I have searched for hours now and haven't found a solution for my problem. I have a NSString which looks like the following:
"spacer": ["value1", "value2"], "spacer": ["value1", "value2"], ...
What I want to do is to remove the [ ] characters from the string. It's seems to be impossible with objective-c. Most other programming languages offer functions like strpos or indexOf which allow me to search for a character or string and locate the position of it. But there seems nothing to be like this in objective-c.
Does anyone has an idea on how to remove these characters?
Additionally there are [] characters in the string which should remain, so I can't just use NSMutableString stringByReplacingOccurencesOfString:withString. I need to search first for the spacer string and then remove only the next two [] chars.
Thank you for helping me.
To find occurrences of a string within a string, use the rangeOfXXX methods in the NSString class. Then you can construct NSRanges to extract substrings, etc.
This example removes only the first set of open/close brackets in your sample string...
NSString *original = #"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSLog(#"%#", original);
NSRange startRange = [original rangeOfString:#"\["];
NSRange endRange = [original rangeOfString:#"]"];
NSRange searchRange = NSMakeRange(0, endRange.location);
NSString *noBrackets = [original stringByReplacingOccurrencesOfString:#"\[" withString:#"" options:0 range:searchRange];
noBrackets = [noBrackets stringByReplacingOccurrencesOfString:#"]" withString:#"" options:0 range:searchRange];
NSLog(#"{%#}", noBrackets);
The String Programming Guide has more details.
You might alternatively also be able to use the NSScanner class.
This worked for me to extract everything to the index of a character, in this case '[':
NSString *original = #"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSRange range = [original rangeOfString:#"\["];
NSString *toBracket = [NSString stringWithString :[original substringToIndex:range.location] ];
NSLog(#"toBracket: %#", toBracket);
Related
I am trying to see if there is a short way of doing the following:
let s say I have the following NSString:
#"123456 copy cat";
I need to remove everything before "copy" and the length of the text before it is variable (depends on the user input)
I know that I can probably do an endless for loop to check the substring but I hope that there is a faster way
NSString *string = #"123456 copy cat";
NSRange range = [string rangeOfString:#"copy"];
NSString *newString = [string substringFromIndex:range.location];
That's the simple answer. The problem is that your user may enter the word 'copy', and then what? Even if you get the range this way:
NSRange range = [string rangeOfString:#"copy cat"];
there's still a chance your user may enter that phrase as well. By the time you put in additional checks, it's no longer a "short way."
EDIT
Don't overlook Chuck's excellent observation in the comments for using:
-[NSString rangeOfString:options:]
To remove part of a string you will either have to create a new string containing the part that you want to keep, or create a mutable copy of your string and modify that.
To create a new string containing the part starting with "copy":
NSString *input;
NSString *output;
NSRange copyRange;
input = #"123456 copy cat";
copyRange = [input rangeOfString:#"copy"];
output = [input substringFromIndex:copyRange.location];
To create a mutable string and remove the part up to "copy":
NSString *input;
NSMutableString *output;
NSRange copyRange;
input = #"123456 copy cat";
output = [input mutableCopy];
copyRange = [output rangeOfString:#"copy"];
[output replaceCharactersInRange:NSMakeRange(0, copyRange.location) withString:#""];
[output autorelease]; // depending on what you want to do
Please also see the comments in the other answer about using rangeOfString:withOptions: with NSBackwardsSearch in case the user input also contains the word "input".
I have several NSStrings with a format similar to the one below:
"Hello, how are you?"
How can I break the string into an array of words? For example, for the above sentence I would expect an array consisting of "Hello,", "how", "are", "you?"
Usually I would break the string into words by using the function [NSString componentsSeparatedByCharactersInSet: NSCharacterSet set]
However this won't work in this situation because the spaces between the words are of unequal length. Note I will not be aware of the size of each word and the space between them.
How can I accomplish this? I am working on an app for OSX not iOS.
EDIT: My eventual goal is to retrieve the second word in the sentence. If there is a easier way to do this without breaking the string into an array please feel free to suggest it.
Try this:
NSMutableArray *parts = [NSMutableArray arrayWithArray:[str componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]];
[parts removeObjectIdenticalTo:#""];
NSString *res = [parts objectAtIndex:1]; // The second string
Well, you could actually write a loop to iterate through the characters and find the first non-blank after the first blank, then iterate further to find the ending blank (or end of line). Would probably be about 5x faster (with much fewer object allocations) than using one of the other methods, and could be done in about 10 lines.
If you dont want to use a CharacterSet try this to remove extra spaces:
NSString* string = #"word1, word2 word3 word4";
bool done = false;
do {
NSString tempStr = [string stringByReplacingOccurrencesOfString:#" " withString:#" "];
done = [string isEqualToString:tempStr];
string = tempStr;
} while (!done);
NSLog(#"%#", string);
this will output "word1, word2 word3 word4"
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.
I have an NSString object and I want to make a substring from it, by locating a word.
For example, my string is: "The dog ate the cat", I want the program to locate the word "ate" and make a substring that will be "the cat".
Can someone help me out or give me an example?
Thanks,
Sagiftw
NSRange range = [string rangeOfString:#"ate"];
NSString *substring = [[string substringFromIndex:NSMaxRange(range)] stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];
NSString *str = #"The dog ate the cat";
NSString *search = #"ate";
NSString *sub = [str substringFromIndex:NSMaxRange([str rangeOfString:search])];
If you want to trim whitespace you can do that separately.
What about this way?
It's nearly the same.
But maybe meaning of NSRange easier to understand for beginners, if it's written this way.
At last, it's the same solution of jtbandes
NSString *szHaystack= #"The dog ate the cat";
NSString *szNeedle= #"ate";
NSRange range = [szHaystack rangeOfString:szNeedle];
NSInteger idx = range.location + range.length;
NSString *szResult = [szHaystack substringFromIndex:idx];
Try this one..
BOOL isValid=[yourString containsString:#"X"];
This method return true or false. If your string contains this character it return true, and otherwise it returns false.
NSString *theNewString = [receivedString substringFromIndex:[receivedString rangeOfString:#"Ur String"].location];
You can search for a string and then get the searched string into another string...
-(BOOL)Contains:(NSString *)StrSearchTerm on:(NSString *)StrText
{
return [StrText rangeOfString:StrSearchTerm options:NSCaseInsensitiveSearch].location==NSNotFound?FALSE:TRUE;
}
You can use any of the two methods provided in NSString class, like substringToIndex: and substringFromIndex:. Pass a NSRange to it as your length and location, and you will have the desired output.
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!