小数点后两位的整数

As in most languages the format is

%.2f

you can see more examples here

Edit: I also got this if your concerned about the display of the point in cases of 25.00

{
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

You can use the NSDecimalRound function

[label setText:@"Value: %.2f", myNumber];

You can use the below code to format it to two decimal places

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

Swift 4:

let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.maximumFractionDigits = 2
formatter.roundingMode = .up


let str = String(describing: formatter.string(from: 12.2345)!)


print(str)

I was going to go with Jason's answer but I noticed that in My version of Xcode (4.3.3) I couldn't do that. so after a bit of research I found they had recently changed the class methods and removed all the old ones. so here's how I had to do it:

NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init];


[fmt setMaximumFractionDigits:2];
NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.342]]);

 value = (round(value*100)) / 100.0;

Use NSNumber *aNumber = [NSNumber numberWithDouble:number]; instead of 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;
}

To remove the decimals from your double, take a look at this output

Obj C

double hellodouble = 10.025;
NSLog(@"Your value with 2 decimals: %.2f", hellodouble);
NSLog(@"Your value with no decimals: %.0f", hellodouble);

The output will be:

10.02
10

Swift 2.1 and 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))

The output will be:

3.14
3

In Swift 2.0 and Xcode 7.2:

let pi:Double = 3.14159265358979
String(format:"%.2f", pi)

Example:

You could do this:

NSNumberFormatter* f = [[NSNumberFormatter alloc] init];
[f setNumberStyle:NSNumberFormatterDecimalStyle];
[f setFormat:@0.00"];


// test
NSNumber* a = @12;
NSString* s = [f stringFromNumber:a];
NSLog(@"%@", s);

For Swift there is a simple solution if you can't either import Foundation, use round() and/or does not want a String (usually the case when you're in Playground):

var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0

result: 31.726

I use the solution posted by Umberto Raimondi extending type Double:

extension Double {
func roundTo(places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}