Finding out whether a string is numeric or not
How can we check if a string is made up of numbers only. I am taking out a substring from a string and want to check if it is a numeric substring or not.
NSString *newString = [myString substrin开发者_如何学运维gWithRange:NSMakeRange(2,3)];
Here's one way that doesn't rely on the limited precision of attempting to parse the string as a number:
NSCharacterSet* notDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
if ([newString rangeOfCharacterFromSet:notDigits].location == NSNotFound)
{
// newString consists only of the digits 0 through 9
}
See +[NSCharacterSet decimalDigitCharacterSet]
and -[NSString rangeOfCharacterFromSet:]
.
I'd suggest using the numberFromString:
method from the NSNumberFormatter class, as if the number is not valid, it will return nil; otherwise, it will return you an NSNumber.
NSNumberFormatter *nf = [[[NSNumberFormatter alloc] init] autorelease];
BOOL isDecimal = [nf numberFromString:newString] != nil;
Validate by regular expression, by pattern "^[0-9]+$"
, with following method -validateString:withPattern:
.
[self validateString:"12345" withPattern:"^[0-9]+$"];
- If "123.123" is considered
- With pattern
"^[0-9]+(.{1}[0-9]+)?$"
- With pattern
- If exactly 4 digit numbers, without
"."
.- With pattern
"^[0-9]{4}$"
.
- With pattern
- If digit numbers without
"."
, and the length is between 2 ~ 5.- With pattern
"^[0-9]{2,5}$"
.
- With pattern
- With minus sign:
"^-?\d+$"
The regular expression can be checked in the online web site.
The helper function is as following.
// Validate the input string with the given pattern and
// return the result as a boolean
- (BOOL)validateString:(NSString *)string withPattern:(NSString *)pattern
{
NSError *error = nil;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionCaseInsensitive error:&error];
NSAssert(regex, @"Unable to create regular expression");
NSRange textRange = NSMakeRange(0, string.length);
NSRange matchRange = [regex rangeOfFirstMatchInString:string options:NSMatchingReportProgress range:textRange];
BOOL didValidate = NO;
// Did we find a matching range
if (matchRange.location != NSNotFound)
didValidate = YES;
return didValidate;
}
Swift 3 version:
Test in playground.
import UIKit
import Foundation
func validate(_ str: String, pattern: String) -> Bool {
if let range = str.range(of: pattern, options: .regularExpression) {
let result = str.substring(with: range)
print(result)
return true
}
return false
}
let a = validate("123", pattern: "^-?[0-9]+")
print(a)
You could create an NSScanner and simply scan the string:
NSDecimal decimalValue;
NSScanner *sc = [NSScanner scannerWithString:newString];
[sc scanDecimal:&decimalValue];
BOOL isDecimal = [sc isAtEnd];
Check out NSScanner's documentation for more methods to choose from.
I think the easiest way to check that every character within a given string is numeric is probably:
NSString *trimmedString = [newString stringByTrimmingCharactersInSet:[NSCharacterSet decimalDigitCharacterSet]];
if([trimmedString length])
{
NSLog(@"some characters outside of the decimal character set found");
}
else
{
NSLog(@"all characters were in the decimal character set");
}
Use one of the other NSCharacterSet factory methods if you want complete control over acceptable characters.
Swift 3 solution if need to verify that the string has only digits:
CharacterSet.decimalDigits.isSuperset(of: CharacterSet(charactersIn: myString))
This original question was about Objective-C, but it was also posted years before Swift was announced. So, if you're coming here from Google and are looking for a solution that uses Swift, here you go:
let testString = "12345"
let badCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
if testString.rangeOfCharacterFromSet(badCharacters) == nil {
print("Test string was a number")
} else {
print("Test string contained non-digit characters.")
}
to be clear, this functions for integers in strings.
heres a little helper category based off of John's answer above:
in .h file
@interface NSString (NumberChecking)
+(bool)isNumber:(NSString *)string;
@end
in .m file
#import "NSString+NumberChecking.h"
@implementation NSString (NumberChecking)
+(bool)isNumber {
if([self rangeOfCharacterFromSet:[[NSCharacterSet decimalDigitCharacterSet] invertedSet]].location == NSNotFound) {
return YES;
}else {
return NO;
}
}
@end
usage:
#import "NSString+NumberChecking.h"
if([someString isNumber]) {
NSLog(@"is a number");
}else {
NSLog(@"not a number");
}
Swift 3 solution could be like:
extension String {
var doubleValue:Double? {
return NumberFormatter().number(from:self)?.doubleValue
}
var integerValue:Int? {
return NumberFormatter().number(from:self)?.intValue
}
var isNumber:Bool {
get {
let badCharacters = NSCharacterSet.decimalDigits.inverted
return (self.rangeOfCharacter(from: badCharacters) == nil)
}
}
}
An extension of @John Calsbeek's answer, and clarification of @Jeff and @gyratory circus's comments.
+ (BOOL)doesContainDigitsOnly:(NSString *)string
{
NSCharacterSet *nonDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
BOOL containsDigitsOnly = [string rangeOfCharacterFromSet:nonDigits].location == NSNotFound;
return containsDigitsOnly;
}
+ (BOOL)doesContainNonDigitsOnly:(NSString *)string
{
NSCharacterSet *digits = [NSCharacterSet decimalDigitCharacterSet];
BOOL containsNonDigitsOnly = [string rangeOfCharacterFromSet:digits].location == NSNotFound;
return containsNonDigitsOnly;
}
The following can be added as category methods for NSString
- (BOOL)doesContainDigitsOnly
{
NSCharacterSet *nonDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
BOOL containsDigitsOnly = [self rangeOfCharacterFromSet:nonDigits].location == NSNotFound;
return containsDigitsOnly;
}
- (BOOL)doesContainNonDigitsOnly
{
NSCharacterSet *digits = [NSCharacterSet decimalDigitCharacterSet];
BOOL containsNonDigitsOnly = [self rangeOfCharacterFromSet:digits].location == NSNotFound;
return containsNonDigitsOnly;
}
John Calsbeek's answer is nearly correct but omits some Unicode edge cases.
Per the documentation for decimalDigitCharacterSet
, that set includes all characters categorized by Unicode as Nd
. Thus their answer will accept, among others:
१
(U+0967 DEVANAGARI DIGIT ONE)᠑
(U+1811 MONGOLIAN DIGIT ONE)
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