I'm writing a regular expression in Objective-C.
The escape sequence \w is illegal and emits a warning, so the regular expression /\w/ must be written as #"\\w"; the escape sequence \? is valid, apparently, and doesn't emit a warning, so the regular expression /\?/ must be written as #"\\?" (i.e., the backslash must be escaped).
Question marks aren't invisible like \t or \n, so why is \? a valid escape sequence?
Edit: To clarify, I'm not asking about the quantifier, I'm asking about a string escape sequence. That is, this doesn't emit a warning:
NSString *valid = #"\?";
By contrast, this does emit a warning ("Unknown escape sequence '\w'"):
NSString *invalid = #"\w";
It specifies a literal question mark. It is needed because of a little-known feature called trigraphs, where you can write a three-character sequence starting with question marks to substitute another character. If you have trigraphs enabled, in order to write "??" in a string, you need to write it as "?\?" in order to prevent the preprocessor from trying to read it as the beginning of a trigraph.
(If you're wondering "Why would anybody introduce a feature like this?": Some keyboards or character sets didn't include commonly used symbols like {. so they introduced trigraphs so you could write ??< instead.)
? in regex is a quantifier, it means 0 or 1 occurences. When appended to the + or * quantifiers, it makes it "lazy".
For example, applying the regex o? to the string foo? would match o.
However, the regex o\? in foo? would match o?, because it is searching for a literal question mark in the string, instead of an arbitrary quantifier.
Applying the regex o*? to foo? would match oo.
More info on quantifiers here.
Related
I am a complete Reg-exp noob, so please bear with me. Tried to google this, but haven't found it yet.
What would be an appropriate way of writing a Regular expression matching files starting with a dot, such as .buildpath or .htaccess?
Thanks a lot!
In most regex languages, ^\. or ^[.] will match a leading dot.
The ^ matches the beginning of a string in most languages. This will match a leading .. You need to add your filename expression to it.
^\.
Likewise, $ will match the end of a string.
You may need to substitute the \ for the respective language escape character. However, under Powershell the Regex I use is: ^(\.)+\/
Test case:
"../NameOfFile.txt" -match '^(\\.)+\\\/'
works, while
"_./NameOfFile.txt" -match '^(\\.)+\\\/'
does not.
Naturally, you may ask, well what is happening here?
The (\\.) searches for the literal . followed by a +, which matches the previous character at least once or more times.
Finally, the \\\/ ensures that it conforms to a Window file path.
It depends a bit on the regular expression library you use, but you can do something like this:
^\.\w+
The ^ anchors the match to the beginning of the string, the \. matches a literal period (since an unescaped . in a regular expression typically matches any character), and \w+ matches 1 or more "word" characters (alphanumeric plus _).
See the perlre documentation for more info on Perl-style regular expressions and their syntax.
It depends on what characters are legal in a filename, which depends on the OS and filesystem.
For example, in Windows that would be:
^\.[^<>:"/\\\|\?\*\x00-\x1f]+$
The above expression means:
Match a string starting with the literal character .
Followed by at least one character which is not one of (whole class of invalid chars follows)
I used this as reference regarding which chars are disallowed in filenames.
To match the string starting with dot in java you will have to write a simple expression
^\\..*
^ means regular expression is to be matched from start of string
\. means it will start with string literal "."
.* means dot will be followed by 0 or more characters
Hi may i know what does the below query means?
REGEXP_REPLACE(number,'[^'' ''-/0-9:-#A-Z''[''-`a-z{-~]', 'xy') ext_number
part 1
In terms of explaining what the function function call is doing:
It is a function call to analyse an input string 'number' with a regex (2nd argument) and replace any parts of the string which match a specific string. As for the name after the parenthesis I am not sure, but the documentation for the function is here
part 2
Sorry to be writing a question within an answer here but I cannot respond in comments yet (not enough rep)
Does this regex work? Unless sql uses different syntax this would appear to be a non-functional regex. There are some red flags, e.g:
The entire regex is wrapped in square parenthesis, indicating a set of characters but seems to predominantly hold an expression
There is a range indicator between a single quote and a character (invalid range: if a dash was required in the match it should be escaped with a '\' (backslash))
One set of square brackets is never closed
After some minor tweaks this regex is valid syntax:
^'' ''\-\/0-9:-#A-Z''[''-a-z{-~]`, but does not match anything I can think of, it is important to know what string is being examined/what the context is for the program in order to identify what the regex might be attempting to do
It seems like it is meant to replaces all ASCII control characters in the column or variable number with xy.
[] encloses a class of characters. Any character in that class matches. [^] negates that, hence all characters match, that are not in the class.
- is a range operator, e.g. a-z means all characters from a to z, like abc...xyz.
It seams like characters enclosed in ' should be escaped (The second ' is to escape the ' in the string itself.) At least this would make some sense. (But for none of the DBMS I found having a regexp_replace() function (Postgres, Oracle, DB2, MariaDB, MySQL), I found something in the docs, that would indicate this escape mechanism. They all use \, but maybe I missed something? Unfortunately you didn't tag which DBMS you're actually using!)
Now if you take an ASCII table you'll see, that the ranges in the expression make up all printable characters (counting space as printable) in groups from space to /, 0 to 9, : to #, etc.. Actually it might have been shorter to express it as '' ''-~, space to ~.
Given the negation, all these don't match. The ones left are from NUL to US and DEL. These match and get replaced by xy one by one.
How to rewrite the [a-zA-Z0-9!$* \t\r\n] pattern to match hyphen along with the existing characters ?
The hyphen is usually a normal character in regular expressions. Only if it’s in a character class and between two other characters does it take a special meaning.
Thus:
[-] matches a hyphen.
[abc-] matches a, b, c or a hyphen.
[-abc] matches a, b, c or a hyphen.
[ab-d] matches a, b, c or d (only here the hyphen denotes a character range).
Escape the hyphen.
[a-zA-Z0-9!$* \t\r\n\-]
UPDATE:
Never mind this answer - you can add the hyphen to the group but you don't have to escape it. See Konrad Rudolph's answer instead which does a much better job of answering and explains why.
It’s less confusing to always use an escaped hyphen, so that it doesn't have to be positionally dependent. That’s a \- inside the bracketed character class.
But there’s something else to consider. Some of those enumerated characters should possibly be written differently. In some circumstances, they definitely should.
This comparison of regex flavors says that C♯ can use some of the simpler Unicode properties. If you’re dealing with Unicode, you should probably use the general category \p{L} for all possible letters, and maybe \p{Nd} for decimal numbers. Also, if you want to accomodate all that dash punctuation, not just HYPHEN-MINUS, you should use the \p{Pd} property. You might also want to write that sequence of whitespace characters simply as \s, assuming that’s not too general for you.
All together, that works out to apattern of [\p{L}\p{Nd}\p{Pd}!$*] to match any one character from that set.
I’d likely use that anyway, even if I didn’t plan on dealing with the full Unicode set, because it’s a good habit to get into, and because these things often grow beyond their original parameters. Now when you lift it to use in other code, it will still work correctly. If you hard‐code all the characters, it won’t.
[-a-z0-9]+,[a-z0-9-]+,[a-z-0-9]+ and also [a-z-0-9]+ all are same.The hyphen between two ranges considered as a symbol.And also [a-z0-9-+()]+ this regex allow hyphen.
use "\p{Pd}" without quotes to match any type of hyphen. The '-' character is just one type of hyphen which also happens to be a special character in Regex.
Is this what you are after?
MatchCollection matches = Regex.Matches(mystring, "-");
I have a localized string that looks something like this in English:
"
5 Mile(s)
5,252 Step(s)
"
My app is localized both in left-to-right and right-to-left languages so I don't want to make assumptions either about the ordering of the step(s) or about the formatting of the number (e.g. 5,252 can be 5.252 depending on user locale). So I need to account for possibilities that can include things like
Step(s) 5.252
as well as what's above.
A few other caveats
All I know is that if the Step(s) line is in there, it will be on its own line (hence in my regex I require \n at each end of the string)
No guarantee that the Mile(s) information will be in the string at all, let alone whether it will be before or after Step(s)
Here's my attempt at pattern extraction:
NSString *patternString = [NSString stringWithFormat:#"\\n(([0-9,\\.]*)\s*%#|%#\s*([0-9,\\.]*))\\n",
NSLocalizedString(#"Step(s)",nil), NSLocalizedString(#"Step(s)",nil)];
There appear to be two problems with this:
XCode is indicating Unknown escape sequence '\s' for the second \s in the pattern string above
No matches are being found even for strings like the following:
0.2 Mile(s)
1,482 Step(s)
Ideally I would extract the 1,482 out of this string in a way that is localization friendly. How should I modify my regex?
as far as the regex, perhaps this approach might work - it simply matches (with named groups) each couplet of numbers in sequence, with the assumption the first is miles and the second is steps. Decimals in the . or , form are optional:
(?<miles>\d+(?:[.,]\d+)?).*?(?<steps>\d+(?:[.,]\d+)?)
(and i think it should be \\s) - i'm not an ios guy, but if you can use a regex literal it would be way more readable.
regular expression demo
First I'd like to ask - Why is Mile(s) mentioned in the question at all?
And now to my two bits - you could simply use a positive look-ahead:
^(?=.*Step\(s\))[^\d]*(\d+(?:[.,]\d+)?)
It makes sure the expected word is present on the line, and then captures the number on it, allowing for localized, optional, decimal separator and decimals. This way it doesn't matter if the numer is before, or after, the "word".
It doesn't take localization of the "word" into account, but that you seem to have handled by yourself ;)
See it here at regex101.
Your regex is close, although in Obj-C you need to double-escape the \s and (s):
^(([0-9,.]*)\\s*%#|%#\\s*([0-9,.]*))$
In your NSLocalizedString you likely also need to escape the parentheses enclosing (s):
NSString *patternString = [NSString stringWithFormat:#"^(([\\d,.]+)\\s%#|%#\\s([\\d,.]+))$",
NSLocalizedString(#"Step\\(s\\)",nil), NSLocalizedString(#"Step\\(s\\)",nil)];
If you don't escape (s) then the regex engine is probably going to interpret it as a capture group.
Looking at NSLog you can see what the pattern actually reads like:
NSLog(#"patternString: %#", patternString);
Output:
patternString: ^(([\d,.]+)\sStep\(s\)|Step\(s\)\s([\d,.]+))$
Since you mentioned the Mile(s) part may not be in the string at all I'm assuming it isn't relevant to the regular expression. As I understand from the question, you just need to capture the number of steps and nothing else. On this basis, here's a modified version of your existing regex:
NSString *patternString =
[NSString stringWithFormat:#"^(?:([0-9,.]*)\\s*%#|%#\\s*([0-9,.]*))$",
NSLocalizedString(#"Step\\(s\\)",nil), NSLocalizedString(#"Step\\(s\\)",nil)];
Demo:
https://www.regex101.com/r/Q6ff1b/1
This is based on the following tips/modifications:
Use the m (= UREGEX_MULTILINE) flag option when creating the regex to specify that ^ and $ match the start and end of each line. This is more sophisticated than using \n as it will also handle the start and end of the string where this might not be present. See here.
Always use a double backslash (\\) for regex escaping - otherwise NSString will interpret the single backslash to be escaping the next character and convert it before it gets to the regex.
Literal parentheses need to be escaped - e.g. Step\\(s\\) instead of Step(s).
Characters within a character class (i.e. anything within the [] square brackets) don't need to be escaped - so it would be . rather than \\. - the latter.
If you are using (x|y|...) as a choice and don't need it to be a capturing group, use ?: after the first parenthesis to ensure it doesn't get captured - i.e. (?:x|y|...).
Alright, I'm trying to write some code that removes words that contain an apostrophe from an NSString. To do this, I've decided to use regular expressions, and I wrote one, that I tested using this website: http://rubular.com/r/YTV90BcgoQ
Here, the expression is: \S*'+\S
As shown on the website, the words containing an apostrophe are matched. But for some reason, in the application I'm writing, using this code:
sourceString = [sourceString stringByReplacingOccurrencesOfRegex:#"\S*'+\S" withString:#""];
Doesn't return any positive result. By NSLogging the 'sourceString', I notice that words like 'Don't' and 'Doesn't' are still present in the output.
It doesn't seem like my expression is the problem, but maybe RegexKitLite doesn't accept certain types of expressions? If someone knows what's going on here, please enlighten me !
Literal NSStrings use \ as an escape character so that you can put things like newlines \n into them. Regexes also use backslashes as an escape character for character classes like \S. When your literal string gets run through the compiler, the backslashes are treated as escape characters, and don't make it to the regex pattern.
Therefore, you need to escape the backslashes themselves in your literal NSString, in order to end up with backslashes in the string that is used as the pattern: #"\\S*'+\\S".
You should have seen a compiler warning about "Unknown escape sequence" -- don't ignore those warnings!