ios - 小数点后的 NSDecimalNumber 零（iPhone SDK）

Question

我有以下代码

(...)
numberStyle = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundPlain
                                                                     scale:2
                                                          raiseOnExactness:NO
                                                           raiseOnOverflow:NO
                                                          raiseOnUnderflow:NO
                                                       raiseOnDivideByZero:NO];

strThree = textFieldThree.text;
strThree = [strThree stringByReplacingOccurrencesOfString:@"." withString:@""];

fieldOneDNa = (NSDecimalNumber *)[NSDecimalNumber decimalNumberWithString:strThree];
oneHundred = (NSDecimalNumber *)[NSDecimalNumber decimalNumberWithString:@"100"];
fieldOneDN = [fieldOneDNa decimalNumberByDividingBy:oneHundred
                                       withBehavior:numberStyle];

firstStepReturn = [fieldOneDN stringValue];
textFieldThree.text = firstStepReturn;
(...)

我希望能够让 firstStepReturn 在小数点后有 2 位数字，即使它们是 .00 或 .*0。例如，如果将 323.390 输入到文本字段“textFieldThree”中，它将被处理并作为 3233.90 返回到文本字段。目前，它显示为 3233.9。

任何帮助将不胜感激。

Answer 1

NSDecimalNumber 子类 NSNumber 所以你只需要使用 NSNumberFormatter

 NSDecimalNumber * dn = [NSDecimalNumber decimalNumberWithString:@"1234.5"];
 NSNumberFormatter * nf = [[NSNumberFormatter alloc] init];
 [nf setMinimumFractionDigits:2];
 [nf setMaximumFractionDigits:2];
 NSString *ns  = [nf stringFromNumber:dn];

简单的

Answer 2

Why not use textFieldThree.text=[NSString stringWithFormat:@"%.2f",[fieldOneDN floatvalue]];

(or something similar - sorry I'm on a PC right now so can't test this)