New*_*bee 23 iphone objective-c ios
我有一个NSString,这有多个空格,我想修剪那些空格,并为@"如何.....你......"制作一个单独的空间进入@"你好吗".(点只是空间)
我试过了
NSString *trimmedString = [user_ids stringByTrimmingCharactersInSet:
[NSCharacterSet whitespaceCharacterSet]];
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它似乎不起作用.任何的想法.
Mad*_*nRP 46
您可以使用正则表达式来完成此任务:
NSError *error = nil;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@" +" options:NSRegularExpressionCaseInsensitive error:&error];
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该模式是一个空格,后跟一个或多个空格.将其替换为字符串中的单个空格:
NSString *trimmedString = [regex stringByReplacingMatchesInString:user_ids options:0 range:NSMakeRange(0, [user_ids length]) withTemplate:@" "];
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Rav*_*rma 11
user1587011
NSString *trimmedString = [user_ids stringByTrimmingCharactersInSet:
[NSCharacterSet whitespaceCharacterSet]];
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这个字符串方法用于删除字符串的开头和结尾处的空格.
尝试这个它给你你想要的东西: -
NSString *theString = @" Hello this is a long string! ";
NSCharacterSet *whitespaces = [NSCharacterSet whitespaceCharacterSet];
NSPredicate *noEmptyStrings = [NSPredicate predicateWithFormat:@"SELF != ''"];
NSArray *parts = [theString componentsSeparatedByCharactersInSet:whitespaces];
NSArray *filteredArray = [parts filteredArrayUsingPredicate:noEmptyStrings];
theString = [filteredArray componentsJoinedByString:@" "];
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Wol*_*ink 11
当我尝试时,这有效:
NSString *trimmedString = @"THIS IS A TEST S STRING S D D F ";
while ([trimmedString rangeOfString:@" "].location != NSNotFound) {
trimmedString = [trimmedString stringByReplacingOccurrencesOfString:@" " withString:@" "];
}
NSLog(@"%@", trimmedString);
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