How to avoid rounding off in NSNumberFormatter

2019-04-06 02:16发布

问题:

I am trying to have a number string with maximum 2 decimals precision, while rest decimals just trimmed off instead of rounding them up. For example:

I have: 123456.9964

I want: 123456.99 -> Just want to trim rest of the decimal places

What I have tried is:

NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle: NSNumberFormatterDecimalStyle];
[numberFormatter setMaximumFractionDigits:2];
NSString *numberAsString = [numberFormatter stringFromNumber:[NSNumber numberWithFloat: 123456.9964]];
 NSLog(@"%@", numberAsString);

There is nothing to set rounding mode as "none". What else can I do to maintain Decimal style formatting along with trimmed decimal digits? Any help will be appreciated. Thanks.

回答1:

The following works for me:

NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:NSNumberFormatterDecimalStyle];
[numberFormatter setMaximumFractionDigits:2];
// optional - [numberFormatter setMinimumFractionDigits:2];
[numberFormatter setRoundingMode:NSNumberFormatterRoundDown];
NSNumber *num = @(123456.9964);
NSString *numberAsString = [numberFormatter stringFromNumber:num];
NSLog(@"%@", numberAsString);

The output is: 123,456.99

Part of your problem is the use of numberWithFloat: instead of numberWithDouble:. Your number has too many digits for float.



回答2:

you can use this

NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setRoundingMode:NSNumberFormatterRoundDown];
[numberFormatter setMaximumFractionDigits:2];
NSString *numberAsString = [numberFormatter stringFromNumber:[NSNumber numberWithDouble: 123456.9964]];
NSLog(@"%@", numberAsString);


回答3:

[numberFormatter setRoundingMode: NSNumberFormatterRoundDown];

Use this code.



回答4:

you can also just use round() and then convert it into a string afterward

var y = round(100 * 123456.8864) / 100

var x:String = String(format:"%.2f", y)
println("x: \(x)")

would print x: 123456.89, rounding the 8 to a 9.