Question

我有一个字符串@"DidNotTurnUp"。我想写一个函数，它将此字符串作为输入，并应输出@"Did Not Turn Up"。最好的方法是什么？如何基于大写字母分隔字符串中的单词？帮助赞赏。

Answer 1

这是NSString上的一个类别，可以做你想要的。

@implementation NSString (SeparateCapitalizedWords)

-(NSString*)stringBySeparatingCapitalizedWords
{
    static NSRegularExpression * __regex ;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        NSError * error = nil ;
        __regex = [ NSRegularExpression regularExpressionWithPattern:@"[\\p{Uppercase Letter}]" options:0 error:&error ] ;
        if ( error ) { @throw error ; }
    });

    NSString * result = [ __regex stringByReplacingMatchesInString:self options:0 range:(NSRange){ 1, self.length - 1 } withTemplate:@" $0" ] ;
    return result ;
}

@end

Answer 2

试试这个

NSString *YourString=@"YouAreAGoodBoy";
NSString *outString;
NSString *string=@"";
NSUInteger length = [YourString length];
for (NSUInteger i = 0; i < length; i++)
{
    char c = [YourString characterAtIndex:i];
    if (i > 0 && c >= 65 && c <=90)
    {
        outString=[NSString stringWithFormat:@"%C", c];
        string = [NSString stringWithFormat:@"%@%@%@",string,@" ",outString];
    }
    else
    {
        outString=[NSString stringWithFormat:@"%C", c];
        string = [NSString stringWithFormat:@"%@%@",string,outString];

    }
}

NSLog(@"%@",string);

希望此代码对您有用。

Answer 3

NSString *string = @"ThisStringIsJoined";
NSRegularExpression *regexp = [NSRegularExpression 
    regularExpressionWithPattern:@"([a-z])([A-Z])" 
    options:0 
    error:NULL];
NSString *newString = [regexp 
    stringByReplacingMatchesInString:string 
    options:0 
    range:NSMakeRange(0, string.length) 
    withTemplate:@"$1 $2"];
NSLog(@"Changed '%@' -> '%@'", string, newString);

从https://stackoverflow.com/a/7322720/2842382复制，试试这个

目标C：在字符串中分隔单词并再次加入

3 个答案: