Replace multiple characters in a string in Objective-C?

Question

In PHP I can do this:

$new = str_replace(array('/', ':', '.'), '', $new);

...to replace all instances of the characters / : . with a blank string (to remove them)

Can I do this easily in Objective-C? Or do I have to roll my own?

Currently I am doing multiple calls to stringByReplacingOccurrencesOfString:

strNew = [strNew stringByReplacingOccurrencesOfString:@"/" withString:@""];
strNew = [strNew stringByReplacingOccurrencesOfString:@":" withString:@""];
strNew = [strNew stringByReplacingOccurrencesOfString:@"." withString:@""];

Thanks,
matt

Nicholas Riley · Accepted Answer · 2009-04-03 13:55:09Z

122

A somewhat inefficient way of doing this:

NSString *s = @"foo/bar:baz.foo";
NSCharacterSet *doNotWant = [NSCharacterSet characterSetWithCharactersInString:@"/:."];
s = [[s componentsSeparatedByCharactersInSet: doNotWant] componentsJoinedByString: @""];
NSLog(@"%@", s); // => foobarbazfoo

Look at NSScanner and -[NSString rangeOfCharacterFromSet: ...] if you want to do this a bit more efficiently.

answered Apr 3, 2009 at 13:55

Nicholas Riley

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Comments

nikkumang · Accepted Answer · 2009-06-02 14:01:12Z

27

There are situations where your method is good enough I think matt.. BTW, I think it's better to use

[strNew setString: [strNew stringByReplacingOccurrencesOfString:@":" withString:@""]];

rather than

strNew = [strNew stringByReplacingOccurrencesOfString:@"/" withString:@""];

as I think you're overwriting an NSMutableString pointer with an NSString which might cause a memory leak.

answered Jun 2, 2009 at 14:01

nikkumang

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2 Comments

gotnull Over a year ago

NSString will not respond to setString.

user102008 Over a year ago

right, assuming that strNew is an NSMutableString. But then if you're going to use NSMutableString, then you might as well do [strNew replaceOccurrencesOfString:@":" withString:@"" options:0 range:NSMakeRange(0, [strNew length])]

Tommie C. · Accepted Answer · 2014-09-06 22:21:52Z

Had to do this recently and wanted to share an efficient method:

(assuming someText is a NSString or text attribute)

NSString* someText = @"1232342jfahadfskdasfkjvas12!";

(this example will strip numbers from a string)

[someText stringByReplacingOccurrencesOfString:@"[^0-9]" withString:@"" options:NSRegularExpressionSearch range:NSMakeRange(0, [someText length])];

Keep in mind that you will need to escape regex literal characters using Obj-c escape character:

(obj-c uses a double backslash to escape special regex literals)

...stringByReplacingOccurrencesOfString:@"[\\\!\\.:\\/]"

What makes this interesting is that NSRegularExpressionSearch option is little used but can lead to some very powerful controls:

You can find a nice iOS regex tutorial here and more on regular expressions at regex101.com

Stonz2 · Accepted Answer · 2014-01-22 16:34:48Z

7

Essentially the same thing as Nicholas posted above, but if you want to remove everything EXCEPT a set of characters (say you want to remove everything that isn't in the set "ABCabc123") then you can do the following:

NSString *s = @"A567B$%C^.123456abcdefg";
NSCharacterSet *doNotWant = [[NSCharacterSet characterSetWithCharactersInString:@"ABCabc123"] invertedSet];
s = [[s componentsSeparatedByCharactersInSet: doNotWant] componentsJoinedByString: @""];
NSLog(@"%@", s); // => ABC123abc

Useful in stripping out symbols and such, if you only want alphanumeric.

answered Jan 22, 2014 at 16:34

Stonz2

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Comments

Josh Brown · Accepted Answer · 2011-03-03 07:35:53Z

+ (NSString*) decodeHtmlUnicodeCharactersToString:(NSString*)str
{
    NSMutableString* string = [[NSMutableString alloc] initWithString:str];  // #&39; replace with '
    NSString* unicodeStr = nil;
    NSString* replaceStr = nil;
    int counter = -1;

    for(int i = 0; i < [string length]; ++i)
    {
        unichar char1 = [string characterAtIndex:i]; 
        for (int k = i + 1; k < [string length] - 1; ++k)
        {
            unichar char2 = [string characterAtIndex:k]; 

            if (char1 == '&'  && char2 == '#' ) 
            { 
                ++counter;
                unicodeStr = [string substringWithRange:NSMakeRange(i + 2 , 2)]; // read integer value i.e, 39
                replaceStr = [string substringWithRange:NSMakeRange (i, 5)];  // #&39;
                [string replaceCharactersInRange: [string rangeOfString:replaceStr] withString:[NSString stringWithFormat:@"%c",[unicodeStr intValue]]];
                break;
            }
        }
    }

    [string autorelease];

    if (counter > 1)
        return [self decodeHtmlUnicodeCharactersToString:string]; 
    else
        return string;
}

Matt Sephton · Accepted Answer · 2016-11-10 18:49:47Z

2

Here is an example in Swift 3 using the regularExpression option of replacingOccurances.

Use replacingOccurrences along with a the String.CompareOptions.regularExpression option.

Example (Swift 3):

var x = "<Hello, [play^ground+]>"
let y = x.replacingOccurrences(of: "[\\[\\]^+<>]", with: "7", options: .regularExpression, range: nil)
print(y)

Output:

7Hello, 7play7ground777

edited Nov 10, 2016 at 18:49

Matt Sephton

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answered Nov 10, 2016 at 15:36

Michael Peterson

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Comments

Gerard · Accepted Answer · 2012-01-18 21:39:17Z

1

If the characters you wish to remove were to be adjacent to each other you could use the

stringByReplacingCharactersInRange:(NSRange) withString:(NSString *)

Other than that, I think just using the same function several times isn't that bad. It is much more readable than creating a big method to do the same in a more generic way.

answered Jan 18, 2012 at 21:39

Gerard

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Comments

Matt Green · Accepted Answer · 2016-11-18 13:41:24Z

1

Create an extension on String...

extension String {

   func replacingOccurrences(of strings:[String], with replacement:String) -> String {
      var newString = self
      for string in strings {
          newString = newString.replacingOccurrences(of: string, with: replacement)
      }
      return newString
   }

}

Call it like this:

aString = aString.replacingOccurrences(of:['/', ':', '.'], with:"")

answered Nov 18, 2016 at 13:41

Matt Green

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