我有一个双倍值22.368511 我想把它舍入到2位小数。即它应该返回22.37
我该怎么做?
答案 0 :(得分:186)
与大多数语言一样,格式为
%.2f
您可以看到更多示例here
编辑 :如果你担心在25.00的情况下显示该点,我也得到了这个
{
NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init];
[fmt setPositiveFormat:@"0.##"];
NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.342]]);
NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.3]]);
NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.0]]);
}
2010-08-22 15:04:10.614 a.out[6954:903] 25.34 2010-08-22 15:04:10.616 a.out[6954:903] 25.3 2010-08-22 15:04:10.617 a.out[6954:903] 25
答案 1 :(得分:68)
您可以使用以下代码将其格式化为两位小数
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
[formatter setNumberStyle:NSNumberFormatterDecimalStyle];
[formatter setMaximumFractionDigits:2];
[formatter setRoundingMode: NSNumberFormatterRoundUp];
NSString *numberString = [formatter stringFromNumber:[NSNumber numberWithFloat:22.368511]];
NSLog(@"Result...%@",numberString);//Result 22.37
斯威夫特4:
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.maximumFractionDigits = 2
formatter.roundingMode = .up
let str = String(describing: formatter.string(from: 12.2345)!)
print(str)
答案 2 :(得分:27)
value = (round(value*100)) / 100.0;
答案 3 :(得分:15)
[label setText:@"Value: %.2f", myNumber];
答案 4 :(得分:11)
您可以使用NSDecimalRound功能
答案 5 :(得分:11)
在Swift 2.0和Xcode 7.2中:
CORBA MARSHAL 0x4942f896 No; nested exception is:
org.omg.CORBA.MARSHAL: Unable to read value from underlying bridge : purge_calls:1927 Reason: CONN_ABORT (1), State: ABORT (5) vmcid: IBM minor code: 896 completed: No
示例:
答案 6 :(得分:5)
要从double中删除小数,请查看此输出
对象C
double hellodouble = 10.025;
NSLog(@"Your value with 2 decimals: %.2f", hellodouble);
NSLog(@"Your value with no decimals: %.0f", hellodouble);
输出将是:
10.02
10
Swift 2.1和Xcode 7.2.1
let hellodouble:Double = 3.14159265358979
print(String(format:"Your value with 2 decimals: %.2f", hellodouble))
print(String(format:"Your value with no decimals: %.0f", hellodouble))
输出将是:
3.14
3
答案 7 :(得分:4)
我打算用Jason的答案,但我注意到在我的Xcode版本(4.3.3)中,我不能这样做。经过一些研究后,我发现他们最近改变了类方法并删除了所有旧方法。所以这就是我必须要做的事情:
NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init];
[fmt setMaximumFractionDigits:2];
NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.342]]);
答案 8 :(得分:2)
使用NSNumber * aNumber = [NSNumber numberWithDouble:number];而不是NSNumber * aNumber = [NSNumber numberWithFloat:number];
+(NSString *)roundToNearestValue:(double)number
{
NSNumber *aNumber = [NSNumber numberWithDouble:number];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:NSNumberFormatterDecimalStyle];
[numberFormatter setUsesGroupingSeparator:NO];
[numberFormatter setMaximumFractionDigits:2];
[numberFormatter setMinimumFractionDigits:0];
NSString *string = [numberFormatter stringFromNumber:aNumber];
return string;
}
答案 9 :(得分:2)
对于Swift,有一个简单的解决方案,如果你不能导入Foundation,使用round()和/或不需要String(通常是你在Playground中的情况):
var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0
结果:31.726
答案 10 :(得分:1)
我使用Umberto Raimondi发布的扩展Double类型的解决方案:
extension Double {
func roundTo(places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
答案 11 :(得分:0)
你可以这样做:
{{1}}