NSString每4个字符附加一次WhiteSpace

时间:2014-09-30 08:32:44

标签: ios objective-c xcode nsstring nsmutablestring

我有一个像@" 1234123412341234"这样的字符串,我需要在每4个字符之间追加空格。

@"1234 1234 1234 1234"

即我需要NSString类似Visa卡类型。我试过这样但我没有得到我的结果。

-(void)resetCardNumberAsVisa:(NSString*)aNumber
{
  NSMutableString *s = [aNumber mutableCopy];

  for(int p=0; p<[s length]; p++)
  {
    if(p%4==0)
    {
        [s insertString:@" " atIndex:p];
    }
  }
  NSLog(@"%@",s);
}

6 个答案:

答案 0 :(得分:7)

以下是NSString

上的类别的unicode感知实现
@interface NSString (NRStringFormatting)
- (NSString *)stringByFormattingAsCreditCardNumber;
@end

@implementation NSString (NRStringFormatting)
- (NSString *)stringByFormattingAsCreditCardNumber
{
    NSMutableString *result = [NSMutableString string];
    __block NSInteger count = -1;
    [self enumerateSubstringsInRange:(NSRange){0, [self length]}
                             options:NSStringEnumerationByComposedCharacterSequences
                          usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
                              if ([substring rangeOfCharacterFromSet:[NSCharacterSet whitespaceCharacterSet]].location != NSNotFound)
                                  return;
                              count += 1;
                              if (count == 4) {
                                  [result appendString:@" "];
                                  count = 0;
                              }
                              [result appendString:substring];
                          }];
    return result;
}
@end

尝试使用此测试字符串:

NSString *string = @"ab  132487 387  e e e ";
NSLog(@"%@", [string stringByFormattingAsCreditCardNumber]);

该方法适用于非BMP字符(即表情符号)并处理现有的空格。

答案 1 :(得分:5)

你应该这样做:

- (NSString *)resetCardNumberAsVisa:(NSString*)originalString {
    NSMutableString *resultString = [NSMutableString string];

    for(int i = 0; i<[originalString length]/4; i++)
    {
        NSUInteger fromIndex = i * 4;
        NSUInteger len = [originalString length] - fromIndex;
        if (len > 4) {
            len = 4;
        }

        [resultString appendFormat:@"%@ ",[originalString substringWithRange:NSMakeRange(fromIndex, len)]];
    }
    return resultString;
}

<强>更新

您的代码将在第一个插入空间字符:

这是您的originalString

Text:     123412341234
Location: 012345678901

根据您的代码,在第一个插入空格字符时,您将插入&#34; 1&#34; 位置为4

之后,你的字符串是:

Text:     1234 12341234
Location: 0123456789012

所以,你看,现在你必须在 9 (9%4 != 0)的位置插入第二个空间字符

希望您自己修复代码!

答案 2 :(得分:3)

您的代码非常接近,但该方法的更好的语义是为任何给定的输入字符串返回新的NSString

-(NSString *)formatStringAsVisa:(NSString*)aNumber
{
    NSMutableString *newStr = [NSMutableString new];
    for (NSUInteger i = 0; i < [aNumber length]; i++)
    {
        if (i > 0 && i % 4 == 0)
           [newStr appendString:@" "];
        unichar c = [aNumber characterAtIndex:i];
        [newStr appendString:[[NSString alloc] initWithCharacters:&c length:1]];
    }
    return newStr;
}

答案 3 :(得分:1)

here的代码段可以做你想做的事情:

- (NSString *)insertSpacesEveryFourDigitsIntoString:(NSString *)string
              andPreserveCursorPosition:(NSUInteger *)cursorPosition 
{
    NSMutableString *stringWithAddedSpaces = [NSMutableString new];
    NSUInteger cursorPositionInSpacelessString = *cursorPosition;
    for (NSUInteger i=0; i<[string length]; i++) {
        if ((i>0) && ((i % 4) == 0)) {
            [stringWithAddedSpaces appendString:@" "];
            if (i < cursorPositionInSpacelessString) {
                (*cursorPosition)++;
            }
        }
        unichar characterToAdd = [string characterAtIndex:i];
        NSString *stringToAdd = 
            [NSString stringWithCharacters:&characterToAdd length:1];

        [stringWithAddedSpaces appendString:stringToAdd];
    }

    return stringWithAddedSpaces;
}

答案 4 :(得分:0)

基于Droppy的swift3

func codeFormat(_ code: String) -> String {
    let newStr = NSMutableString()
    for i in 0..<code.characters.count {
        if (i > 0 && i % 4 == 0){
            newStr.append(" ")
        }
            var c = (code as NSString).character(at: i)
            newStr.append(NSString(characters: &c, length: 1) as String)
    }
    return newStr as String
}

答案 5 :(得分:0)

请确保您的字符串长度应为4。
该解决方案将首先插入右侧。

- (NSString*) fillWhiteGapWithString:(NSString*)source
{
    NSInteger dl = 4;
    NSMutableString* result = [NSMutableString stringWithString:source];
    for(NSInteger cnt = result.length - dl ; cnt > 0 ; cnt -= dl)
    {
        [result insertString:@" " atIndex:cnt];
    }
    return result;
}