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In a regular expression, I need to know how to match one thing or another, or both (in order). But at least one of the things needs to be there.
For example, the following regular expression
/^([0-9]+|\.[0-9]+)$/
will match
234
and
.56
but not
234.56
While the following regular expression
/^([0-9]+)?(\.[0-9]+)?$/
will match all three of the strings above, but it will also match the empty string, which we do not want.
I need something that will match all three of the strings above, but not the empty string. Is there an easy way to do that?
UPDATE:
Both Andrew's and Justin's below work for the simplified example I provided, but they don't (unless I'm mistaken) work for the actual use case that I was hoping to solve, so I should probably put that in now. Here's the actual regexp I'm using:
/^\s*-?0*(?:[0-9]+|[0-9]{1,3}(?:,[0-9]{3})+)(?:\.[0-9]*)?(\s*|[A-Za-z_]*)*$/
This will match
45
45.988
45,689
34,569,098,233
567,900.90
-9
-34 banana fries
0.56 points
but it WON'T match
.56
and I need it to do this.
The fully general method, given regexes /^A$/ and /^B$/ is:
/^(A|B|AB)$/
i.e.
/^([0-9]+|\.[0-9]+|[0-9]+\.[0-9]+)$/
Note the others have used the structure of your example to make a simplification. Specifically, they (implicitly) factorised it, to pull out the common [0-9]* and [0-9]+ factors on the left and right.
The working for this is:
all the elements of the alternation end in [0-9]+, so pull that out: /^(|\.|[0-9]+\.)[0-9]+$/
Now we have the possibility of the empty string in the alternation, so rewrite it using ? (i.e. use the equivalence (|a|b) = (a|b)?): /^(\.|[0-9]+\.)?[0-9]+$/
Again, an alternation with a common suffix (\. this time): /^((|[0-9]+)\.)?[0-9]+$/
the pattern (|a+) is the same as a*, so, finally: /^([0-9]*\.)?[0-9]+$/
Nice answer by huon (and a bit of brain-twister to follow it along to the end). For anyone looking for a quick and simple answer to the title of this question, 'In a regular expression, match one thing or another, or both', it's worth mentioning that even (A|B|AB) can be simplified to:
A|A?B
Handy if B is a bit more complex.
Now, as c0d3rman's observed, this, in itself, will never match AB. It will only match A and B. (A|B|AB has the same issue.) What I left out was the all-important context of the original question, where the start and end of the string are also being matched. Here it is, written out fully:
^(A|A?B)$
Better still, just switch the order as c0d3rman recommended, and you can use it anywhere:
A?B|A
Yes, you can match all of these with such an expression:
/^[0-9]*\.?[0-9]+$/
Note, it also doesn't match the empty string (your last condition).
Sure. You want the optional quantifier, ?.
/^(?=.)([0-9]+)?(\.[0-9]+)?$/
The above is slightly awkward-looking, but I wanted to show you your exact pattern with some ?s thrown in. In this version, (?=.) makes sure it doesn't accept an empty string, since I've made both clauses optional. A simpler version would be this:
/^\d*\.?\d+$/
This satisfies your requirements, including preventing an empty string.
Note that there are many ways to express this. Some are long and some are very terse, but they become more complex depending on what you're trying to allow/disallow.
Edit:
If you want to match this inside a larger string, I recommend splitting on and testing the results with /^\d*\.?\d+$/. Otherwise, you'll risk either matching stuff like aaa.123.456.bbb or missing matches (trust me, you will. JavaScript's lack of lookbehind support ensures that it will be possible to break any pattern I can think of).
If you know for a fact that you won't get strings like the above, you can use word breaks instead of ^$ anchors, but it will get complicated because there's no word break between . and (a space).
/(\b\d+|\B\.)?\d*\b/g
That ought to do it. It will block stuff like aaa123.456bbb, but it will allow 123, 456, or 123.456. It will allow aaa.123.456.bbb, but as I've said, you'll need two steps if you want to comprehensively handle that.
Edit 2: Your use case
If you want to allow whitespace at the beginning, negative/positive marks, and words at the end, those are actually fairly strict rules. That's a good thing. You can just add them on to the simplest pattern above:
/^\s*[-+]?\d*\.?\d+[a-z_\s]*$/i
Allowing thousands groups complicates things greatly, and I suggest you take a look at the answer I linked to. Here's the resulting pattern:
/^\s*[-+]?(\d+|\d{1,3}(,\d{3})*)?(\.\d+)?\b(\s[a-z_\s]*)?$/i
The \b ensures that the numeric part ends with a digit, and is followed by at least one whitespace.
Maybe this helps (to give you the general idea):
(?:((?(digits).^|[A-Za-z]+)|(?<digits>\d+))){1,2}
This pattern matches characters, digits, or digits following characters, but not characters following digits.
The pattern matches aa, aa11, and 11, but not 11aa, aa11aa, or the empty string.
Don't be puzzled by the ".^", which means "a character followd by line start", it is intended to prevent any match at all.
Be warned that this does not work with all flavors of regex, your version of regex must support (?(named group)true|false).
Is it possible to design a regular expression in a way that a part of it is dependent on another section of the same regular expression?
Consider the following example:
(ABCHEHG)[HGE]{5,1230}(EEJOPK)[DM]{5}
I want to continue this regex, and at some point I will have a section where the result of that section should depend on the result of [DM]{5}.
For example, D will be complemented by C, and M will be complemented by N.
(ABCHEHG)[HGHE]{5,1230}(EEJOPK)[DM]{5}[ACF]{1,1000}(BBBA)[CU]{2,5}[D'M']{5}
By D' I mean C, and by M' I mean N.
So a resulting string that matches the above regex, if it has DDDMM matching to the section [DM]{5}, it should necessarily have CCCNN matching to [D'M']{5}. Therefore, the result of [D'M']{5} always depends on [DM]{5}, or in other words, what matches to [DM]{5} always dictates what will match to [D'M']{5}.
Is it possible to do such a thing with regex?
Please note that, in this example I have extremely over-simplified the problem. The regex pattern I currently have is really much more complex and longer and my actual pattern includes about 5-6 of such dependent sections.
I cannot think of a way you can do this in pure regex. I would run 2 regex expressions. The first regex to extract the [DM]{5} string, such as
(ABCHEHG)[HGHE]{5,1230}(EEJOPK)[DM]{5}
And take the last 5 characters. Now replace the characters, for example in C# it would be result = result.Substring(result.Length - 5, 5).Replace('D', 'C').Replace('M', 'N'), and then concatenate like
(ABCHEHG)[HGHE]{5,1230}(EEJOPK)[DM]{5}[ACF]{1,1000}(BBBA)[CU]{2,5} + result
This is pretty easy to do in Perl:
m{
ABCHEHG
[HGHE]{5,1230}
EEJOPK
( [DM]{5} )
[ACF]{1,1000}
BBBA
[CU]{2,5}
(??{ $1 =~ tr/DM/CN/r })
}x
I've added the x modifier and whitespace for better readability. I've also removed the capturing groups around the fixed strings (they're fixed strings; you already know what they're going to capture).
The crucial part is that we capture the string that was actually matched by [DM]{5} (in $1), which we then use at the end to dynamically generate a subpattern by replacing all D by C and M by N in $1.
This sounds like bioinformatics in python. Do 2-stage filtering, at regex level and at app level.
Wildcard the DM portions, so the regex is permissive in what it accepts. Bury the regex in a token generator that yields several matching sections. Have your app iterate through the generator's results, discarding any result rejected by your business logic, such as finding that one token is not the complement of another token.
Alternatively, you might push some of that work down into a complex generated regex, which likely will perform worse and will be harder to debug. Your DDDMM example might be summarized as D+M+, or [DM]+, not sure if sequence matters. The complement might be C+N+, or [CN]+. Apparently there's two cases here. So start assembling a regex: stuff1 [DM]+ stuff2 [CN]+ stuff3. Then tack on '|' for alternation, and tack on the other case: stuff1 [CN]+ stuff2 [DM]+ stuff3 (or factor out suffix and prefix so alternation starts after stuff1). I can't imagine you'll be happy with such an approach, as the combinatorics get ugly, and the regex engine is forced to do lots of scanning and backtracking. And recompiling additional regexes on the fly doesn't come for free. Instead you should use the regex engine for the simple things that it's good at, and delegate complex business logic decisions to your app.
I'm pulling car submodels from the DB and I'm building my regular expression on the fly.
Here is an example of a search string:
EX-L Sedan 4-Door
Here is my regular expression:
preg_match("/LX|EX|EX-L|LX-P|LX-S/Ui", $input_line, $output_array);
For some reason the output is EX and not EX-L as it supposed to be. Can someone explain why?
Your pattern is unanchored and thus the first alternative that matches a substring makes the regex engine stop processing the whole group. This is a common behavior with NFA regexes.
Also, there are no quantifiers in your pattern, thus the /U modifier is redundant.
So, you can use
/EX-L|LX-P|LX-S|LX|EX/i
It is a readable form. However, best practice with regexes is to make sure no alternative branch can match at the same location as another. That means you can use
/EX(-L)?|LX(-[PS])?/i
As others have pointed out, the reason for this undesired outcome is because the regex engine is happy to have the first alternative and run for the door since your pattern has no anchors (like: ^, $, and some other lesser known ones). This is the same short-circuiting behavior you'd see in php's if($x || $y) conditions; if $x is true there is no need to evaluate further. But enough about that...
I would like to offer some additional logic that I think is relevant to your case/question.
You say your regex is built on the fly, so I am assuming your method goes something like this:
A user identifies which substrings/keywords they want to search for.
$strings=array('LX','EX','EX-L','LX-P','LX-S');
// array of substrings in any order
As mentioned earlier, you need longer strings to precede shorter ones with identical starting characters.
rsort($strings);
// sort DESC, longer strings precede shorter strings when leading characters match
Pipe all strings into a single regex pattern with implode().
$piped_regex='/\b(?:'.implode('|',$array).')\b/i';
// word boundaries ensure the string is not part of a larger word; remove if not desired
// pattern: /\b(?:LX-S|LX-P|LX|EX-L|EX)\b/i
While programmatically condensing your similar strings into a concise pattern as Wiktor recommended is possible, it's probably not worth the effort with your on-the-fly patterns.
Finally run preg_match() as normal.
$input_line='EX-L Sedan 4-Door';
if(preg_match($piped_regex,$input_line,$output_array)){
var_export($output_array);
}
// output: array(0=>'EX-L')
I hope stepping out this method is helpful to you and future SO readers.
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?
The Greedy Option of Regex is really needed?
Lets say I have following texts, I like to extract texts inside [Optionx] and [/Optionx] blocks
[Option1]
Start=1
End=10
[/Option1]
[Option2]
Start=11
End=20
[/Option2]
But with Regex Greedy Option, its give me
Start=1
End=10
[/Option1]
[Option2]
Start=11
End=20
Anybody need like that? If yes, could you let me know?
If I understand correctly, the question is “why (when) do you need greedy matching?”
The answer is – almost always. Consider a regular expression that matches a sequence of arbitrary – but equal – characters, of length at least two. The regular expression would look like this:
(.)\1+
(\1 is a back-reference that matches the same text as the first parenthesized expression).
Now let’s search for repeats in the following string: abbbbbc. What do we find? Well, if we didn’t have greedy matching, we would find bb. Probably not what we want. In fact, in most application s we would be interested in finding the whole substring of bs, bbbbb.
By the way, this is a real-world example: the RLE compression works like that and can be easily implemented using regex.
In fact, if you examine regular expressions all around you will see that a lot of them use quantifiers and expect them to behave greedily. The opposite case is probably a minority. Often, it makes no difference because the searched expression is inside guard clauses (e.g. a quoted string is inside the quote marks) but like in the example above, that’s not always the case.
Regular expressions can potentially match multiple portion of a text.
For example consider the expression (ab)*c+ and the string "abccababccc". There are many portions of the string that can match the regular expressions:
(abc)cababccc
(abcc)ababccc
abcc(ababccc)
abccab(abccc)
ab(c)cababccc
ab(cc)ababccc
abcabab(c)ccc
....
some regular expressions implementation are actually able to return the entire set of matches but it is most common to return a single match.
There are many possible ways to determine the "winning match". The most common one is to take the "longest leftmost match" which results in the greedy behaviour you observed.
This is tipical of search and replace (a la grep) when with a+ you probably mean to match the entire aaaa rather than just a single a.
Choosing the "shortest non-empty leftmost" match is the usual non-greedy behaviour. It is the most useful when you have delimiters like your case.
It all depends on what you need, sometimes greedy is ok, some other times, like the case you showed, a non-greedy behaviour would be more meaningful. It's good that modern implementations of regular expressions allow us to do both.
If you're looking for text between the optionx blocks, instead of searching for .+, search for anything that's not "[\".
This is really rough, but works:
\[[^\]]+]([^(\[/)]+)
The first bit searches for anything in square brackets, then the second bit searches for anything that isn't "[\". That way you don't have to care about greediness, just tell it what you don't want to see.
One other consideration: In many cases, greedy and non-greedy quantifiers result in the same match, but differ in performance:
With a non-greedy quantifier, the regex engine needs to backtrack after every single character that was matched until it finally has matched as much as it needs to. With a greedy quantifier, on the other hand, it will match as much as possible "in one go" and only then backtrack as much as necessary to match any following tokens.
Let's say you apply a.*c to
abbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbc. This finds a match in 5 steps of the regex engine. Now apply a.*?c to the same string. The match is identical, but the regex engine needs 101 steps to arrive at this conclusion.
On the other hand, if you apply a.*c to abcbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb, it takes 101 steps whereas a.*?c only takes 5.
So if you know your data, you can tailor your regex to match it as efficiently as possible.
just use this algorithm which you can use in your fav language. No need regex.
flag=0
open file for reading
for each line in file :
if check "[/Option" in line:
flag=0
if check "[Option" in line:
flag=1
continue
if flag:
print line.strip()
# you can store the values of each option in this part