re compile error: sre_constants.error: bad character range [duplicate] - regex

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, "-");

Related

What do we need Lookahead/Lookbehind Zero Width Assertions for?

I've just learned about these two concepts in more detail. I've always been good with RegEx, and it seems I've never seen the need for these 2 zero width assertions.
I'm pretty sure I'm wrong, but I do not see why these constructs are needed. Consider this example:
Match a 'q' which is not followed by a 'u'.
2 strings will be the input:
Iraq
quit
With negative lookahead, the regex looks like this:
q(?!u)
Without it, it looks like this:
q[^u]
For the given input, both of these regex give the same results (i.e. matching Iraq but not quit) (tested with perl). The same idea applies to lookbehinds.
Am I missing a crucial feature that makes these assertions more valuable than the classic syntax?
Why your test probably worked (and why it shouldn't)
The reason you were able to match Iraq in your test might be that your string contained a \n at the end (for instance, if you read it from the shell). If you have a string that ends in q, then q[^u] cannot match it as the others said, because [^u] matches a non-u character - but the point is there has to be a character.
What do we actually need lookarounds for?
Obviously in the above case, lookaheads are not vital. You could workaround this by using q(?:[^u]|$). So we match only if q is followed by a non-u character or the end of the string. There are much more sophisticated uses for lookaheads though, which become a pain if you do them without lookaheads.
This answer tries to give an overview of some important standard situations which are best solved with lookarounds.
Let's start with looking at quoted strings. The usual way to match them is with something like "[^"]*" (not with ".*?"). After the opening ", we simply repeat as many non-quote characters as possible and then match the closing quote. Again, a negated character class is perfectly fine. But there are cases, where a negated character class doesn't cut it:
Multi-character delimiters
Now what if we don't have double-quotes to delimit our substring of interest, but a multi-character delimiter. For instance, we are looking for ---sometext---, where single and double - are allowed within sometext. Now you can't just use [^-]*, because that would forbid single -. The standard technique is to use a negative lookahead at every position, and only consume the next character, if it is not the beginning of ---. Like so:
---(?:(?!---).)*---
This might look a bit complicated if you haven't seen it before, but it's certainly nicer (and usually more efficient) than the alternatives.
Different delimiters
You get a similar case, where your delimiter is only one character but could be one of two (or more) different characters. For instance, say in our initial example, we want to allow for both single- and double-quoted strings. Of course, you could use '[^']*'|"[^"]*", but it would be nice to treat both cases without an alternative. The surrounding quotes can easily be taken care of with a backreference: (['"])[^'"]*\1. This makes sure that the match ends with the same character it began with. But now we're too restrictive - we'd like to allow " in single-quoted and ' in double-quoted strings. Something like [^\1] doesn't work, because a backreference will in general contain more than one character. So we use the same technique as above:
(['"])(?:(?!\1).)*\1
That is after the opening quote, before consuming each character we make sure that it is not the same as the opening character. We do that as long as possible, and then match the opening character again.
Overlapping matches
This is a (completely different) problem that can usually not be solved at all without lookarounds. If you search for a match globally (or want to regex-replace something globally), you may have noticed that matches can never overlap. I.e. if you search for ... in abcdefghi you get abc, def, ghi and not bcd, cde and so on. This can be problem if you want to make sure that your match is preceded (or surrounded) by something else.
Say you have a CSV file like
aaa,111,bbb,222,333,ccc
and you want to extract only fields that are entirely numerical. For simplicity, I'll assume that there is no leading or trailing whitespace anywhere. Without lookarounds, we might go with capturing and try:
(?:^|,)(\d+)(?:,|$)
So we make sure that we have the start of a field (start of string or ,), then only digits, and then the end of a field (, or end of string). Between that we capture the digits into group 1. Unfortunately, this will not give us 333 in the above example, because the , that precedes it was already part of the match ,222, - and matches cannot overlap. Lookarounds solve the problem:
(?<=^|,)\d+(?=,|$)
Or if you prefer double negation over alternation, this is equivalent to
(?<![^,])\d+(?![^,])
In addition to being able to get all matches, we get rid of the capturing which can generally improve performance. (Thanks to Adrian Pronk for this example.)
Multiple independent conditions
Another very classic example of when to use lookarounds (in particular lookaheads) is when we want to check multiple conditions on an input at the same time. Say we want to write a single regex that makes sure our input contains a digit, a lower case letter, an upper case letter, a character that is none of those, and no whitespace (say, for password security). Without lookarounds you'd have to consider all permutations of digit, lower case/upper case letter, and symbol. Like:
\S*\d\S*[a-z]\S*[A-Z]\S*[^0-9a-zA_Z]\S*|\S*\d\S*[A-Z]\S*[a-z]\S*[^0-9a-zA_Z]\S*|...
Those are only two of the 24 necessary permutations. If you also want to ensure a minimum string length in the same regex, you'd have to distribute those in all possible combinations of the \S* - it simply becomes impossible to do in a single regex.
Lookahead to the rescue! We can simply use several lookaheads at the beginning of the string to check all of these conditions:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^0-9a-zA-Z])(?!.*\s)
Because the lookaheads don't actually consume anything, after checking each condition the engine resets to the beginning of the string and can start looking at the next one. If we wanted to add a minimum string length (say 8), we could simply append (?=.{8}). Much simpler, much more readable, much more maintainable.
Important note: This is not the best general approach to check these conditions in any real setting. If you are making the check programmatically, it's usually better to have one regex for each condition, and check them separately - this let's you return a much more useful error message. However, the above is sometimes necessary, if you have some fixed framework that lets you do validation only by supplying a single regex. In addition, it's worth knowing the general technique, if you ever have independent criteria for a string to match.
I hope these examples give you a better idea of why people would like to use lookarounds. There are a lot more applications (another classic is inserting commas into numbers), but it's important that you realise that there is a difference between (?!u) and [^u] and that there are cases where negated character classes are not powerful enough at all.
q[^u] will not match "Iraq" because it will look for another symbol.
q(?!u) however, will match "Iraq":
regex = /q[^u]/
/q[^u]/
regex.test("Iraq")
false
regex.test("Iraqf")
true
regex = /q(?!u)/
/q(?!u)/
regex.test("Iraq")
true
Well, another thing along with what others mentioned with the negative lookahead, you can match consecutive characters (e.g. you can negate ui while with [^...], you cannot negate ui but either u or i and if you try [^ui]{2}, you will also negate uu, ii and iu.
The whole point is to not "consume" the next character(s), so that it can be e.g. captured by another expression that comes afterwards.
If they're the last expression in the regex, then what you've shown are equivalent.
But e.g. q(?!u)([a-z]) would let the non-u character be part of the next group.

A pattern matching an expression that doesn't end with specific sequence

I need a regex pattern which matches such strings that DO NOT end with such a sequence:
\.[A-z0-9]{2,}
by which I mean the examined string must not have at its end a sequence of a dot and then two or more alphanumeric characters.
For example, a string
/home/patryk/www
and also
/home/patryk/www/
should match desired pattern and
/home/patryk/images/DSC002.jpg should not.
I suppose this has something to do with lookarounds (look aheads) but still I have no idea how to make it.
Any help appreciated.
Old Answer
You can use a negative lookbehind at the end if your regex flavor supports it:
^.*+(?<!\.\w{2,})$
This will match a string that has an end anchor not preceded by the icky sequence you don't want.
Note that as m.buettner has pointed out, this uses an indefinite length lookbehind, which is a feature unique to .NET
New Answer
After a bit of digging around, however, I've found that variable length look-aheads are pretty widely supported, so here is a version that uses those:
^(?:(?!\.\w{2,}$).)++$
In a comment on an answer, you have stated you wanted to not match strings with forward slashes at the end, which is accomplished by simply adding a forward slash to the lookahead.
^(?:(?!(\.\w{2,}|/)$).)++$
Note that I am using \w for succinctness, but it lets underscores through. If this is important, you could replace it with [^\W_].
Asad's version is very convenient, but only .NET's regex engine supports variable-length lookbehinds (which is one of the many reasons why every regex question should include the language or tool used).
We can reduce this to a fixed-length lookbehind (which is supported in most engines except for JavaScrpit) if we think about the possible cases which should match. That would be either one or zero letters/digits at the end (whether preceded by . or not) or two or more letters/digits that are not preceded by a dot.
^.*(?:(?<![a-zA-Z0-9])[a-zA-Z0-9]?|(?<![a-zA-Z0-9.])[a-zA-Z0-9]{2,})$
This should do it:
^(?:[^.]+|\.(?![A-Za-z0-9]{2,}$))+$
It alternates between matching one or more of anything except a dot, or a dot if it's not followed by two or more alphanumeric characters and the end of the string.
EDIT: Upgrading it to meet the new requirement is just more of the same:
^(?:[^./]+|/(?=.)|\.(?![A-Za-z0-9]{2,}$))+$
Breaking that down, we have:
[^./]+ # one or more of any characters except . or /
/(?=.) # a slash, as long as there's at least one character following it
\.(?![A-Za-z0-9]{2,}$) # a dot, unless it's followed by two or more alphanumeric characters followed by the end of the string
On another note: [A-z] is an error. It matches all the uppercase and lowercase ASCII letters, but it also matches the characters [, ], ^, _, backslash and backtick, whose code points happen to lie between Z and a.
Variable length look behinds are rarely supported, but you don't need one:
^.*(?<!\.[A-z0-9][A-z0-9]?)$

Regex doesn't recognize underscore as special character

/(?=^.{8,}$)(?=.*[_!##$%^&*-])(?=.*\d)(?=.*\W+)(?![.\n])(?=.*[a-z])(?=.*[A-Z]).*$/
I'm trying to make a regex for password validation such that the password must be at least 8 chars and include one uppercase, one lowercase, one number, and one special char. It works fine except it won't recognize the underscore (_) as a special character. I.e., Pa$$w0rd matches, but Pass_w0rd doesn't. Thoughts?
This portion of the regex seems to be looking for special characters:
(?=.*[!##$%^&*-])
Note that the character class does not include an underscore, try changing this to the following:
(?=.*[_!##$%^&*-])
You will also need to modify or remove this portion of the regex:
(?=.*\W+)
\W is equivalent to [^a-zA-Z0-9_], so if an underscore is your only special character this portion of the regex will cause it to fail. Instead, change it to the following (or remove it, it is redundant since you already check for special characters earlier):
(?=.*[^\w_])
Complete regex:
/(?=^.{8,}$)(?=.*[_!##$%^&*-])(?=.*\d)(?=.*[^\w_])(?![.\n])(?=.*[a-z])(?=.*[A-Z]).*$/
This one here works as well. It defines a special character as by excluding alphanumerical characters and whitespace, so it includes the underscore:
(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[\d])(?=.*?[^\sa-zA-Z0-9]).{8,}
The problem is that the only thing that could possibly satisfy the \W, by definition, is something other than [a-zA-Z0-9_]. The underscore is specifically not matched by \W, and in Pass_w0rd, nothing else is matched by it, either.
I suspect that having both your specific list of special characters and the \W is overkill. Pick one and you're likely to be happier. I also recommend splitting this whole thing up into several separate tests for much better maintainability.
A much simpler regex that works for you is this:
/(?=.*[_!##$%^&*-])(?=.*\d)(?!.*[.\n])(?=.*[a-z])(?=.*[A-Z])^.{8,}$/
There were few mistakes in your original regex eg:
You don't need to use lookahead for making sure there are 8 chars in input
negative lookahead [.\n] was missing .*
(?=.*\W+) is superfluous and probably not serving any purpose

Regex only allow letters and some characters

I am attempting to create a regex that only allows letters upper or lowercase, and the characters of space, '-', ',' '.', '(', and ')'. This is what I have so far but for some reason it is still letting me enter numbers
^[a-zA-Z -,.()]*$
any help would be great! Thanks.
- is special in character class. It is used to define a range as you've done with a-z.
To match a literal - you need to either escape it or place it such that it'll not function as range operator:
^[a-zA-Z \-,.()]*$
^^ escaping \
or
^[-a-zA-Z ,.()]*$
^ placing it at the beginning.
or
^[a-zA-Z -,.()-]*$
^ placing it at the end.
and interestingly
^[a-z-A-Z -,.()]*$
^ placing in the middle of two ranges.
In the final case - is place between a-z and A-Z since both the characters surrounding the -(the one which we want to treat literally) that is z and A are already involved in ranges, the - is treated literally again.
Of all the mentioned methods, the escaping method is recommended as it makes your code easier to read and understand. Anyone seeing the \ would expect that an escape is intended. Placing the - at the beginning(end) will create problems if you later add a character before(after) it in the character class without escaping the - thus forming a range.
Well, there is an issue in that -, is being interpreted as a range, like a-z, allowing all characters from space to comma. Escape that and at least some of the bugs should be fixed.
^[a-zA-Z \-,.()]*$
Strictly speaking, you should probably also escape the . and (), too, since those have special meaning in regular expressions. The Javascript regex engine (where I was testing) seems to interpret them literally within a [] context, anyway, but it's always far better to be explicit.
^[a-zA-Z \-,\.\(\)]*$
However, this still shouldn't be allowing 0-9 digits, so your actual code that uses this regular expression probably has an issue, as well.
The  -, in [a-zA-Z -,.()] describes a range from   (0x20) to , (0x2C). And that is equivalent to [ !"#$%'()*+,]. You should either escape the - or place it somewhere else where it is not interpreted as a range indicator.
But that’s not the cause of this issue as the digits are from 0x30 to 0x39.
I tried that with javascript and it works fine. The others are correct, though. If in javascript, check if everything works fine or else the check will not happen at all.

Regexp Question - Negating a captured character

I'm looking for a regular expression that allows for either single-quoted or double-quoted strings, and allows the opposite quote character within the string. For example, the following would both be legal strings:
"hello 'there' world"
'hello "there" world'
The regexp I'm using uses negative lookahead and is as follows:
(['"])(?:(?!\1).)*\1
This would work I think, but what about if the language didn't support negative lookahead. Is there any other way to do this? Without alternation?
EDIT:
I know I can use alternation. This was more of just a hypothetical question. Say I had 20 different characters in the initial character class. I wouldn't want to write out 20 different alternations. I'm trying to actually negate the captured character, without using lookahead, lookbehind, or alternation.
This is actually much simpler than you may have realized. You don't really need the negative look-ahead. What you want to do is a non-greedy (or lazy) match like this:
(['"]).*?\1
The ? character after the .* is the important part. It says, consume the minimum possible characters before hitting the next part of the regex. So, you get either kind of quote, and then you go after 0-M characters until you encounter a character matching whichever quote you first ran into. You can learn more about greedy matching vs. non-greedy here and here.
Sure:
'([^']*)'|"([^"]*)"
On a successful match, the $+ variable will hold the contents of whichever alternate matched.
In the general case, regexps are not really the answer. You might be interested in something like Text::ParseWords, which tokenizes text, accounting for nested quotes, backslashed quotes, backslashed spaces, and other oddities.