我需要从字符串的末尾删除空格。我该怎么做呢? 示例:如果字符串是“Hello”,它必须变成“Hello”


当前回答

NSString* NSStringWithoutSpace(NSString* string)
{
    return [string stringByReplacingOccurrencesOfString:@" " withString:@""];
}

其他回答

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。

在Swift中删除字符串开头和结尾的空格:

斯威夫特3

string.trimmingCharacters(in: .whitespacesAndNewlines)

Swift以前的版本

string.stringByTrimmingCharactersInSet(.whitespaceAndNewlineCharacterSet()))

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
}

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"

解决方案在这里描述:如何从NSString的右端删除空白?

添加以下类别到NSString:

- (NSString *)stringByTrimmingTrailingCharactersInSet:(NSCharacterSet *)characterSet {
    NSRange rangeOfLastWantedCharacter = [self rangeOfCharacterFromSet:[characterSet invertedSet]
                                                               options:NSBackwardsSearch];
    if (rangeOfLastWantedCharacter.location == NSNotFound) {
        return @"";
    }
    return [self substringToIndex:rangeOfLastWantedCharacter.location+1]; // non-inclusive
}

- (NSString *)stringByTrimmingTrailingWhitespaceAndNewlineCharacters {
    return [self stringByTrimmingTrailingCharactersInSet:
            [NSCharacterSet whitespaceAndNewlineCharacterSet]];
}

你可以这样使用它:

[yourNSString stringByTrimmingTrailingWhitespaceAndNewlineCharacters]