Question

是否有一种干净的方法来获取仅包含允许字符的字符串？

例如：

NSString* myStr = @"a5&/Öñ33";
NSString* allowedChars = @"abcdefghijklmnopqrstuvwxyz0123456789";
NSString* result = [myStr stringWIthAllowedChrs:allowedChars];

结果现在应为@"a533";

Answer 1

这不是最干净的，但您可以使用字符集分隔字符串，然后使用空字符串组合生成的数组。

// Create a character set with every character not in allowedChars
NSCharacterSet *charSet = [[NSCharacterSet characterSetWithCharactersInString:allowedChars] invertedSet];
// Split the original string at any occurrence of those characters
NSArray *splitString = [myStr componentsSeparatedByCharactersInSet:charSet];
// Combine the result into a string
NSString *result = [splitString componentsJoinedByString:@""];

Answer 2

简单易用的定制和理解方法：

NSString* myStr = @"a5&/Öñ33";
NSString* allowedChars = @"abcdefghijklmnopqrstuvwxyz0123456789";
NSCharacterSet *set = [[NSCharacterSet characterSetWithCharactersInString:allowedChars] invertedSet];
NSString *result = myStr;
NSRange range = [result rangeOfCharacterFromSet:set];
while (range.location != NSNotFound)
{
    result = [result stringByReplacingCharactersInRange:range withString:@""];
    range = [result rangeOfCharacterFromSet:set];
}
NSLog(@"%@", result);

Answer 3

最简单的一个 -

NSString* result = @"";

for(NSUInteger i = 0; i < [myStr length]; i++)
{
    unichar charArr[1] = {[myStr characterAtIndex:i]};
    NSString* charString = [NSString stringWithCharacters:charArr length:1];

    if([allowedChars rangeOfString:charString].location != NSNotFound)
        result = [result stringByAppendingString:charString];
}

return result;

带有允许字符的字符串

3 个答案: