStart && i <= charLowerEnd)) {
[chars appendFormat:@"\n%c", (char) i];
}
}
NSLog(@"chars: %@", chars);
// END log
// Generate a uid of 20 characters that chooses from our desired range.
int uidLength = 20;
NSMutableString *uid = [NSMutableString stringWithCapacity:uidLength];
for (int i = 0; i < uidLength; i++) {
// Generate a random number within our character range.
int randomNum = arc4random() % amountOfChars;
// Add the lowest value number to line this up with a desirable character.
randomNum += charNumStart;
// if the number is in the letter range, skip over the characters between the numbers and letters.
if (randomNum > charNumEnd) {
randomNum += firstGap;
}
// if the number is in the lowercase letter range, skip over the characters between the uppercase and lowercase letters.
if (randomNum > charCapitalEnd) {
randomNum += secondGap;
}
// append the chosen character.
[uid appendFormat:@"%c", (char) randomNum];
}
NSLog(@"uid: %@", uid);
// Generate a UID that selects any kind of character, including a lot of punctuation. It's a bit easier to do it this way.
int amountOfAnyCharacters = charLowerEnd - charNumStart; // A new range of characters.
NSMutableString *multiCharUid = [NSMutableString stringWithCapacity:uidLength];
for (int i = 0; i < uidLength; i++) {
// Generate a random number within our new character range.
int randomNum = arc4random() % amountOfAnyCharacters;
// Add the lowest value number to line this up with our range of characters.
randomNum += charNumStart;
// append the chosen character.
[multiCharUid appendFormat:@"%c", (char) randomNum];
}
NSLog(@"multiCharUid: %@", multiCharUid);
当我进行随机字符生成时，我更喜欢直接使用整数并将它们转换，而不是写出我想要从中提取的字符列表。在顶部声明变量使其更加独立于系统，但此代码假定数字的值低于字母，并且大写字母的值低于小写字母。
于 2012-09-12T23:24:18.980 回答
3
Swift 中的替代解决方案
func generateString(len: Int) -> String {
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let lettersLength = UInt32(countElements(letters))
let result = (0..<len).map { _ -> String in
let idx = Int(arc4random_uniform(lettersLength))
return String(letters[advance(letters.startIndex, idx)])
}
return "".join(result)
}
于 2015-02-27T12:30:30.233 回答
2
除了 Melvin 给出的好答案之外，这是我制作的一个函数（在 SWIFT 中！）以获取随机字符串：
func randomStringOfLength(length:Int)->String{
var wantedCharacters:NSString="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXZY0123456789"
var s=NSMutableString(capacity: length)
for (var i:Int = 0; i < length; i++) {
let r:UInt32 = arc4random() % UInt32( wantedCharacters.length)
let c:UniChar = wantedCharacters.characterAtIndex( Int(r) )
s.appendFormat("%C", c)
}
return s
}
这是调用的测试结果randomStringOfLength(10)： uXa0igA8wm
于 2014-11-14T15:28:49.567 回答
2
生成具有给定长度的小写字母数字随机字符串：
-(NSString*)randomStringWithLength:(NSUInteger)length
{
NSMutableString* random = [NSMutableString stringWithCapacity:length];
for (NSUInteger i=0; i<length; i++)
{
char c = '0' + (unichar)arc4random()%36;
if(c > '9') c += ('a'-'9'-1);
[random appendFormat:@"%c", c];
}
return random;
}
于 2015-01-21T13:02:16.240 回答
2
这里修改了一些想法，并在 Swift 4.0 中完成
extension String
{
subscript (i: Int) -> Character
{
return self[index(startIndex, offsetBy:i)]
}
static func Random(length:Int=32, alphabet:String="ABCDEF0123456789") -> String
{
let upperBound = UInt32(alphabet.count)
return String((0..<length).map { _ -> Character in
return alphabet[Int(arc4random_uniform(upperBound))]
})
}
}
用法：
let myHexString = String.Random()
let myLongHexString = String.Random(length:64)
let myLettersString = String.Random(length:32, alphabet:"ABCDEFGHIJKLMNOPQRSTUVWXYZ")
于 2018-03-06T15:30:45.133 回答
1
static NSUInteger length = 32;
static NSString *letters = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
NSMutableString * randomString = [NSMutableString stringWithCapacity:length];
for (NSInteger i = 0; i < length; ++i) {
[randomString appendFormat: @"%C", [letters characterAtIndex:(NSUInteger)arc4random_uniform((u_int32_t)[letters length])]];
}
于 2014-12-16T13:54:05.443 回答
1
我使用简单char[]的而不是NSString *字母表来做到这一点。我将此添加到 NSString 类别中。
static const char __alphabet[] =
"0123456789"
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"abcdefghijklmnopqrstuvwxyz";
+ (NSString *)randomString:(int)length
{
NSMutableString *randomString = [NSMutableString stringWithCapacity:length];
u_int32_t alphabetLength = (u_int32_t)strlen(__alphabet);
for (int i = 0; i < length; i++) {
[randomString appendFormat:@"%c", __alphabet[arc4random_uniform(alphabetLength)]];
}
return randomString;
}
于 2014-08-07T22:54:25.757 回答
1
如果你想要一个随机的 unicode 字符串，你可以创建随机字节，然后使用有效的字节。
OSStatus sanityCheck = noErr;
uint8_t * randomBytes = NULL;
size_t length = 200; // can of course be variable
randomBytes = malloc( length * sizeof(uint8_t) );
memset((void *)randomBytes, 0x0, length);
sanityCheck = SecRandomCopyBytes(kSecRandomDefault, length, randomBytes);
if (sanityCheck != noErr) NSLog(@"Error generating random bytes, OSStatus == %ld.", sanityCheck);
NSData* randomData = [[NSData alloc] initWithBytes:(const void *)randomBytes length: length];
if (randomBytes) free(randomBytes);
NSString* dataString = [[NSString alloc] initWithCharacters:[randomData bytes] length:[randomData length]]; // create an NSString from the random bytes
NSData* tempData = [dataString dataUsingEncoding:NSUTF8StringEncoding allowLossyConversion:YES]; // remove illegal characters from string
NSString* randomString = [[NSString alloc] initWithData:tempData encoding:NSUTF8StringEncoding];
从 NSString 到 NSData 并返回的转换对于获得有效的 UTF-8 字符串是必要的。请注意，长度不一定是最终创建的 NSString 的长度。
于 2014-02-20T17:23:27.203 回答
1
调用方法：
NSString *string = [self stringWithRandomSuffixForFile:@"file.pdf" withLength:4]
方法：
- (NSString *)stringWithRandomSuffixForFile:(NSString *)file withLength:(int)length
{
NSString *alphabet = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
NSString *fileExtension = [file pathExtension];
NSString *fileName = [file stringByDeletingPathExtension];
NSMutableString *randomString = [NSMutableString stringWithFormat:@"%@_", fileName];
for (int x = 0; x < length; x++) {
[randomString appendFormat:@"%C", [alphabet characterAtIndex: arc4random_uniform((int)[alphabet length]) % [alphabet length]]];
}
[randomString appendFormat:@".%@", fileExtension];
NSLog(@"## randomString: %@ ##", randomString);
return randomString;
}
结果：
## randomString: file_Msci.pdf ##
## randomString: file_xshG.pdf ##
## randomString: file_abAD.pdf ##
## randomString: file_HVwV.pdf ##
于 2014-09-16T05:07:17.677 回答
1
斯威夫特 3.0
func randomString(_ length: Int) -> String {
let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let len = UInt32(letters.length)
var randomString = ""
for _ in 0 ..< length {
let rand = arc4random_uniform(len)
var nextChar = letters.character(at: Int(rand))
randomString += NSString(characters: &nextChar, length: 1) as String
}
return randomString
}
于 2016-09-19T14:00:07.140 回答
0
Modification for keithyip's answer:
+ (NSString *)randomAlphanumericStringWithLength:(NSInteger)length
{
static NSString * const letters = @"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^{
srand(time(NULL));
});
NSMutableString *randomString = [NSMutableString stringWithCapacity:length];
for (int i = 0; i < length; i++) {
[randomString appendFormat:@"%C", [letters characterAtIndex:arc4random() % [letters length]]];
}
return randomString;
}
于 2017-12-08T13:11:07.130 回答
0
#define ASCII_START_NUMERS 0x30
#define ASCII_END_NUMERS 0x39
#define ASCII_START_LETTERS_A 0x41
#define ASCII_END_LETTERS_Z 0x5A
#define ASCII_START_LETTERS_a 0x61
#define ASCII_END_LETTERS_z 0x5A
-(NSString *)getRandomString:(int)length {
NSMutableString *result = [[NSMutableString alloc]init];
while (result.length != length) {
NSMutableData* data = [NSMutableData dataWithLength:1];
SecRandomCopyBytes(kSecRandomDefault, 1, [data mutableBytes]);
Byte currentChar = 0;
[data getBytes:¤tChar length:1];
NSString *s = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
if (currentChar > ASCII_START_NUMERS && currentChar < ASCII_END_NUMERS) { // 0 to 0
[result appendString:s];
continue;
}
if (currentChar > ASCII_START_LETTERS_A && currentChar < ASCII_END_LETTERS_Z) { // 0 to 0
[result appendString:s];
continue;
}
if (currentChar > ASCII_START_LETTERS_a && currentChar < ASCII_END_LETTERS_z) { // 0 to 0
[result appendString:s];
continue;
}
}
return result;
}
于 2015-12-14T12:57:59.067 回答