在objective-c中将“ThisStringIsJoined”等字符串转换为“This String is Joined”的最有效方法是什么?
我从一个不受我控制的网络服务中收到这样的字符串,我想将数据呈现给用户,所以我想通过在每个大写单词的前面添加空格来整理它。字符串的格式始终为每个单词以大写字母开头。
我对Objective-c很新,所以不能真正想出这个。
由于
答案 0 :(得分:37)
实现这一目标的一种方法如下:
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);
这种情况下的输出是:
'ThisStringIsJoined' -> 'This String Is Joined'
您可能希望根据自己的需要调整正则表达式。您可能希望将其转换为NSString上的类别。
答案 1 :(得分:9)
NSRegularExpression
是要走的路,但作为琐事,NSCharacterSet也很有用:
- (NSString *)splitString:(NSString *)inputString {
int index = 1;
NSMutableString* mutableInputString = [NSMutableString stringWithString:inputString];
while (index < mutableInputString.length) {
if ([[NSCharacterSet uppercaseLetterCharacterSet] characterIsMember:[mutableInputString characterAtIndex:index]]) {
[mutableInputString insertString:@" " atIndex:index];
index++;
}
index++;
}
return [NSString stringWithString:mutableInputString];
}
答案 2 :(得分:1)
这是NSString上的一个类别,可以做你想要的。这将处理非ASCII字母。它还将正确拆分“IDidAGoodThing”。
@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
答案 3 :(得分:1)
这是Swift Code(webstersx的目标c代码),谢谢!
var str: NSMutableString = "iLoveSwiftCode"
var str2: NSMutableString = NSMutableString()
for var i:NSInteger = 0 ; i < str.length ; i++ {
var ch:NSString = str.substringWithRange(NSMakeRange(i, 1))
if(ch .rangeOfCharacterFromSet(NSCharacterSet.uppercaseLetterCharacterSet()).location != NSNotFound) {
str2 .appendString(" ")
}
str2 .appendString(ch)
}
println("\(str2.capitalizedString)")
}
输出:我喜欢Swift Code
答案 4 :(得分:0)
对于那些来到这里寻找类似问题的人来说,在Swift中回答: 也许更干净(添加到Sankalp的答案),以及更多&#39; Swifty &#39;的方法:
var string = "iLoveSwiftCode"
addSpaces(to: string)
//Result: string = "i Love Swift Code"
你可以像这样调用函数:
extension String{
mutating func seperatedWithSpaces(){
//indexOffset is needed because each time replaceSubrange is called, the resulting count is incremented by one (owing to the fact that a space is added to every capitalised letter)
var indexOffset = 0
for (index, character) in characters.enumerated(){
let stringCharacter = String(character)
if stringCharacter.lowercased() != stringCharacter{
guard index != 0 else { continue } //"ILoveSwift" should not turn into " I Love Swift"
let stringIndex = self.index(self.startIndex, offsetBy: index + indexOffset)
let endStringIndex = self.index(self.startIndex, offsetBy: index + 1 + indexOffset)
let range = stringIndex..<endStringIndex
indexOffset += 1
self.replaceSubrange(range, with: " \(stringCharacter)")
}
}
}
}
或者,如果您更喜欢扩展名:
var string = "iLoveSwiftCode"
string.seperatedWithSpaces()
//Result: string = "i Love Swift Code"
从字符串中调用方法:
IsoFields.QUARTER_YEARS
答案 5 :(得分:-1)
您可以尝试创建一个新字符串,该字符串是原始字符串的小写副本。然后比较两个字符串并在字符不同的地方插入空格。
使用NSString方法转为小写。
- (NSString *)lowercaseString