Anyone knows how to do it?
So we split that based on space, and then capitalize each, and then move them back right?
You can use either of these methods to accomplish that:
-[NSString capitalizedString]
Or:
-[NSString capitalizedStringWithLocale:]
Just use the capitalizedString method.
Return Value
A string with the first character from each word in the receiver changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values.
To capitalize use :
string=[string capitalizedString];
Gives each word of the sentence in capital.
So we split that based on space, and then capitalize each, and then
move them back right?
This is not required. No need to break in words and individually capitalize each one of them.
Related
Like for example "sampletext^p^p^p" where "^p" is the carriage return. I want to move from the last "^p" to the left until it reach any character from A to Z like for the example above is "t". Is it possible? And I also want it be extend so it would select from the last "^p" to "t".
I tried to use .MoveUntil cset:="?", count:=wdBackward since that in wildcard, ? is considered as any single character but it only move backward until it find the ? character.
These "Move" methods don't support wild cards. The value you hand to CSet has to contain the literal string characters. There are (at least) two ways you can accomplish what you postulate:
Range.MoveStartUntil Cset:="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz", Count:=wdBackward
OR
Range.MoveStartWhile cSet:=vbCr, count:=wdbackward
Note: also works with the Selection object.
I'm new to objective-c and I need help with this.
A phone number in this country has 10 digits like 6195946191 or 619JYDN191. It is hard to read a phone number formatted like that. One common format is 619-594-6191. Add the method phoneFormat to the NSString class. The method phoneFormat converts strings like
#"6195946191", #"619 594 6191", #"619 5946191" and #"619-594-6191" to #"619-594-6191".
That is all the methods below will return #"619-594-6191"
[#"6195946191" phoneFormat]
[#"619 594 6191" phoneFormat]
[#"619 5946191" phoneFormat]
[#"619-594-6191" phoneFormat]
Can anyone show me how to do this?
Do it in two stages. First remove all undesirable characters (everything that is not a digit? depends what input you are expecting). Personally, I would probably use NSScanner for this, though I can think of other ways.
Now you are guaranteed you have 10 digits in succession, so insert a hyphen after the 3rd and 6th characters and you're done.
Let's say, I have a regular expression that checks the validation of the input value as a whole. For example, it is an email input box and when user hits enter, I check it against ^[A-Z0-9._%+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}$ to see if it is a valid email address.
What I want to achieve is, I want to intercept the character input too, and check every single input character to see if that character is also a valid character. I can do this by adding an extra regular expression, e.g. [A-Z0-9._%+-] but that is not what I want.
Is there a way to extract the widest possible range of acceptable characters from a given regular expression? So in the example above, can I extract all the valid characters that are defined by the original regular expression (i.e. ^[A-Z0-9._%+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}$) programmatically?
I would appreciate any help or hint.
P.S. This is project for iOS written in Objective-C.
If you don't mind writing half a regex parser, certainly. You would have to be able to distinguish literals from meta-characters and to unroll/merge all character classes (including negated character classes, and nested negated character classes, if you regex flavor supports them).
If NSRegularExpressions doesn't come with some convenience method, I cannot imagine how it would be possible otherwise. Just think about ^. When it is outside of a character class, it's a meta-character that you can ignore. If it is inside a character class, it's a meta-character, that negates the character class unless it is not the first character. - is a meta-character inside character classes, unless it is the first character, the last character, or right after another character range (depending on regex flavor). And I'm not even speaking about escaped characters.
I don't know about NSRegularExpressions, but some flavors also support nested character classes (like [a-z[^aeiou]] for all consonants). I think you get where I am going with this.
I'm working on a translator that will take English language text (as user input into a UITextView) and (with a button press) replace specific words with alternatives. I have both the English words in scope plus their alternatives in separate Arrays (englishArray and alternativeArray), indexed correspondingly.
My challenge is finding an algorithm that will allow me to identify a word in the input text (a UITextView) ignoring characters like <",.()>, lookup the word in englishArray (case insensitive), locate the corresponding word in alternativeArray and then use that word in place of the original - writing it back to the UITextView.
Any help greatly appreciated.
NB. I have created a Category extending the NSArray functionality with a indexOfCaseInsensitiveString method that ignores case when doing an indexOfObject type lookup if that helps.
Tony.
I think that using an NSScanner would be best to parse the string into separate words which you could then pass to your indexOfCaseInsensitiveString method. scanCharactersFromSet:intoString: using a set of all the characters you want to ignore, including whitespace and newline characters should get you to the start of a word, and then you could use scanUpToCharactersFromSet:intoString: using the same set to scan to the end of the word. Using scanLocation at the beginning and end of each scan should allow you to get the range of that word, so if you find a match in your array, you will know where in your string to make the replacement.
Thanks for your suggestion. It's working with one exception.
I want to capture all punctuation so I can recreate the original input but with the substituted words. Even though I have a 'space' in my Character Set, the scanner is not putting the spaces into the 'intoString'. Other characters I specify in the Character Set such as '(' and ';' are represented in the 'intoString'.
Net is that when I recreate the input, it's perfect except that I get individual words running into each other.
UPDATE: I fixed that issue by including:
[theScanner setCharactersToBeSkipped:nil];
Thanks again.
I'd like to create a regular expression such that when I compare the a string against an array of strings, matches are returned with the regex ignoring certain characters.
Here's one example. Consider the following array of names:
{
"Andy O'Brien",
"Bob O'Brian",
"Jim OBrien",
"Larry Oberlin"
}
If a user enters "ob", I'd like the app to apply a regex predicate to the array and all of the names in the above array would match (e.g. the ' is ignored).
I know I can run the match twice, first against each name and second against each name with the ignored chars stripped from the string. I'd rather this by done by a single regex so I don't need two passes.
Is this possible? This is for an iOS app and I'm using NSPredicate.
EDIT: clarification on use
From the initial answers I realized I wasn't clear. The example above is a specific one. I need a general solution where the array of names is a large array with diverse names and the string I am matching against is entered by the user. So I can't hard code the regex like [o]'?[b].
Also, I know how to do case-insensitive searches so don't need the answer to focus on that. Just need a solution to ignore the chars I don't want to match against.
Since you have discarded all the answers showing the ways it can be done, you are left with the answer:
NO, this cannot be done. Regex does not have an option to 'ignore' characters. Your only options are to modify the regex to match them, or to do a pass on your source text to get rid of the characters you want to ignore and then match against that. (Of course, then you may have the problem of correlating your 'cleaned' text with the actual source text.)
If I understand correctly, you want a way to match the characters "ob" 1) regardless of capitalization, and 2) regardless of whether there is an apostrophe in between them. That should be easy enough.
1) Use a case-insensitivity modifier, or use a regexp that specifies that the capital and lowercase version of the letter are both acceptable: [Oo][Bb]
2) Use the ? modifier to indicate that a character may be present either one or zero times. o'?b will match both "o'b" and "ob". If you want to include other characters that may or may not be present, you can group them with the apostrophe. For example, o['-~]?b will match "ob", "o'b", "o-b", and "o~b".
So the complete answer would be [Oo]'?[Bb].
Update: The OP asked for a solution that would cause the given character to be ignored in an arbitrary search string. You can do this by inserting '? after every character of the search string. For example, if you were given the search string oleary, you'd transform it into o'?l'?e'?a'?r'?y'?. Foolproof, though probably not optimal for performance. Note that this would match "o'leary" but also "o'lea'r'y'" if that's a concern.
In this particular case, just throw the set of characters into the middle of the regex as optional. This works specifically because you have only two characters in your match string, otherwise the regex might get a bit verbose. For example, match case-insensitive against:
o[']*b
You can add more characters to that character class in the middle to ignore them. Note that the * matches any number of characters (so O'''Brien will match) - for a single instance, change to ?:
o[']?b
You can make particular characters optional with a question mark, which means that it will match whether they're there or not, e.g:
/o\'?b/
Would match all of the above, add .+ to either side to match all other characters, and a space to denote the start of the surname:
/.+? o\'?b.+/
And use the case-insensitivity modifier to make it match regardless of capitalisation.