Question

是否有办法以零头和小ci为主编NSNumber? 例如,我需要有能力撰写4.5和000字。目前,我在那里会允许诽谤,但不会领导零。

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterNoStyle;

NSString *myString = [f numberFromString:@"4.5"];
NSLog(@"myString: %@",myString);

NSString *myOtherString = [f numberFromString:@"000"];
NSLog(@"myOtherString:%@",myOtherString);

上述产出是:我的缩略语:4.5和我的其他成员:0。作为产出,我需要能够做到4.5和000。

我对“催化格式指南”进行了研究,但没有取得多大成功。

Answer 1

注:[f numberFromString:@“4.5”] 回归代码>。 NSNumber* not a NSString*

You:

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterNoStyle;

NSNumber *myNumber;
NSString *myString;

myNumber = [f numberFromString:@"4.5"];
[f setNumberStyle:kCFNumberFormatterDecimalStyle];
myString = [f stringFromNumber:myNumber];
NSLog(@"myString: %@",myString);

myNumber = [f numberFromString:@"000"]; // Note that the extra zeros are useless
[f setFormatWidth:3];
[f setPaddingCharacter:@"0"];
myString = [f stringFromNumber:myNumber];
NSLog(@"myString: %@",myString);

安保记录仪产出:

myString: 4.5
myString: 000

如果你不敢动地开始仅仅创造像:

myNumber = [NSNumber numberWithFloat:4.5];
myNumber = [NSNumber numberWithInt:0];

或只是使用标准格式:

myString = [NSString stringWithFormat:@"%.1f", [myNumber floatValue]];
myString = [NSString stringWithFormat:@"%03d", [myNumber intValue]];

或如果你不需要<代码>。 NSNumber 代表仅使用标准格式:

myString = [NSString stringWithFormat:@"%.1f", 4.5];
myString = [NSString stringWithFormat:@"%03d", 0];

Answer 2

You could try something like:

NSString *myString = [NSString stringWithFormat:@"%03f", [myNSNumber floatValue]];

在印本之后,将印刷至少3位数,并用0个空位印制。

Answer 3

How about this as a variation on theme for the 000 s

 NSNumber *myNumber;
NSString *myString =@"000" ;
NSString * myStringResult;
NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterNoStyle;
[f setHasThousandSeparators:FALSE]; //-- remove seperator
[f  setMinimumIntegerDigits:[myString length ]]; //-- set minimum number of digits to display using the string length.



myNumber = [f numberFromString:myString];



myStringResult = [f stringFromNumber:myNumber];
NSLog(@"myStringResult: %@",myStringResult);

Answer 4

既然经常有人问这个问题, Apple果docs,那是人们要找的答案。以下链接有两个解决办法。 1个使用<代码> WithFormat: and the other using NSNumberFormatter.

https://stackoverflow.com/a/11131497/1058199

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