I'm having some difficulties with the regex boundary \b character. I need to search for an exact keyword inside some loaded text(either plain textual data or Xml). Because the need for exact matches I use the \bkeyword\b pattern but I get a different behaviour than what I expect when the keyword starts with a special character. For example the pattern \b€ 3,5\b doesn't match in I have € 3,5 to spend!. This is the case with any special characters.
I've search around but came up with no solution. Is there some mechanism that acts like a \b but for special characters ? Also note that i cannot alter the keyword.
Any help would be appreciated.
You can perhaps make use of a positive lookbehind:
(?<=^|\s)€ 3,5\b
The positive lookbehind will match either the beginning of the string or a \s, without including them in the match itself.
Related
I m trying to matching a string which will not allow same special character at same time
my regular expression is:
[RegularExpression(#"^+[a-zA-Z0-9]+[a-zA-Z0-9.&' '-]+[a-zA-Z0-9]$")]
this solve my all requirement except the below two issues
this is my string : bracks
acceptable :
bra-cks, b-r-a-c-ks, b.r.a.c.ks, bra cks (by the way above regular expression solved this)
not acceptable:
issue 1: b.. or bra..cks, b..racks, bra...cks (two or more any special character together),
issue 2: bra cks (two ore more white space together)
You can use a negative lookahead to invalidate strings containing two consecutive special characters:
^(?!.*[.&' -]{2})[a-zA-Z0-9.&' -]+$
Demo: https://regex101.com/r/7j14bu/1
The goal
From what i can tell by your description and pattern, you are trying to match text, which start and end with alphanumeric (due to ^+[a-zA-Z0-9] and [a-zA-Z0-9]$ inyour original pattern), and inside, you just don't want to have any two consecuive (adjacent) special characters, which, again, guessing from the regex, are . & ' -
What was wrong
^+ - i think here you wanted to assure that match starts at the beginning of the line/string, so you don't need + here
[a-zA-Z0-9.&' '-] - in this character class you doubled ' which is totally unnecessary
Solution
Please try pattern
^[a-zA-Z0-9](?:(?![.& '-]{2,})[a-zA-Z0-9.& '-])*[a-zA-Z0-9]$
Pattern explanation
^ - anchor, match the beginning of the string
[a-zA-Z0-9] - character class, match one of the characters inside []
(?:...) - non capturing group
(?!...) - negative lookahead
[.& '-]{2,} - match 2 or more of characters inside character class
[a-zA-Z0-9.& '-] - character class, match one of the characters inside []
* - match zero or more text matching preceeding pattern
$ - anchor, match the end of the string
Regex demo
Some remarks on your current regex:
It looks like you placed the + quantifiers before the pattern you wanted to quantify, instead of after. For instance, ^+ doesn't make much sense, since ^ is just the start of the input, and most regex engines would not even allow that.
The pattern [a-zA-Z0-9.&' '-]+ doesn't distinguish between alphanumerical and other characters, while you want the rules for them to be different. Especially for the other characters you don't want them to repeat, so that + is not desired for those.
In a character class it doesn't make sense to repeat the same character, like you have a repeat of a quote ('). Maybe you wanted to somehow delimit the space, but realise that those quotes are interpreted literally. So probably you should just remove them. Or if you intended to allow for a quote, only list it once.
Here is a correction (add the quote if you still need it):
^[a-zA-Z0-9]+(?:[.& -][a-zA-Z0-9]+)*$
Follow-up
Based on a comment, I suspect you would allow a non-alphanumerical character to be surrounded by single spaces, even if that gives a sequence of more than one non-alphanumerical character. In that case use this:
^[a-zA-Z0-9]+(?:(?:[ ]|[ ]?[.&-][ ]?)[a-zA-Z0-9]+)*$
So here the space gets a different role: it can optionally occur before and after a delimiter (one of ".&-"), or it can occur on its own. The brackets around the spaces are not needed, but I used them to stress that the space is intended and not a typo.
[SOME_WORDS:200:1000]
Trying to match just the last 1000 part. Both numbers are variable and can contain an unknown number of characters (although they are expected to contain digits, I cannot rule out that they may also contain other characters). The SOME_WORDS part is known and does not change.
So I begin by doing a positive lookbehind for [SOME_WORDS: followed by a positive lookahead for the trailing ]
That gives us the pattern (?<=\[SOME_WORDS:).*(?=])
And captures the part 200:1000
Now because I don't know how many characters are after SOME_WORDS:, but I know that it ends with another : I use .*: to indicate any character any amount of time followed by :
That gives us the pattern (?<=\[SOME_WORDS:.*:).*(?=])
However at this point the pattern no longer matches anything and this is where I become confused. What am I doing wrong here?
If I assume that the first number will always be 3 characters long I can replace .* with ... to get the pattern (?<=\[SOME_WORDS:...:).*(?=]) and this correctly captures just the 1000 part. However I don't understand why replacing ... with .* makes the pattern not capture anything.
EDIT:
It seems like the online tool I was using to test the regex pattern wasn't working correctly. The pattern (?<=\[SOME_WORDS:.*:).*(?=]) matches the 1000 with no issues when actually done in .net
You usually cannot use a + or a * in a lookbehind, only in a lookahead.
If c# does allow these than you could use a .*? instead of a .* as the .* will eat the second :
Try this:
(?<=\[SOME_WORDS:)(?=\d+:(\d+)])
The match wil be in the first capture group
Quote from http://www.regular-expressions.info/lookaround.html
The bad news is that most regex flavors do not allow you to use just any regex inside a lookbehind, because they cannot apply a regular expression backwards. The regular expression engine needs to be able to figure out how many characters to step back before checking the lookbehind. When evaluating the lookbehind, the regex engine determines the length of the regex inside the lookbehind, steps back that many characters in the subject string, and then applies the regex inside the lookbehind from left to right just as it would with a normal regex.
As Robert Smit mentions this is due to the * being a greedy operator. Greedy operators consume as many characters as they possibly can when they are matched first. They only give up characters if the match fails. If you make the greedy operator lazy(*?), then matching consumes as little number of characters as possible for the match to succeed, so the : is not consumed by *. You can also use [^:]* which is match any character other than :.
I know the regex for excluding words, roughly anyway, It would be (!?wordToIgnore|wordToIgnore2|wordToIgnore3)
But I have an existing, complicated regex that I need to add this to, and I am a bit confused about how to go about that. I'm still pretty new to regex, and it took me a very long time to make this particular one, but I'm not sure where to insert it or how ...
The regex I have is ...
^(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:/\p{L}'-]{1,64}$)$
This should only allow the person typing to insert between 1 and 64 letters that match that pattern, cannot start with a space, quote, double quote, special character, a dash, an escape character, etc, and only allows a-z both upper and lowercase, can include a space, ":", a dash, and a quote anywhere but the beginning.
But I want to forbid them from using certain words, so I have this list of words that I want to be forbidden, I just cannot figure out how to get that to fit into here.. I tried just pasting the whole .. "block" in, and that didn't work.
?!the|and|or|a|given|some|that|this|then|than
Has anyone encountered this before?
ciel, first off, congratulations for getting this far trying to build your regex rule. If you want to read something detailed about all kinds of exclusions, I suggest you have a look at Match (or replace) a pattern except in situations s1, s2, s3 etc
Next, in your particular situation, here is how we could approach your regex.
For consision, let's make all the negative lookarounds more compact, replacing them with a single (?!.*(?: |-|'){2})
In your character class, the \: just escapes the colon, needlessly so as : is enough. I assume you wanted to add a backslash character, and if so we need to use \\
\p{L} includes [a-zA-Z], so you can drop [a-zA-Z]. But are you sure you want to match all letters in any script? (Thai etc). If so, remember to set the u flag after the regex string.
For your "bad word exclusion" applying to the whole string, place it at the same position as the other lookarounds, i.e., at the head of the string, but using the .* as in your other exclusions: (?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3)) It does not matter which lookahead comes first because lookarounds do not change your position in the string. For more on this, see Mastering Lookahead and Lookbehind
This gives us this glorious regex (I added the case-insensitive flag):
^(?i)(?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3))(?!.*(?: |-|'){2})(?:[\\0-9 :/\p{L}'-]{1,64}$)$
Of course if you don't want unicode letters, replace \p{L} with a-z
Also, if you want to make sure that the wordToIgnore is a real word, as opposed to an embedded string (for instance you don't want cat but you are okay with catalog), add boundaries to the lookahead rule: (?!.*\b(?:wordToIgnore|wordToIgnore2|wordToIgnore3)\b)
use this:
^(?!.*(the|and|or|a|given|some|that|this|then|than))(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:\p{L}'-]{1,64}$)$
see demo
After extensive search, I am unable to find an explanation for the need to use .* in regex. For example, MSDN suggests a password regex of
#\"(?=.{6,})(?=(.*\d){1,})(?=(.*\W){1,})"
for length >= 6, 1+ digit and 1+ special character.
Why can't I just use:
#\"(?=.{6,})(?=(\d){1,})(?=(\W){1,})"
.* just means "0 or more of any character"
It's broken down into two parts:
. - a "dot" indicates any character
* - means "0 or more instances of the preceding regex token"
In your example above, this is important, since they want to force the password to contain a special character and a number, while still allowing all other characters. If you used \d instead of .*, for example, then that would restrict that portion of the regex to only match decimal characters (\d is shorthand for [0-9], meaning any decimal). Similarly, \W instead of .*\W would cause that portion to only match non-word characters.
A good reference containing many of these tokens for .NET can be found on the MSDN here: Regular Expression Language - Quick Reference
Also, if you're really looking to delve into regex, take a look at http://www.regular-expressions.info/. While it can sometimes be difficult to find what you're looking for on that site, it's one of the most complete and begginner-friendly regex references I've seen online.
Just FYI, that regex doesn't do what they say it does, and the way it's written is needlessly verbose and confusing. They say it's supposed to match more than seven characters, but it really matches as few as six. And while the other two lookaheads correctly match at least one each of the required character types, they can be written much more simply.
Finally, the string you copied isn't just a regex, it's an XML attribute value (including the enclosing quotes) that seems to represent a C# string literal (except the closing quote is missing). I've never used a Membership object, but I'm pretty sure that syntax is faulty. In any case, the actual regex is:
(?=.{6,})(?=(.*\d){1,})(?=(.*\W){1,})
..but it should be:
(?=.{8,})(?=.*\d)(?=.*\W)
The first lookahead tries to match eight or more of any characters. If it succeeds, the match position (or cursor, if you prefer) is reset to the beginning and the second lookahead scans for a digit. If it finds one, the cursor is reset again and the third lookahead scans for a special character. (Which, by the way, includes whitespace, control characters, and a boatload of other esoteric characters; probably not what the author intended.)
If you left the .* out of the latter two lookaheads, you would have (?=\d) asserting that the first character is a digit, and (?=\W) asserting that it's not a digit. (Digits are classed as word characters, and \W matches anything that's not a word character.) The .* in each lookahead causes it to initially gobble up the whole string, then backtrack, giving back one character at a time until it reaches a spot where the \d or \W can match. That's how they can match the digit and the special character anywhere in the string.
The .* portion just allows for literally any combination of characters to be entered. It's essentially allowing for the user to add any level of extra information to the password on top of the data you are requiring
Note: I don't think that MSDN page is actually suggesting that as a password validator. It is just providing an example of a possible one.
What is the regular exp for a text that can't contain any special characters except space?
Because Prajeesh only wants to match spaces, \s will not suffice as it matches all whitespace characters including line breaks and tabs.
A character set that should universally work across all RegEx parsers is:
[a-zA-Z0-9 ]
Further control depends on your needs. Word boundaries, multi-line support, etc... I would recommend visiting Regex Library which also has some links to various tutorials on how Regular Expression Parsing works.
[\w\s]*
\w will match [A-Za-z0-9_] and the \s will match whitespaces.
[\w ]* should match what you want.
Assuming "special characters" means anything that's not a letter or digit, and "space" means the space character (ASCII 32):
^[A-Za-z0-9 ]+$
You need #"^[A-Za-z0-9 ]+$". The \s character class matches things other than space (such as tab) and you since you want to match sure that no part of the string has other characters you should anchor it with ^ and $.
If you just want alphabets and spaces then you can use: #"[A-Za-z\s]+" to match at least one character or space. You could also use #"[A-Za-z ]+" instead without explicitly denoting the space.
Otherwise please clarify.
In C#, I'd believe it's ^(\w|\s)*$