[英]Objective-C: how to round of double to 2 decimal places if there is some value after decimal otherwise the value should be rounded to 0 decimal places
如何格式化实际值,如下所示:
/// variable
double value = 9.99999
/// conversion
NSSting* str = [NSString stringWithFormat:@"%.2f", value];
/// result
9.99
但是当值是9.00000时返回9.00
我应该如何格式化将9.0000 as 9
显示9.0000 as 9
并将9.99999 as 9.99
的字符串?
您应该看看NSNumberFormatter。 我在Swift中添加了代码,但是您可以轻松地将其转换为Objective-C:
let formatter = NSNumberFormatter()
formatter.numberStyle = NSNumberFormatterStyle.DecimalStyle
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = 2
formatter.roundingMode = NSNumberFormatterRoundingMode.RoundFloor
print("\(formatter.stringFromNumber(9.00))")
// Objective-C(信用归NSNoob所有)
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
[formatter setNumberStyle: NSNumberFormatterDecimalStyle];
formatter.minimumFractionDigits = 0;
formatter.maximumFractionDigits = 2;
[formatter setRoundingMode:NSNumberFormatterRoundFloor];
NSString* formattedStr = [formatter stringFromNumber:[NSNumber numberWithFloat:9.00]];
NSLog(@"number: %@",formattedStr);
根据此相关问题的答案 ,您可以尝试做:
// Usage for Output 1 — 1.23
[NSString stringWithFormat:@"%@ — %@", [self stringWithFloat:1],
[self stringWithFloat:1.234];
// Checks if it's an int and if not displays 2 decimals.
+ (NSString*)stringWithFloat:(CGFloat)_float
{
NSString *format = (NSInteger)_float == _float ? @"%.0f" : @"%.2f";
return [NSString stringWithFormat:format, _float];
}
在Swift中,您可以使用“ %g
”(“使用最短表示形式”)格式说明符:
import UIKit
let firstFloat : Float = 1.0
let secondFloat : Float = 1.2345
var outputString = NSString(format: "first: %0.2g second: %0.2g", firstFloat, secondFloat)
print("\(outputString)")
返回:
"first: 1 second: 1.2"
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