How can I convert a NSString
containing a number of any primitive data type (e.g. int
, float
, char
, unsigned int
, etc.)? The problem is, I don't know which number type the string will contain at runtime.
I have an idea how to do it, but I'm not sure if this works with any type, also unsigned and floating point values:
long long scannedNumber;
NSScanner *scanner = [NSScanner scannerWithString:aString];
[scanner scanLongLong:&scannedNumber];
NSNumber *number = [NSNumber numberWithLongLong: scannedNumber];
Thanks for the help.
Use an NSNumberFormatter
:
NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
NSNumber *myNumber = [f numberFromString:@"42"];
If the string is not a valid number, then myNumber
will be nil
. If it is a valid number, then you now have all of the NSNumber
goodness to figure out what kind of number it actually is.
You can use -[NSString integerValue]
, -[NSString floatValue]
, etc. However, the correct (locale-sensitive, etc.) way to do this is to use -[NSNumberFormatter numberFromString:]
which will give you an NSNumber converted from the appropriate locale and given the settings of the NSNumberFormatter
(including whether it will allow floating point values).
@"2000"
to an int
and [@"2000" integerValue]
worked nicely and is a little simpler for my case.
[NSNumber numberWithInteger ...]
Objective-C
(Note: this method doesn't play nice with difference locales, but is slightly faster than a NSNumberFormatter
)
NSNumber *num1 = @([@"42" intValue]);
NSNumber *num2 = @([@"42.42" floatValue]);
Swift
Simple but dirty way
// Swift 1.2
if let intValue = "42".toInt() {
let number1 = NSNumber(integer:intValue)
}
// Swift 2.0
let number2 = Int("42')
// Swift 3.0
NSDecimalNumber(string: "42.42")
// Using NSNumber
let number3 = NSNumber(float:("42.42" as NSString).floatValue)
The extension-way This is better, really, because it'll play nicely with locales and decimals.
extension String {
var numberValue:NSNumber? {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
return formatter.number(from: self)
}
}
Now you can simply do:
let someFloat = "42.42".numberValue
let someInt = "42".numberValue
,
and .
the other way around (eg 42.000,42
. Something floatValue
probably does not account for?
@([@"42" intValue])
worked perfectly. Not worried about Locale.
For strings starting with integers, e.g., @"123"
, @"456 ft"
, @"7.89"
, etc., use -[NSString integerValue]
.
So, @([@"12.8 lbs" integerValue])
is like doing [NSNumber numberWithInteger:12]
.
You can also do this:
NSNumber *number = @([dictionary[@"id"] intValue]]);
Have fun!
If you know that you receive integers, you could use:
NSString* val = @"12";
[NSNumber numberWithInt:[val intValue]];
Here's a working sample of NSNumberFormatter reading localized number NSString (xCode 3.2.4, osX 10.6), to save others the hours I've just spent messing around. Beware: while it can handle trailing blanks ("8,765.4 " works), this cannot handle leading white space and this cannot handle stray text characters. (Bad input strings: " 8" and "8q" and "8 q".)
NSString *tempStr = @"8,765.4";
// localization allows other thousands separators, also.
NSNumberFormatter * myNumFormatter = [[NSNumberFormatter alloc] init];
[myNumFormatter setLocale:[NSLocale currentLocale]]; // happen by default?
[myNumFormatter setFormatterBehavior:NSNumberFormatterBehavior10_4];
// next line is very important!
[myNumFormatter setNumberStyle:NSNumberFormatterDecimalStyle]; // crucial
NSNumber *tempNum = [myNumFormatter numberFromString:tempStr];
NSLog(@"string '%@' gives NSNumber '%@' with intValue '%i'",
tempStr, tempNum, [tempNum intValue]);
[myNumFormatter release]; // good citizen
I wanted to convert a string to a double. This above answer didn't quite work for me. But this did: How to do string conversions in Objective-C?
All I pretty much did was:
double myDouble = [myString doubleValue];
Thanks All! I am combined feedback and finally manage to convert from text input ( string ) to Integer. Plus it could tell me whether the input is integer :)
NSNumberFormatter * f = [[NSNumberFormatter alloc] init];
[f setNumberStyle:NSNumberFormatterDecimalStyle];
NSNumber * myNumber = [f numberFromString:thresholdInput.text];
int minThreshold = [myNumber intValue];
NSLog(@"Setting for minThreshold %i", minThreshold);
if ((int)minThreshold < 1 )
{
NSLog(@"Not a number");
}
else
{
NSLog(@"Setting for integer minThreshold %i", minThreshold);
}
[f release];
I think NSDecimalNumber will do it:
Example:
NSNumber *theNumber = [NSDecimalNumber decimalNumberWithString:[stringVariable text]]];
NSDecimalNumber is a subclass of NSNumber, so implicit casting allowed.
What about C's standard atoi
?
int num = atoi([scannedNumber cStringUsingEncoding:NSUTF8StringEncoding]);
Do you think there are any caveats?
You can just use [string intValue]
or [string floatValue]
or [string doubleValue]
etc
You can also use NSNumberFormatter
class:
you can also do like this code 8.3.3 ios 10.3 support
[NSNumber numberWithInt:[@"put your string here" intValue]]
NSDecimalNumber *myNumber = [NSDecimalNumber decimalNumberWithString:@"123.45"];
NSLog(@"My Number : %@",myNumber);
Try this
NSNumber *yourNumber = [NSNumber numberWithLongLong:[yourString longLongValue]];
Note - I have used longLongValue as per my requirement. You can also use integerValue, longValue, or any other format depending upon your requirement.
Worked in Swift 3
NSDecimalNumber(string: "Your string")
I know this is very late but below code is working for me.
Try this code
NSNumber *number = @([dictionary[@"keyValue"] intValue]]);
This may help you. Thanks
extension String {
var numberValue:NSNumber? {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
return formatter.number(from: self)
}
}
let someFloat = "12.34".numberValue
Success story sharing
+currentLocale
, but you're correct; one must always consider the locale when dealing with converting stuff to-and-from human-readable form.