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
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。
其他回答
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()))
NSString* NSStringWithoutSpace(NSString* string)
{
return [string stringByReplacingOccurrencesOfString:@" " withString:@""];
}
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
}
斯威夫特版本
只修整字符串末尾的空格:
private func removingSpacesAtTheEndOfAString(var str: String) -> String {
var i: Int = countElements(str) - 1, j: Int = i
while(i >= 0 && str[advance(str.startIndex, i)] == " ") {
--i
}
return str.substringWithRange(Range<String.Index>(start: str.startIndex, end: advance(str.endIndex, -(j - i))))
}
修整字符串两侧的空格:
var str: String = " Yolo "
var trimmedStr: String = str.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
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]