另一个解决方案涉及创建可变字符串:

//make mutable string
NSMutableString *stringToTrim = [@" i needz trim " mutableCopy];

//pass it by reference to CFStringTrimSpace
CFStringTrimWhiteSpace((__bridge CFMutableStringRef) stringToTrim);

//stringToTrim is now "i needz trim"

在Objective-C中只修剪一端而不是两端的简单解决方案:

@implementation NSString (category)

/// trims the characters at the end
- (NSString *)stringByTrimmingSuffixCharactersInSet:(NSCharacterSet *)characterSet {
    NSUInteger i = self.length;
    while (i > 0 && [characterSet characterIsMember:[self characterAtIndex:i - 1]]) {
        i--;
    }
    return [self substringToIndex:i];
}

@end

和一个对称的实用工具，仅用于修剪开头:

@implementation NSString (category)

/// trims the characters at the beginning
- (NSString *)stringByTrimmingPrefixCharactersInSet:(NSCharacterSet *)characterSet {
    NSUInteger i = 0;
    while (i < self.length && [characterSet characterIsMember:[self characterAtIndex:i]]) {
        i++;
    }
    return [self substringFromIndex:i];
}

@end

要修剪所有尾随的空格字符(我猜这实际上是你的意图)，下面是一种相当干净简洁的方法。

斯威夫特5:

let trimmedString = string.replacingOccurrences(of: "\\s+$", with: "", options: .regularExpression)

objective - c:

NSString *trimmedString = [string stringByReplacingOccurrencesOfString:@"\\s+$" withString:@"" options:NSRegularExpressionSearch range:NSMakeRange(0, string.length)];

一行，加上一段正则表达式。

我想出了这个函数，它的行为基本类似于Alex的答案:

-(NSString*)trimLastSpace:(NSString*)str{
    int i = str.length - 1;
    for (; i >= 0 && [str characterAtIndex:i] == ' '; i--);
    return [str substringToIndex:i + 1];
}

whitespaceCharacterSet除了空格本身还包括制表符，在我的情况下不能出现。所以我想一个简单的比较就足够了。

let string = "测试修剪过的字符串"

对于删除空白和新行使用以下代码:-

let str_trim = yourString。trimmingCharacters (.whitespacesAndNewlines):

对于仅从字符串中删除空格使用以下代码:-

let str_trim = yourString。trimmingCharacters (.whitespaces):

这将只删除您选择的尾随字符。

func trimRight(theString: String, charSet: NSCharacterSet) -> String {

    var newString = theString

    while String(newString.characters.last).rangeOfCharacterFromSet(charSet) != nil {
        newString = String(newString.characters.dropLast())
    }

    return newString
}