Remove all non-numeric characters from an NSString, keeping spaces

Question

I am trying to remove all of the non-numeric characters from an NSString, but I also need to keep the spaces. Here is what I have been using.

NS         


        

Answer 1

try using NSScanner

NSString *originalString = @"(123) 123123 abc";
NSMutableString *strippedString = [NSMutableString 
    stringWithCapacity:originalString.length];

NSScanner *scanner = [NSScanner scannerWithString:originalString];
NSCharacterSet *numbers = [NSCharacterSet 
    characterSetWithCharactersInString:@"0123456789 "];

while ([scanner isAtEnd] == NO) {
    NSString *buffer;
    if ([scanner scanCharactersFromSet:numbers intoString:&buffer]) {
        [strippedString appendString:buffer];
    } else {
        [scanner setScanLocation:([scanner scanLocation] + 1)];
    }
}

NSLog(@"%@", strippedString); // "123123123"

Answer 2

Easily done by creating a character set of characters you want to keep and using invertedSet to create an "all others" set. Then split the string into an array separated by any characters in this set and reassemble the string again. Sounds complicated but very simple to implement:

NSCharacterSet *setToRemove =   
    [NSCharacterSet characterSetWithCharactersInString:@"0123456789 "];
NSCharacterSet *setToKeep = [setToRemove invertedSet];

NSString *newString = 
        [[someString componentsSeparatedByCharactersInSet:setToKeep]
            componentsJoinedByString:@""];

result: 333 9599 99

Answer 3

In brief, you can use NSCharacterSet to examine only those chars that are interesting to you and ignore the rest.

- (void) stripper {

    NSString *inString = @"A1 B2 C3 D4";
    NSString *outString = @"";

    for (int i = 0; i < inString.length; i++) {

        if ([[NSCharacterSet whitespaceCharacterSet] characterIsMember:[inString characterAtIndex:i]] || [[NSCharacterSet decimalDigitCharacterSet] characterIsMember:[inString characterAtIndex:i]]) {

            outString = [outString stringByAppendingString:[NSString stringWithFormat:@"%c",[inString characterAtIndex:i]]];
        }
    }
}

Answer 4

You could alter your first regex to include a space after the 9:

In swift:

var str = "test Test 333 9599 999";
val strippedStr = str.stringByReplacingOccurrencesOfString("[^0-9 ]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range:nil);
// strippedStr = " 33 9599 999"

While this leaves the leading space, you could apply a whitespace trimming to deal with that:

strippedStr.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
// strippedStr = "33 9599 999"

Answer 5

NSMutableString strippedBbox = [_bbox mutableCopy];
NSCharacterSet* charSet = [NSCharacterSet characterSetWithCharactersInString:@"1234567890 "].invertedSet;

NSUInteger start = 0;
NSUInteger length = _bbox.length;

while(length > 0)
{
    NSRange range = [strippedBbox rangeOfCharacterFromSet:charSet options:0 range:NSMakeRange(start, length)];

    if(range.location == NSNotFound)
    {
        break;
    }

    start += (range.location + range.length);
    length -= range.length;

    [strippedBbox replaceCharactersInRange:range withString:@""];
}

Answer 6

For Swift 3.0.1 folks

var str = "1 3 6 .599.188-99 "
    str.replacingOccurrences(of: "[^0-9]", with: "", options: .regularExpression, range: nil)

Output: "13659918899"

This also trim spaces from string