Ton*_*lon 32 iphone xcode objective-c
如何在目标c中填充一个字符串.例如,如果我的整数值为6,我希望它显示为06.
我使用stringByPaddingToLength:它,但它像60一样填充到右边.
非常感谢您的帮助.
nev*_*ing 78
这个用10个零填充左边.
NSString *padded = [NSString stringWithFormat:@"Padded left with zeros: %010d", 65];
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aim*_*ess 20
+ (NSString *)formatValue:(int)value forDigits:(int)zeros {
NSString *format = [NSString stringWithFormat:@"%%0%dd", zeros];
return [NSString stringWithFormat:format,value];
}
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dre*_*lax 13
笨拙,但它会完成这项工作.
@implementation NSString (LeftPadding)
- (NSString *)stringByPaddingTheLeftToLength:(NSUInteger)newLength withString:(NSString *)padString startingAtIndex:(NSUInteger)padIndex
{
if ([self length] <= newLength)
return [[@"" stringByPaddingToLength:newLength - [self length] withString:padString startingAtIndex:padIndex] stringByAppendingString:self];
else
return [[self copy] autorelease];
}
@end
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然后你可以这样做:
NSString *test1 = [@"6" stringByPaddingTheLeftToLength:10 withString:@"0" startingAtIndex:0];
// test1 = "0000000006"
NSString *test2 = [@"asdf" stringByPaddingTheLeftToLength:10 withString:@"qwer" startingAtIndex:0];
// test2 = "qwerqwasdf"
NSString *test3 = [@"More than ten characters" stringByPaddingTheLeftToLength:10 withString:@"bamboo" startingAtIndex:0];
// test3 = "More than ten characters"
NSString *test4 = [test3 stringByPaddingTheLeftToLength:100 withString:test2 startingAtIndex:0];
// test4 = "qwerqwasdfqwerqwasdfqwerqwasdf...qwerqMore than ten characters"
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老线程,我知道,但这是一个方便的左边垫.我有我的Util课程.如果需要,您可以将其调整到右侧垫.最好的部分是它适用于你想要的任何填充.
+ (NSString *) leftPadString:(NSString *)s withPadding:(NSString *)padding {
NSString *padded = [padding stringByAppendingString:s];
return [padded substringFromIndex:[padded length] - [padding length]];
}
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ARC和非ARC兼容.:)
要打电话,请使用[Util leftPadString:@"42" withPadding:@"0000"];.
您也可以将它放在NSString的类别中,以便更轻松地调用[@"42" stringByLeftPadding:@"0000"].
两者都给你0042.
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