How to convert an NSString into an NSNumber

问题:

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.

回答1:

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.

回答2:

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).

回答3:

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 = NSNumberFormatter()
        formatter.numberStyle = .DecimalStyle
        return formatter.number(from: self)
    }
}

Now you can simply do:

let someFloat = \"42.42\".numberValue
let someInt = \"42\".numberValue

回答4:

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].

回答5:

You can also do this:

NSNumber *number = @([dictionary[@\"id\"] intValue]]);

Have fun!

回答6:

If you know that you receive integers, you could use:

NSString* val = @\"12\";
[NSNumber numberWithInt:[val intValue]];

回答7:

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

回答8:

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];

回答9:

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];

回答10:

I think NSDecimalNumber will do it:

Example:

NSNumber *theNumber = [NSDecimalNumber decimalNumberWithString:[stringVariable text]]];

NSDecimalNumber is a subclass of NSNumber, so implicit casting allowed.

回答11:

What about C\'s standard atoi?

int num = atoi([scannedNumber cStringUsingEncoding:NSUTF8StringEncoding]);

Do you think there are any caveats?

回答12:

You can just use [string intValue] or [string floatValue] or [string doubleValue] etc

You can also use NSNumberFormatter class:

回答13:

you can also do like this code 8.3.3 ios 10.3 support

[NSNumber numberWithInt:[@\"put your string here\" intValue]]

回答14:

NSDecimalNumber *myNumber = [NSDecimalNumber decimalNumberWithString:@\"123.45\"];
NSLog(@\"My Number : %@\",myNumber);

回答15:

Worked in Swift 3

NSDecimalNumber(string: \"Your string\") 

回答16:

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

回答17:

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.

回答18:

extension String {

    var numberValue:NSNumber? {
        let formatter = NumberFormatter()
        formatter.numberStyle = .decimal
        return formatter.number(from: self)
    }
}

let someFloat = \"12.34\".numberValue