I have a test string such as: The Sun and the Moon together, forever
I want to be able to type a few characters or words and be able to match this string if the characters appear in the correct sequence together, even if there are missing words. For example, the following search word(s) should all match against this string:
The Moon
Sun tog
Tsmoon
The get ever
What regex pattern should I be using for this? I should add that the supplied test strings are going to be dynamic within an app, and so I'd like to be able to use a pattern based on the search string.
From your example Tsmoon you show partial words (T), ignoring case (s, m) and allow anything between each entered character. So as a first attempt you can:
Set the ignore case option
Between each chapter input insert the regular expression to match zero or more of anything. You can choose whether to match the shortest or longest run.
Try that, reading the documentation for NSRegularExpression if you're stuck, and see how it goes. If you get stuck ask a new question showing your code and the RE constructed and explain what happens/doesn't work as expected.
HTH
Related
My question is somewhat specific, I'm not using any kind of code compiler to achieve the result in the title, I am using a IRC Client that allows the use of "Quirks" so the users can have specific mannerisms when chatting, like starting every word with an uppercase, or changing every "s" into a "2".
Problem is that I can't see the whole code so even though I'm not familiar with REGEXP_REPLACE it makes things harder to learn.
The client simplifies the whole coding process, here's a screenshot of the
interface
Filling the text boxes with "^(\w)" and "upper(\1)" respectively makes the first character capitalized, "(\w)$" and "upper(\1)" does the same with the last character.
I've discovered that "\b(\w)" will uppercase the first character of every word, i've tried "\b(\w)%" for the last character but it didn't work, as there is some syntax error, probably...
So, how do I get every last character capitalized?
1:
I need to match two ipaddress/hostname with a regular expression:
Like 20.20.20.20
should match with 20.20.20.20
should match with [http://20.20.20.20/abcd]
should not match with 20.20.20.200
should not match with [http://20.20.20.200/abcd]
should not match with [http://120.20.20.20/abcd]
should match with AB_20.20.20.20
should match with 20.20.20.20_AB
At present i am using something like this regular expression: "(.*[^(\w)]|^)20.20.20.20([^(\w)].*|$)"
But it is not working for the last two cases. As the "\w" is equal to [a-zA-Z0-9_]. Here I also want to eliminate the "_" underscore. I tried different combination but not able to succeed. Please help me with this regular expression.
(.*[_]|[^(\w)]|^)10.10.10.10([_]|[^(\w)].*|$)
I spent some more time on this.This regular expression seems to work.
I don't know which language you're using, but with Perl-like regular expressions you could use the following, shorter expression:
(?:\b|\D)20\.20\.20\.20(?:\b|\D)
This effectively says:
Match word boundary (\b, here: the start of the word) or a non-digit (\D).
Match IP address.
Match word boundary (\b, here: the end of the word) or a non-digit (\D).
Note 1: ?: causes the grouping (\b|\D) not to create a backreference, i.e. to store what it has found. You probably don't need the word boundaries/non-digits to be stored. If you actually need them stored, just remove the two ?:s.
Note 2: This might be nit-picking, but you need to escape the dots in the IP address part of the regular expression, otherwise you'd also match any other character at those positions. Using 20.20.20.20 instead of 20\.20\.20\.20, you might for example match a line carrying a timestamp when you're searching through a log file...
2012-07-18 20:20:20,20 INFO Application startup successful, IP=20.20.20.200
...even though you're looking for IP addresses and that particular one (20.20.20.200) explicitly shouldn't match, according to your question. Admittedly though, this example is quite an edge case.
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.
I have url, for example:
http://i.myhost.com/myimage.jpg
I want to change this url to
http://i.myhost.com/myimageD.jpg.
(Add D after image name and before point)
i.e I want add some words after image name and before point using regex.
What is the best way do it using regex?
Try using ^(.*)\.([a-zA-Z]{3,5}) and replacing with \1D\2. I'm assuming the extension is 3-5 alphanumeric numbers but you can modify it to suit. E.g. if it's just jpg images then you can put that instead of the [a-zA-Z]{3,5}.
Sounds like a homework question given the solution must use a regex, on that assumption here is an outline to get you going.
If all you have is a URL then #mathematical.coffee's solution will suit. However if you have a chunk of text within which is one or more URLs and you have to locate and change just those then you'll need something a little more involved.
Look at the structure of a URL: {protocol}{address}{item}; where
{protocol} is "http://", "ftp://" etc.;
{address} is a name, e.g. "www.google.com", or a number, e.g. "74.125.237.116" - there will always be at least one dot in the address; and
{item} is "/name" where name is quite flexible - there will be zero or more items, you can think of them as directories and a file but this isn't strictly true. Also the sequence of items can end in a "/" (including when there are zero of them).
To make a regex which matches a URL start by matching each part. In the case of the items you'll want to match the last in the sequence separately - you'll have zero or more "directories" and one "file", the latter must be of the form "name.extension".
Once you have regexes for each part you just concatenate them to produce a regex for the whole. To form the replacement pattern you can surround parts of your regex with parentheses and refer to those parts using \number in the replacement string - see #mathematical.coffee's solution for an example.
The best way to learn regexs is to use an editor which supports them and just experiment. The exact syntax may not be the same as NSRegularExpression but they are mostly pretty similar for the basic stuff and you can translate from one to another easily.