I have been solving old question from stack so that I can improve my regex knowledge. As I have a basic knowledge of regex, most of them were easy but this question regex problem is tough.
It asks for a regex that extracts from this kind of string ou=persons,ou=(.*),dc=company,dc=org the last string immediately preceded by a comma not followed by (.*). In the last case, this should give dc=company,dc=org.
The solution is (?<=,(?!.*\Q(.*)\E)).* but I cannot understand its flow. I understood (?!.*\Q(.*)\E) portion but other are still mystery to me. Specially ?<= which is a positive look-behind. Does it search from end of string? Can anyone explain it to me like I am a 7 year old kid — and please http://regex101.com/ is not helping.
The RegEx (?<=,(?!.*\Q(.*)\E)).* look-behind potion works like this:
Start at the beginning of the string at first character.
Can we match the the thing we are looking for? ,(?!.*\Q(.*)\E)
If we can't: Move forward one character, Go To 2. and check match again.
If a match is found: Capture all the remaining characters until we can't find any .* (or generally then try the matching the remaining RegEx).
For a more wordly explaination consider reading Lookahead and Lookbehind Zero-Length Assertions.
A lookbehind allows you to specify a context just before the actual match.
You can say ,(dc=) and only return the capture group, or ,\Kdc=, or (?<=,)dc= to return the match on dc= but require that the comma is present just before the match.
The facility also allows for multiple lookbehinds, so you could do (?<=a.*)(?<=b.*)c to match c only if it is preceded by both a and b somewhere in the input.
A lookbehind is basically syntactic sugar, in that you can usually rephrase your conditions using some other regex construct. It can be really handy when you have multiple unanchored constraints, like in the last example
Related
I have the below patterns to be excluded.
make it cheaper
make it cheapere
makeitcheaper.com.au
makeitcheaper
making it cheaper
www.make it cheaper
ww.make it cheaper.com
I've created a regex to match any of these. However, I want to get everything else other than these. I am not sure how to inverse this regex I've created.
mak(e|ing) ?it ?cheaper
Above pattern matches all the strings listed. Now I want it to match everything else. How do I do it?
From the search, it seems I need something like negative lookahead / look back. But, I don't really get it. Can some one point me in the right direction?
You can just put it in a negative look-ahead like so:
(?!mak(e|ing) ?it ?cheaper)
Just like that isn't going to work though since, if you do a matches1, it won't match since you're just looking ahead, you aren't actually matching anything, and, if you do a find1, it will match many times, since you can start from lots of places in the string where the next characters doesn't match the above.
To fix this, depending on what you wish to do, we have 2 choices:
If you want to exclude all strings that are exactly one of those (i.e. "make it cheaperblahblah" is not excluded), check for start (^) and end ($) of string:
^(?!mak(e|ing) ?it ?cheaper$).*
The .* (zero or more wild-cards) is the actual matching taking place. The negative look-ahead checks from the first character.
If you want to exclude all strings containing one of those, you can make sure the look-ahead isn't matched before every character we match:
^((?!mak(e|ing) ?it ?cheaper).)*$
An alternative is to add wild-cards to the beginning of your look-ahead (i.e. exclude all strings that, from the start of the string, contain anything, then your pattern), but I don't currently see any advantage to this (arbitrary length look-ahead is also less likely to be supported by any given tool):
^(?!.*mak(e|ing) ?it ?cheaper).*
Because of the ^ and $, either doing a find or a matches will work for either of the above (though, in the case of matches, the ^ is optional and, in the case of find, the .* outside the look-ahead is optional).
1: Although they may not be called that, many languages have functions equivalent to matches and find with regex.
The above is the strictly-regex answer to this question.
A better approach might be to stick to the original regex (mak(e|ing) ?it ?cheaper) and see if you can negate the matches directly with the tool or language you're using.
In Java, for example, this would involve doing if (!string.matches(originalRegex)) (note the !, which negates the returned boolean) instead of if (string.matches(negLookRegex)).
The negative lookahead, I believe is what you're looking for. Maybe try:
(?!.*mak(e|ing) ?it ?cheaper)
And maybe a bit more flexible:
(?!.*mak(e|ing) *it *cheaper)
Just in case there are more than one space.
I'm trying to write a very simple regular expression that matches any file name that doesn't end in .php. I came up with the following...
(.*?)(?!\.php)$
...however this matches all filenames. If someone could point me in the right direction I'd be very grateful.
Almost:
.*(?!\.php)....$
The last four dots make sure that there is something to look ahead at, when the look-ahead is checked.
The outer parentheses are unnecessary since you are interested in the entire match.
The reluctant .*? is unnecessary, since backtracking four steps is more efficient than checking the following condition with every step.
Instead of using negative lookahead, sometimes it's easier to use the negation outside the regex at the hosting language level. In many languages, the boolean complement operator is the unary !.
So you can write something like this:
! str.hasMatch(/\.php$/)
Depending on language, you can also skip regex altogether and use something like (e.g. Java):
! str.endsWith(".php")
As for the problem with the original pattern itself:
(.*?)(?!\.php)$ // original pattern, doesn't work!
This matches, say, file.php, because the (.*?) can capture file.php, and looking ahead, you can't match \.php, but you can match a $, so altogether it's a match! You may want to use look behind, or if it's not supported, you can lookahead at the start of the string.
^(?!.*\.php$).*$ // negative lookahead, works
This will match all strings that does not end with ".php" using negative lookahead.
References
regular-expressions.info/Lookarounds
Related questions
How does the regular expression (?<=#)[^#]+(?=#) work?
You are at the end of the string and looking ahead. What you want is a look behind instead:
(.*)$(?<!\.php)
Note that not all regular expression engines support lookbehind assertions.
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.
I know it is quite some weird goal here but for a quick and dirty fix for one of our system we do need to not filter any input and let the corruption go into the system.
My current regex for this is "\^.*"
The problem with that is that it does not match characters as planned ... but for one match it does work. The string that make it not work is ^#jj (basically anything that has ^ ... ).
What would be the best way to not match any characters now ? I was thinking of removing the \ but only doing this will transform the "not" into a "start with" ...
The ^ character doesn't mean "not" except inside a character class ([]). If you want to not match anything, you could use a negative lookahead that matches anything: (?!.*).
A simple and cheap regex that will never match anything is to match against something that is simply unmatchable, for example: \b\B.
It's simply impossible for this regex to match, since it's a contradiction.
References
regular-expressions.info\Word Boundaries
\B is the negated version of \b. \B matches at every position where \b does not.
Another very well supported and fast pattern that would fail to match anything that is guaranteed to be constant time:
$unmatchable pattern $anything goes here etc.
$ of course indicates the end-of-line. No characters could possibly go after $ so no further state transitions could possibly be made. The additional advantage are that your pattern is intuitive, self-descriptive and readable as well!
tldr; The most portable and efficient regex to never match anything is $- (end of line followed by a char)
Impossible regex
The most reliable solution is to create an impossible regex. There are many impossible regexes but not all are as good.
First you want to avoid "lookahead" solutions because some regex engines don't support it.
Then you want to make sure your "impossible regex" is efficient and won't take too much computation steps to match... nothing.
I found that $- has a constant computation time ( O(1) ) and only takes two steps to compute regardless of the size of your text (https://regex101.com/r/yjcs1Z/3).
For comparison:
$^ and $. both take 36 steps to compute -> O(1)
\b\B takes 1507 steps on my sample and increase with the number of character in your string -> O(n)
Empty regex (alternative solution)
If your regex engine accepts it, the best and simplest regex to never match anything might be: an empty regex .
Instead of trying to not match any characters, why not just match all characters? ^.*$ should do the trick. If you have to not match any characters then try ^\j$ (Assuming of course, that your regular expression engine will not throw an error when you provide it an invalid character class. If it does, try ^()$. A quick test with RegexBuddy suggests that this might work.
^ is only not when it's in class (such as [^a-z] meaning anything but a-z). You've turned it into a literal ^ with the backslash.
What you're trying to do is [^]*, but that's not legal. You could try something like
" {10000}"
which would match exactly 10,000 spaces, if that's longer than your maximum input, it should never be matched.
((?iLmsux))
Try this, it matches only if the string is empty.
Interesting ... the most obvious and simple variant:
~^
.
https://regex101.com/r/KhTM1i/1
requiring usually only one computation step (failing directly at the start and being computational expensive only if the matched string begins with a long series of ~) is not mentioned among all the other answers ... for 12 years.
You want to match nothing at all? Neg lookarounds seems obvious, but can be slow, perhaps ^$ (matches empty string only) as an alternative?
Is is possible to construct a regular expression that rejects all input strings?
Probably this:
[^\w\W]
\w - word character (letter, digit, etc)
\W - opposite of \w
[^\w\W] - should always fail, because any character should belong to one of the character classes - \w or \W
Another snippets:
$.^
$ - assert position at the end of the string
^ - assert position at the start of the line
. - any char
(?#it's just a comment inside of empty regex)
Empty lookahead/behind should work:
(?<!)
The best standard regexs (i.e., no lookahead or back-references) that reject all inputs are (after #aku above)
.^
and
$.
These are flat contradictions: "a string with a character before its beginning" and "a string with a character after its end."
NOTE: It's possible that some regex implementations would reject these patterns as ill-formed (it's pretty easy to check that ^ comes at the beginning of a pattern and $ at the end... with a regular expression), but the few I've checked do accept them. These also won't work in implementations that allow ^ and $ to match newlines.
(?=not)possible
?= is a positive lookahead. They're not supported in all regexp flavors, but in many.
The expression will look for "not", then look for "possible" starting at the same position (since lookaheads don't move forward in the string).
One example of why such thing could possibly be needed is when you want to filter some input with regexes and you pass regex as an argument to a function.
In spirit of functional programming, for algebraic completeness, you may want some trivial primary regexes like "everything is allowed" and "nothing is allowed".
To me it sounds like you're attacking a problem the wrong way, what exactly
are you trying to solve?
You could do a regular expression that catches everything and negate the result.
e.g in javascript:
if (! str.match( /./ ))
but then you could just do
if (!foo)
instead, as #[jan-hani] said.
If you're looking to embed such a regex in another regex, you
might be looking for $ or ^ instead, or use lookaheads like #[henrik-n] mentioned.
But as I said, this looks like a "I think I need x, but what I really need is y" problem.
Why would you even want that? Wouldn't a simple if statment do the trick? Something along the lines of:
if ( inputString != "" )
doSomething ()
[^\x00-\xFF]
It depends on what you mean by "regular expression". Do you mean regexps in a particular programming language or library? In that case the answer is probably yes, and you can refer to any of the above replies.
If you mean the regular expressions as taught in computer science classes, then the answer is no. Every regular expression matches some string. It could be the empty string, but it always matches something.
In any case, I suggest you edit the title of your question to narrow down what kinds of regular expressions you mean.
[^]+ should do it.
In answer to aku's comment attached to this, I tested it with an online regex tester (http://www.regextester.com/), and so assume it works with JavaScript. I have to confess to not testing it in "real" code. ;)
EDIT:
[^\n\r\w\s]
Well,
I am not sure if I understood, since I always thought of regular expression of a way to match strings. I would say the best shot you have is not using regex.
But, you can also use regexp that matches empty lines like ^$ or a regexp that do not match words/spaces like [^\w\s] ...
Hope it helps!