How to check if NSString is contains a numeric value?

I have a string that is being generate from a formula, however I only want to use the string as long as all of its characters are numeric, if not that I want to do something different for instance display an error message.

I have been having a look round but am finding it hard to find anything that works along the lines of what I am wanting to do. I have looked at NSScanner but I am not sure if its checking the whole string and then I am not actually sure how to check if these characters are numeric

- (void)isNumeric:(NSString *)code{

    NSScanner *ns = [NSScanner scannerWithString:code];
    if ( [ns scanFloat:NULL] ) //what can I use instead of NULL?
    {
        NSLog(@"INSIDE IF");
    }
    else {
    NSLog(@"OUTSIDE IF");
    }
}

So after a few more hours searching I have stumbled across an implementation that dose exactly what I am looking for.

so if you are looking to check if their are any alphanumeric characters in your NSString this works here

-(bool) isNumeric:(NSString*) hexText
{

    NSNumberFormatter* numberFormatter = [[[NSNumberFormatter alloc] init] autorelease];

    NSNumber* number = [numberFormatter numberFromString:hexText];

    if (number != nil) {
        NSLog(@"%@ is numeric", hexText);
        //do some stuff here      
        return true;
    }

    NSLog(@"%@ is not numeric", hexText);
    //or do some more stuff here
    return false;
}

hope this helps.

Something like this would work:

@interface NSString (usefull_stuff)
- (BOOL) isAllDigits;
@end

@implementation NSString (usefull_stuff)

- (BOOL) isAllDigits
{
    NSCharacterSet* nonNumbers = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
    NSRange r = [self rangeOfCharacterFromSet: nonNumbers];
    return r.location == NSNotFound && self.length > 0;
}

@end

then just use it like this:

NSString* hasOtherStuff = @"234 other stuff";
NSString* digitsOnly = @"123345999996665003030303030";

BOOL b1 = [hasOtherStuff isAllDigits];
BOOL b2 = [digitsOnly isAllDigits];

You don't have to wrap the functionality in a private category extension like this, but it sure makes it easy to reuse..

I like this solution better than the others since it wont ever overflow some int/float that is being scanned via NSScanner - the number of digits can be pretty much any length.

Consider NSString integerValue - it returns an NSInteger. However, it will accept some strings that are not entirely numeric and does not provide a mechanism to determine strings which are not numeric at all. This may or may not be acceptable.

For instance, " 13 " -> 13, "42foo" -> 42 and "helloworld" -> 0.

Happy coding.

Now, since the above was sort of a tangent to the question, see determine if string is numeric. Code taken from link, with comments added:

BOOL isNumeric(NSString *s)
{
   NSScanner *sc = [NSScanner scannerWithString: s];
   // We can pass NULL because we don't actually need the value to test
   // for if the string is numeric. This is allowable.
   if ( [sc scanFloat:NULL] )
   {
      // Ensure nothing left in scanner so that "42foo" is not accepted.
      // ("42" would be consumed by scanFloat above leaving "foo".)
      return [sc isAtEnd];
   }
   // Couldn't even scan a float :(
   return NO;
}

The above works with just scanFloat -- e.g. no scanInt -- because the range of a float is much larger than that of an integer (even a 64-bit integer).

This function checks for "totally numeric" and will accept "42" and "0.13E2" but reject " 13 ", "42foo" and "helloworld".

Like this:

- (void)isNumeric:(NSString *)code{

    NSScanner *ns = [NSScanner scannerWithString:code];
    float the_value;
    if ( [ns scanFloat:&the_value] )
    {
        NSLog(@"INSIDE IF");
        // do something with `the_value` if you like
    }
    else {
    NSLog(@"OUTSIDE IF");
    }
}

It's very simple.

+ (BOOL)isStringNumeric:(NSString *)text
{
    NSCharacterSet *alphaNums = [NSCharacterSet decimalDigitCharacterSet];
    NSCharacterSet *inStringSet = [NSCharacterSet characterSetWithCharactersInString:text];        
    return [alphaNums isSupersetOfSet:inStringSet];
}

Faced same problem in Swift.
In Swift you should use this code, according TomSwift's answer:

func isAllDigits(str: String) -> Bool {

    let nonNumbers = NSCharacterSet.decimalDigitCharacterSet()

    if let range = str.rangeOfCharacterFromSet(nonNumbers) {
        return true
    }
    else {
        return false
    }
}

P.S. Also you can use other NSCharacterSets or their combinations to check your string!

For simple numbers like "12234" or "231231.23123" the answer can be simple.

There is a transformation law for int numbers: when string with integer transforms to int (or long) number and then, again, transforms it back to another string these strings will be equal.

In Objective C it will looks like:

NSString *numStr=@"1234",*num2Str=nil;
num2Str=[NSString stringWithFormat:@"%lld",numStr.longlongValue];

if([numStr isEqualToString: num2Str]) NSLog(@"numStr is an integer number!");

By using this transformation law we can create solution
to detect double or long numbers:

NSString *numStr=@"12134.343"
NSArray *numList=[numStr componentsSeparatedByString:@"."];

if([[NSString stringWithFormat:@"%lld", numStr.longLongValue] isEqualToString:numStr]) NSLog(@"numStr is an integer number"); 
else 
if( numList.count==2 &&
               [[NSString stringWithFormat:@"%lld",((NSString*)numList[0]).longLongValue] isEqualToString:(NSString*)numList[0]] &&
               [[NSString stringWithFormat:@"%lld",((NSString*)numList[1]).longLongValue] isEqualToString:(NSString*)numList[1]] )
            NSLog(@"numStr is a double number");
else 
NSLog(@"numStr is not a number");

I did not copy the code above from my work code so can be some mistakes, but I think the main point is clear. Of course this solution doesn't work with numbers like "1E100", as well it doesn't take in account size of integer and fractional part. By using the law described above you can do whatever number detection you need.

C.Johns' answer is wrong. If you use a formatter, you risk apple changing their codebase at some point and having the formatter spit out a partial result. Tom's answer is wrong too. If you use the rangeOfCharacterFromSet method and check for NSNotFound, it'll register a true if the string contains even one number. Similarly, other answers in this thread suggest using the Integer value method. That is also wrong because it will register a true if even one integer is present in the string. The OP asked for an answer that ensures the entire string is numerical. Try this:

NSCharacterSet *searchSet = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];

Tom was right about this part. That step gives you the non-numerical string characters. But then we do this:

NSString *trimmedString = [string stringByTrimmingCharactersInSet:searchSet];

return (string.length == trimmedString.length);

Tom's inverted character set can TRIM a string. So we can use that trim method to test if any non numerals exist in the string by comparing their lengths.

来源：https://stackoverflow.com/questions/6644004/how-to-check-if-nsstring-is-contains-a-numeric-value

标签

iphone

ios

nsstring

numeric