Call a method on every word in NSString - objective-c

I would like to loop through an NSString and call a custom function on every word that has certain criterion (For example, "has 2 'L's"). I was wondering what the best way of approaching that was. Should I use Find/Replace patterns? Blocks?
-(NSString *)convert:(NSString *)wordToConvert{
/// This I have already written
Return finalWord;
}
-(NSString *) method:(NSString *) sentenceContainingWords{
// match every word that meets the criteria (for example the 2Ls) and replace it with what convert: does.
}

To enumerate the words in a string, you should use -[NSString enumerateSubstringsInRange:options:usingBlock:] with NSStringEnumerationByWords and NSStringEnumerationLocalized. All of the other methods listed use a means of identifying words which may not be locale-appropriate or correspond to the system definition. For example, two words separated by a comma but not whitespace (e.g. "foo,bar") would not be treated as separate words by any of the other answers, but they are in Cocoa text views.
[aString enumerateSubstringsInRange:NSMakeRange(0, [aString length])
options:NSStringEnumerationByWords | NSStringEnumerationLocalized
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){
if ([substring rangeOfString:#"ll" options:NSCaseInsensitiveSearch].location != NSNotFound)
/* do whatever */;
}];
As documented for -enumerateSubstringsInRange:options:usingBlock:, if you call it on a mutable string, you can safely mutate the string being enumerated within the enclosingRange. So, if you want to replace the matching words, you can with something like [aString replaceCharactersInRange:substringRange withString:replacementString].

The two ways I know of looping an array that will work for you are as follows:
NSArray *words = [sentence componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]];
for (NSString *word in words)
{
NSString *transformedWord = [obj method:word];
}
and
NSArray *words = [sentence componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]];
[words enumerateObjectsWithOptions:NSEnumerationConcurrent usingBlock:^(id word, NSUInteger idx, BOOL *stop){
NSString *transformedWord = [obj method:word];
}];
The other method, –makeObjectsPerformSelector:withObject:, won't work for you. It expects to be able to call [word method:obj] which is backwards from what you expect.

If you could write your criteria with regular expressions, then you could probably do a regular expression matching to fetch these words and then pass them to your convert: method.
You could also do a split of string into an array of words using componentsSeparatedByString: or componentsSeparatedByCharactersInSet:, then go over the words in the array and detect if they fit your criteria somehow. If they fit, then pass them to convert:.
Hope this helps.

As of iOS 12/macOS 10.14 the recommended way to do this is with the Natural Language framework.
For example:
import NaturalLanguage
let myString = "..."
let tokeniser = NLTokenizer(unit: .word)
tokeniser.string = myString
tokeniser.enumerateTokens(in: myString.startIndex..<myString.endIndex) { wordRange, attributes in
performActionOnWord(myString[wordRange])
return true // or return false to stop enumeration
}
Using NLTokenizer also has the benefit of allowing you to optionally specify the language of the string beforehand:
tokeniser.setLanguage(.hebrew)

I would recommend using a while loop to go through the string like this.
NSRange spaceRange = [sentenceContainingWords rangeOfString:#" "];
NSRange previousRange = (NSRange){0,0};
do {
NSString *wordString;
wordString = [sentenceContainingWord substringWithRange:(NSRange){previousRange.location+1,(spaceRange.location-1)-(previousRange.location+1)}];
//use the +1's to not include the spaces in the strings
[self convert:wordString];
previousRange = spaceRange;
spaceRange = [sentenceContainingWords rangeOfString:#" "];
} while(spaceRange.location != NSNotFound);
This code would probably need to be rewritten because its pretty rough, but you should get the idea.
Edit: Just saw Jacob Gorban's post, you should definitely do it like that.

Related

Create NSArray of keywords starting/ending with % inside a string

I have a long string that contain keywords that start and end with the percent sign. E.g.:
My name is %user_username% and I live at %location_address%. You can
reach me at %user_phone%.
What method would I use to extract all strings that begin and end with % and put those into an NSArray so that I can replace them with their correct text representations?
Assuming that there are no % signs inside your strings of interest (e.g "a%ab%b%c"), you could use the componentsSeparatedByString: or componentsSeparatedByCharactersInSet: to get an array of strings separated by the % sign. From there, it's pretty easy to figure out which strings in that array are between the percent signs, and which are unnecessary.
I think internally though, those methods are likely implemented as something like a loop looking for %s. Maybe they parallelize the search on big strings, or use special knowledge of the internal structure of the string to make things faster -- those are the only ways I can see to speed up the search, assuming that you're stuck with keeping it all in a % delimited string (if speed is really an issue, then the answer is probably to use an alternative representation).
This is what I came up with that works:
- (NSArray *)replaceKeywords:(NSString *)keywordString {
NSString *start = #"%";
NSString *end = #"%";
NSMutableArray* strings = [NSMutableArray arrayWithCapacity:0];
NSRange startRange = [keywordString rangeOfString:start];
for( ;; ) {
if (startRange.location != NSNotFound) {
NSRange targetRange;
targetRange.location = startRange.location + startRange.length;
targetRange.length = [keywordString length] - targetRange.location;
NSRange endRange = [keywordString rangeOfString:end options:0 range:targetRange];
if (endRange.location != NSNotFound) {
targetRange.length = endRange.location - targetRange.location;
[strings addObject:[keywordString substringWithRange:targetRange]];
NSRange restOfString;
restOfString.location = endRange.location + endRange.length;
restOfString.length = [keywordString length] - restOfString.location;
startRange = [keywordString rangeOfString:start options:0 range:restOfString];
} else {
break;
}
} else {
break;
}
}
return strings;
}
I slightly modified the method from Get String Between Two Other Strings in ObjC

Split NSString into words, then rejoin it into original form

I am splitting an NSString like this: (filter string is an nsstring)
seperatorSet = [NSMutableCharacterSet whitespaceAndNewlineCharacterSet];
[seperatorSet formUnionWithCharacterSet:[NSCharacterSet punctuationCharacterSet]];
NSMutableArray *words = [[filterString componentsSeparatedByCharactersInSet:seperatorSet] mutableCopy];
I want to put words back into the form of filter string with the original punctuation and spacing. The reason I want to do this is I want to change some words and put it back together as it was originally.
A more robust way to split by words is to use string enumeration. A space is not always the delimiter and not all languages delimit spaces anyway (e.g. Japanese).
NSString * string = #" \n word1! word2,%$?'/word3.word4 ";
[string enumerateSubstringsInRange:NSMakeRange(0, string.length)
options:NSStringEnumerationByWords
usingBlock:
^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
NSLog(#"Substring: '%#'", substring);
}];
// Logs:
// Substring: 'word1'
// Substring: 'word2'
// Substring: 'word3'
// Substring: 'word4'
NSString *myString = #"Foo Bar Blah B..";
NSArray *myWords = [myString componentsSeparatedByCharactersInSet:
[NSCharacterSet characterSetWithCharactersInString:#" "]
];
NSString* string = [myWords componentsJoinedByString: #" "];
NSLog(#"%#",string);
Since you eliminate the original punctuation, there's no way to turn it back automatically.
The only way is not to use componentsSeparatedByCharactersInSet.
An alternative solution may be to iterate through the string and, for each char, check if it belongs to your character set.
If yes, add the char to a list and the substring to another list (you may use NSMutableArray class).
This way, for example, you know that the punctuation char between the first and the second substring is the first character in your list of separators.
You can use the pathArray componentsJoinedByString: method of the array class to rejoin the words:
NSString *orig = [words pathArray componentsJoinedByString:#" "];
How are you determining which words need to be replaced? Instead of breaking it apart in the first place, perhaps using -stringByReplacingOccurrencesOfString:withString:options:range: would be more suitable.
My guess is you may not be using the best API. If you're really worried about words, you should be using a word-based API. I'm a bit hazy on whether that would be NSDataDetector or something else. (I believe NSRegularExpression can deal with word boundaries in a smarter way.)
If you are using Mac OS X 10.7+ or iOS 4+ you can use NSRegularExpression, The pattern to replace a word is: "\b word \b" - (no spaces around word) \b matches a word boundary. Look at methods replaceMatchesInString:options:range:withTemplate: and stringByReplacingMatchesInString:options:range:withTemplate:.
Under 10.6 pr earlier if you wish to use regular expressions you can wrap the regcomp/regexec C-based functions, they support word boundaries as well. However you may prefer to use one of the other Cocoa options mentioned in other answers for this simple case.

Objective-C: -[NSString wordCount]

What's a simple implementation of the following NSString category method that returns the number of words in self, where words are separated by any number of consecutive spaces or newline characters? Also, the string will be less than 140 characters, so in this case, I prefer simplicity & readability at the sacrifice of a bit of performance.
#interface NSString (Additions)
- (NSUInteger)wordCount;
#end
I found the following solutions:
implementation of -[NSString wordCount]
implementation of -[NSString wordCount] - seems a bit simpler
But, isn't there a simpler way?
Why not just do the following?
- (NSUInteger)wordCount {
NSCharacterSet *separators = [NSCharacterSet whitespaceAndNewlineCharacterSet];
NSArray *words = [self componentsSeparatedByCharactersInSet:separators];
NSIndexSet *separatorIndexes = [words indexesOfObjectsPassingTest:^BOOL(id obj, NSUInteger idx, BOOL *stop) {
return [obj isEqualToString:#""];
}];
return [words count] - [separatorIndexes count];
}
I believe you have identified the 'simplest'. Nevertheless, to answer to your original question - "a simple implementation of the following NSString category...", and have it posted directly here for posterity:
#implementation NSString (GSBString)
- (NSUInteger)wordCount
{
__block int words = 0;
[self enumerateSubstringsInRange:NSMakeRange(0,self.length)
options:NSStringEnumerationByWords
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {words++;}];
return words;
}
#end
There are a number of simpler implementations, but they all have tradeoffs. For example, Cocoa (but not Cocoa Touch) has word-counting baked in:
- (NSUInteger)wordCount {
return [[NSSpellChecker sharedSpellChecker] countWordsInString:self language:nil];
}
It's also trivial to count words as accurately as the scanner simply using [[self componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]] count]. But I've found the performance of that method degrades a lot for longer strings.
So it depends on the tradeoffs you want to make. I've found the absolute fastest is just to go straight-up ICU. If you want simplest, using existing code is probably simpler than writing any code at all.
- (NSUInteger) wordCount
{
NSArray *words = [self componentsSeparatedByString:#" "];
return [words count];
}
Looks like the second link I gave in my question still reigns as not only the fastest but also, in hindsight, a relatively simple implementation of -[NSString wordCount].
A Objective-C one-liner version
NSInteger wordCount = word ? ([word stringByTrimmingCharactersInSet:NSCharacterSet.whitespaceAndNewlineCharacterSet.invertedSet].length + 1) : 0;
Swift 3:
let words: [Any] = (string.components(separatedBy: " "))
let count = words.count

Is there a way to get Spell Check data from an NSString?

I'm writing a simple shift cipher iPhone app as a pet project, and one piece of functionality I'm currently designing is a "universal" decryption of an NSString, that returns an NSArray, all of NSStrings:
- (NSArray*) decryptString: (NSString*)ciphertext{
NSMutableArray* theDecryptions = [NSMutableArray arrayWithCapacity:ALPHABET];
for (int i = 0; i < ALPHABET; ++i) {
NSString* theNewPlainText = [self decryptString:ciphertext ForShift:i];
[theDecryptions insertObject:theNewPlainText
atIndex:i];
}
return theDecryptions;
}
I'd really like to pass this NSArray into another method that attempts to spell check each individual string within the array, and builds a new array that puts the strings with the fewest typo'd words at lower indicies, so they're displayed first. I'd like to use the system's dictionary like a text field would, so I can match against words that have been trained into the phone by its user.
My current guess is to split a given string up into words, then spell check each with NSSpellChecker's -checkSpellingOfString:StartingAt: and using the number of correct words to sort the Array. Is there an existing library method or well-accepted pattern that would help return such a value for a given string?
Well, I found a solution that works using UIKit/UITextChecker. It correctly finds the user's most preferred language dictionary, but I'm not sure if it includes learned words in the actual rangeOfMisspelledWords... method. If it doesn't, calling [UITextChecker hasLearnedWord] on currentWord inside the bottom if statement should be enough to find user-taught words.
As noted in the comments, it may be prudent to call rangeOfMisspelledWords with each of the top few languages in [UITextChecker availableLanguages], to help multilingual users.
-(void) checkForDefinedWords {
NSArray* words = [message componentsSeparatedByString:#" "];
NSInteger wordsFound = 0;
UITextChecker* checker = [[UITextChecker alloc] init];
//get the first language in the checker's memory- this is the user's
//preferred language.
//TODO: May want to search with every language (or top few) in the array
NSString* preferredLang = [[UITextChecker availableLanguages] objectAtIndex:0];
//for each word in the array, determine whether it is a valid word
for(NSString* currentWord in words){
NSRange range;
range = [checker rangeOfMisspelledWordInString:currentWord
range:NSMakeRange(0, [currentWord length])
startingAt:0
wrap:NO
language:preferredLang];
//if it is valid (no errors found), increment wordsFound
if (range.location == NSNotFound) {
//NSLog(#"%# %#", #"Valid Word found:", currentWord);
wordsFound++;
}
else {
//NSLog(#"%# %#", #"Invalid Word found:", currentWord);
}
}
//After all "words" have been searched, save wordsFound to validWordCount
[self setValidWordCount:wordsFound];
[checker release];
}

NSString - Convert to pure alphabet only (i.e. remove accents+punctuation)

I'm trying to compare names without any punctuation, spaces, accents etc.
At the moment I am doing the following:
-(NSString*) prepareString:(NSString*)a {
//remove any accents and punctuation;
a=[[[NSString alloc] initWithData:[a dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES] encoding:NSASCIIStringEncoding] autorelease];
a=[a stringByReplacingOccurrencesOfString:#" " withString:#""];
a=[a stringByReplacingOccurrencesOfString:#"'" withString:#""];
a=[a stringByReplacingOccurrencesOfString:#"`" withString:#""];
a=[a stringByReplacingOccurrencesOfString:#"-" withString:#""];
a=[a stringByReplacingOccurrencesOfString:#"_" withString:#""];
a=[a lowercaseString];
return a;
}
However, I need to do this for hundreds of strings and I need to make this more efficient. Any ideas?
NSString* finish = [[start componentsSeparatedByCharactersInSet:[[NSCharacterSet letterCharacterSet] invertedSet]] componentsJoinedByString:#""];
Before using any of these solutions, don't forget to use decomposedStringWithCanonicalMapping to decompose any accented letters. This will turn, for example, é (U+00E9) into e ‌́ (U+0065 U+0301). Then, when you strip out the non-alphanumeric characters, the unaccented letters will remain.
The reason why this is important is that you probably don't want, say, “dän” and “dün”* to be treated as the same. If you stripped out all accented letters, as some of these solutions may do, you'll end up with “dn”, so those strings will compare as equal.
So, you should decompose them first, so that you can strip the accents and leave the letters.
*Example from German. Thanks to Joris Weimar for providing it.
On a similar question, Ole Begemann suggests using stringByFoldingWithOptions: and I believe this is the best solution here:
NSString *accentedString = #"ÁlgeBra";
NSString *unaccentedString = [accentedString stringByFoldingWithOptions:NSDiacriticInsensitiveSearch locale:[NSLocale currentLocale]];
Depending on the nature of the strings you want to convert, you might want to set a fixed locale (e.g. English) instead of using the user's current locale. That way, you can be sure to get the same results on every machine.
One important precision over the answer of BillyTheKid18756 (that was corrected by Luiz but it was not obvious in the explanation of the code):
DO NOT USE stringWithCString as a second step to remove accents, it can add unwanted characters at the end of your string as the NSData is not NULL-terminated (as stringWithCString expects it).
Or use it and add an additional NULL byte to your NSData, like Luiz did in his code.
I think a simpler answer is to replace:
NSString *sanitizedText = [NSString stringWithCString:[sanitizedData bytes] encoding:NSASCIIStringEncoding];
By:
NSString *sanitizedText = [[[NSString alloc] initWithData:sanitizedData encoding:NSASCIIStringEncoding] autorelease];
If I take back the code of BillyTheKid18756, here is the complete correct code:
// The input text
NSString *text = #"BûvérÈ!#$&%^&(*^(_()-*/48";
// Defining what characters to accept
NSMutableCharacterSet *acceptedCharacters = [[NSMutableCharacterSet alloc] init];
[acceptedCharacters formUnionWithCharacterSet:[NSCharacterSet letterCharacterSet]];
[acceptedCharacters formUnionWithCharacterSet:[NSCharacterSet decimalDigitCharacterSet]];
[acceptedCharacters addCharactersInString:#" _-.!"];
// Turn accented letters into normal letters (optional)
NSData *sanitizedData = [text dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
// Corrected back-conversion from NSData to NSString
NSString *sanitizedText = [[[NSString alloc] initWithData:sanitizedData encoding:NSASCIIStringEncoding] autorelease];
// Removing unaccepted characters
NSString* output = [[sanitizedText componentsSeparatedByCharactersInSet:[acceptedCharacters invertedSet]] componentsJoinedByString:#""];
If you are trying to compare strings, use one of these methods. Don't try to change data.
- (NSComparisonResult)localizedCompare:(NSString *)aString
- (NSComparisonResult)localizedCaseInsensitiveCompare:(NSString *)aString
- (NSComparisonResult)compare:(NSString *)aString options:(NSStringCompareOptions)mask range:(NSRange)range locale:(id)locale
You NEED to consider user locale to do things write with strings, particularly things like names.
In most languages, characters like ä and å are not the same other than they look similar. They are inherently distinct characters with meaning distinct from others, but the actual rules and semantics are distinct to each locale.
The correct way to compare and sort strings is by considering the user's locale. Anything else is naive, wrong and very 1990's. Stop doing it.
If you are trying to pass data to a system that cannot support non-ASCII, well, this is just a wrong thing to do. Pass it as data blobs.
https://developer.apple.com/library/ios/documentation/cocoa/Conceptual/Strings/Articles/SearchingStrings.html
Plus normalizing your strings first (see Peter Hosey's post) precomposing or decomposing, basically pick a normalized form.
- (NSString *)decomposedStringWithCanonicalMapping
- (NSString *)decomposedStringWithCompatibilityMapping
- (NSString *)precomposedStringWithCanonicalMapping
- (NSString *)precomposedStringWithCompatibilityMapping
No, it's not nearly as simple and easy as we tend to think.
Yes, it requires informed and careful decision making. (and a bit of non-English language experience helps)
Consider using the RegexKit framework. You could do something like:
NSString *searchString = #"This is neat.";
NSString *regexString = #"[\W]";
NSString *replaceWithString = #"";
NSString *replacedString = [searchString stringByReplacingOccurrencesOfRegex:regexString withString:replaceWithString];
NSLog (#"%#", replacedString);
//... Thisisneat
Consider using NSScanner, and specifically the methods -setCharactersToBeSkipped: (which accepts an NSCharacterSet) and -scanString:intoString: (which accepts a string and returns the scanned string by reference).
You may also want to couple this with -[NSString localizedCompare:], or perhaps -[NSString compare:options:] with the NSDiacriticInsensitiveSearch option. That could simplify having to remove/replace accents, so you can focus on removing puncuation, whitespace, etc.
If you must use an approach like you presented in your question, at least use an NSMutableString and replaceOccurrencesOfString:withString:options:range: — that will be much more efficient than creating tons of nearly-identical autoreleased strings. It could be that just reducing the number of allocations will boost performance "enough" for the time being.
To give a complete example by combining the answers from Luiz and Peter, adding a few lines, you get the code below.
The code does the following:
Creates a set of accepted characters
Turn accented letters into normal letters
Remove characters not in the set
Objective-C
// The input text
NSString *text = #"BûvérÈ!#$&%^&(*^(_()-*/48";
// Create set of accepted characters
NSMutableCharacterSet *acceptedCharacters = [[NSMutableCharacterSet alloc] init];
[acceptedCharacters formUnionWithCharacterSet:[NSCharacterSet letterCharacterSet]];
[acceptedCharacters formUnionWithCharacterSet:[NSCharacterSet decimalDigitCharacterSet]];
[acceptedCharacters addCharactersInString:#" _-.!"];
// Turn accented letters into normal letters (optional)
NSData *sanitizedData = [text dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *sanitizedText = [NSString stringWithCString:[sanitizedData bytes] encoding:NSASCIIStringEncoding];
// Remove characters not in the set
NSString* output = [[sanitizedText componentsSeparatedByCharactersInSet:[acceptedCharacters invertedSet]] componentsJoinedByString:#""];
Swift (2.2) example
let text = "BûvérÈ!#$&%^&(*^(_()-*/48"
// Create set of accepted characters
let acceptedCharacters = NSMutableCharacterSet()
acceptedCharacters.formUnionWithCharacterSet(NSCharacterSet.letterCharacterSet())
acceptedCharacters.formUnionWithCharacterSet(NSCharacterSet.decimalDigitCharacterSet())
acceptedCharacters.addCharactersInString(" _-.!")
// Turn accented letters into normal letters (optional)
let sanitizedData = text.dataUsingEncoding(NSASCIIStringEncoding, allowLossyConversion: true)
let sanitizedText = String(data: sanitizedData!, encoding: NSASCIIStringEncoding)
// Remove characters not in the set
let components = sanitizedText!.componentsSeparatedByCharactersInSet(acceptedCharacters.invertedSet)
let output = components.joinWithSeparator("")
Output
The output for both examples would be: BuverE!_-48
Just bumped into this, maybe its too late, but here is what worked for me:
// text is the input string, and this just removes accents from the letters
// lossy encoding turns accented letters into normal letters
NSMutableData *sanitizedData = [text dataUsingEncoding:NSASCIIStringEncoding
allowLossyConversion:YES];
// increase length by 1 adds a 0 byte (increaseLengthBy
// guarantees to fill the new space with 0s), effectively turning
// sanitizedData into a c-string
[sanitizedData increaseLengthBy:1];
// now we just create a string with the c-string in sanitizedData
NSString *final = [NSString stringWithCString:[sanitizedData bytes]];
#interface NSString (Filtering)
- (NSString*)stringByFilteringCharacters:(NSCharacterSet*)charSet;
#end
#implementation NSString (Filtering)
- (NSString*)stringByFilteringCharacters:(NSCharacterSet*)charSet {
NSMutableString * mutString = [NSMutableString stringWithCapacity:[self length]];
for (int i = 0; i < [self length]; i++){
char c = [self characterAtIndex:i];
if(![charSet characterIsMember:c]) [mutString appendFormat:#"%c", c];
}
return [NSString stringWithString:mutString];
}
#end
These answers didn't work as expected for me. Specifically, decomposedStringWithCanonicalMapping didn't strip accents/umlauts as I'd expected.
Here's a variation on what I used that answers the brief:
// replace accents, umlauts etc with equivalent letter i.e 'é' becomes 'e'.
// Always use en_GB (or a locale without the characters you wish to strip) as locale, no matter which language we're taking as input
NSString *processedString = [string stringByFoldingWithOptions: NSDiacriticInsensitiveSearch locale: [NSLocale localeWithLocaleIdentifier: #"en_GB"]];
// remove non-letters
processedString = [[processedString componentsSeparatedByCharactersInSet:[[NSCharacterSet letterCharacterSet] invertedSet]] componentsJoinedByString:#""];
// trim whitespace
processedString = [processedString stringByTrimmingCharactersInSet: [NSCharacterSet whitespaceCharacterSet]];
return processedString;
Peter's Solution in Swift:
let newString = oldString.componentsSeparatedByCharactersInSet(NSCharacterSet.letterCharacterSet().invertedSet).joinWithSeparator("")
Example:
let oldString = "Jo_ - h !. nn y"
// "Jo_ - h !. nn y"
oldString.componentsSeparatedByCharactersInSet(NSCharacterSet.letterCharacterSet().invertedSet)
// ["Jo", "h", "nn", "y"]
oldString.componentsSeparatedByCharactersInSet(NSCharacterSet.letterCharacterSet().invertedSet).joinWithSeparator("")
// "Johnny"
I wanted to filter out everything except letters and numbers, so I adapted Lorean's implementation of a Category on NSString to work a little different. In this example, you specify a string with only the characters you want to keep, and everything else is filtered out:
#interface NSString (PraxCategories)
+ (NSString *)lettersAndNumbers;
- (NSString*)stringByKeepingOnlyLettersAndNumbers;
- (NSString*)stringByKeepingOnlyCharactersInString:(NSString *)string;
#end
#implementation NSString (PraxCategories)
+ (NSString *)lettersAndNumbers { return #"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"; }
- (NSString*)stringByKeepingOnlyLettersAndNumbers {
return [self stringByKeepingOnlyCharactersInString:[NSString lettersAndNumbers]];
}
- (NSString*)stringByKeepingOnlyCharactersInString:(NSString *)string {
NSCharacterSet *characterSet = [NSCharacterSet characterSetWithCharactersInString:string];
NSMutableString * mutableString = #"".mutableCopy;
for (int i = 0; i < [self length]; i++){
char character = [self characterAtIndex:i];
if([characterSet characterIsMember:character]) [mutableString appendFormat:#"%c", character];
}
return mutableString.copy;
}
#end
Once you've made your Categories, using them is trivial, and you can use them on any NSString:
NSString *string = someStringValueThatYouWantToFilter;
string = [string stringByKeepingOnlyLettersAndNumbers];
Or, for example, if you wanted to get rid of everything except vowels:
string = [string stringByKeepingOnlyCharactersInString:#"aeiouAEIOU"];
If you're still learning Objective-C and aren't using Categories, I encourage you to try them out. They're the best place to put things like this because it gives more functionality to all objects of the class you Categorize.
Categories simplify and encapsulate the code you're adding, making it easy to reuse on all of your projects. It's a great feature of Objective-C!