I've written this regex
/(?=.*[a-z])(?!.*[A-Z])([\w\_\-\.].{3,10})/g
to check the following conditions
>has minimum of 3 and maximum of 10 characters.
>must contain atleast a lowercase alphabet.
>must contain only lowercase alphabets, '_', '-', '.' and digits.
this works but returnes true even if there is more than 10 characters.
I would like a new or modified regular expression to check the above given conditions.
add hanchors
remove the last dot
the negative lookahead is useless is you use a correct character class
This regex will work:
^(?=.*[a-z])[a-z0-9_.-]{3,10}$
Demo & explanation
You can use this REGEX
REGEX Demo
([a-z]{1}[0-9a-z_.-]{2,9})
, Tried text
username77
usr
username10
user_test
usr.1000
There are many ways of doing this. I believe the common characteristic is they will all have a positive lookahead. Here is another.
^(?=.{3,10}$)[a-z\d_.-]*[a-z][a-z\d_.-]*$
Demo
Notice that [a-z\d_.-]* appears twice. Some regex engines support subroutines (or subexpressions) that allow one to save a repeated part of the regex to a numbered or named capture group for reuse later in the string. When using the PCRE engine, for example, you could write
^(?=.{3,10}$)([a-z\d_.-]*)[a-z](?1)$
Demo
(?1) is replaced by the regex tokens that matched the string saved to capture group 1 ([a-z\d_.-]*), as contrasted with \1, which references the content of capture group 1. The use of subroutines can shorten the regex expression, but more importantly it reduces the chance of errors when changes are made to the regex's tokens that are repeated.
Related
I am trying to make a regex match which is discarding the lookahead completely.
\w+([-+.]\w+)*#\w+([-.]\w+)*\.\w+([-.]\w+)*
This is the match and this is my regex101 test.
But when an email starts with - or _ or . it should not match it completely, not just remove the initial symbols. Any ideas are welcome, I've been searching for the past half an hour, but can't figure out how to drop the entire email when it starts with those symbols.
You can use the word boundary near # with a negative lookbehind to check if we are at the beginning of a string or right after a whitespace, then check if the 1st symbol is not inside the unwanted class [^\s\-_.]:
(?<=^|\s)[^\s\-_.]\w*(?:[-+.]\w+)*\b#\w+(?:[-.]\w+)*\.\w+(?:[-.]\w+)*
See demo
List of matches:
support#github.com
s.miller#mit.edu
j.hopking#york.ac.uk
steve.parker#soft.de
info#company-hotels.org
kiki#hotmail.co.uk
no-reply#github.com
s.peterson#mail.uu.net
info-bg#software-software.software.academy
Additional notes on usage and alternative notation
Note that it is best practice to use as few escaped chars as possible in the regex, so, the [^\s\-_.] can be written as [^\s_.-], with the hyphen at the end of the character class still denoting a literal hyphen, not a range. Also, if you plan to use the pattern in other regex engines, you might find difficulties with the alternation in the lookbehind, and then you can replace (?<=\s|^) with the equivalent (?<!\S). See this regex:
(?<!\S)[^\s_.-]\w*(?:[-+.]\w+)*\b#\w+(?:[-.]\w+)*\.\w+(?:[-.]\w+)*
And last but not least, if you need to use it in JavaScript or other languages not supporting lookarounds, replace the (?<!\S)/(?<=\s|^) with a (non)capturing group (\s|^), wrap the whole email pattern part with another set of capturing parentheses and use the language means to grab Group 1 contents:
(\s|^)([^\s_.-]\w*(?:[-+.]\w+)*\b#\w+(?:[-.]\w+)*\.\w+(?:[-.]\w+)*)
See the regex demo.
I use this for multiple email addresses, separate with ‘;':
([A-Za-z0-9._%-]+#[A-Za-z0-9.-]+\.[A-Za-z]{2,4};)*
For a single mail:
[A-Za-z0-9._%-]+#[A-Za-z0-9.-]+\.[A-Za-z]{2,4}
I'm trying to find a regular expression for a Tokenizer operator in Rapidminer.
Now, what I'm trying to do is to split text in parts of, let's say, two words.
For example, That was a good movie. should result to That was, was a, a good, good movie.
What's special about a regex in a tokenizer is that it plays the role of a delimiter, so you match the splitting point and not what you're trying to keep.
Thus the first thought is to use \s in order to split on white spaces, but that would result in getting each word separately.
So, my question is how could I force the expression to somehow skip one in two whitespaces?
First of all, we can use the \W for identifying the characters that separate the words. And for removing multiple consecutive instances of them, we will use:
\W+
Having that in mind, you want to split every 2 instances of characters that are included in the "\W+" expression. Thus, the result must be strings that have the following form:
<a "word"> <separators that are matched by the pattern "\W+"> <another "word">
This means that each token you get from the split you are asking for will have to be further split using the pattern "\W+", in order to obtain the 2 "words" that form it.
For doing the first split you can try this formula:
\w+\W+\w+\K\W+
Then, for each token you have to tokenize it again using:
\W+
For getting tokens of 3 "words", you can use the following pattern for the initial split:
\w+\W+\w+\W+\w+\K\W+
This approach makes use of the \K feature that removes from the match everything that has been captured from the regex up to that point, and starts a new match that will be returned. So essentially, we do: match a word, match separators, match another word, forget everything, match separators and return only those.
In RapidMiner, this can be implemented with 2 consecutive regex tokenizers, the first with the above formula and the second with only the separators to be used within each token (\W+).
Also note that, the pattern \w selects only Latin characters, so if your documents contain text in a different character set, these characters will be consumed by the \W which is supposed to match the separators. If you want to capture text with non-Latin character sets, like Greek for example, you need to change the formula like this:
\p{L}+\P{L}+\p{L}+\K\P{L}+
Furthermore, if you want the formula to capture text on one language and not on another language, you can modify it accordingly, by specifying {Language_Identifier} in place of {L}. For example, if you only want to capture text in Greek, you will use "{Greek}", or "{InGreek}" which is what RapidMiner supports.
What you can do is use a zero width group (like a positive look-ahead, as shown in example). Regex usually "consumes" characters it checks, but with a positive lookahead/lookbehind, you assert that characters exist without preventing further checks from checking those letters too.
This should work for your purposes:
(\w+)(?=(\W+\w+))
The following pattern matches for each pair of two words (note that it won't match the last word since it does not have a pair). The first word is in the first capture group, (\w+). Then a positive lookahead includes a match for a sequence of non word characters \W+ and then another string of word characters \w+. The lookahead (?=...) the second word is not "consumed".
Here is a link to a demo on Regex101
Note that for each match, each word is in its own capture group (group 1, group 2)
Here is an example solution, (?=(\b[A-Za-z]+\s[A-Za-z]+)) inspired from this SO question.
My question sounds wrong once you understand that is a problem of an overlapping regex pattern.
I hav a list of strings, such as: Ø20X400
I need to extract the first of the numbers - between Ø and X
I've come so far to match the numbers in general with \d+ - as simple as it is...
But I need an expression to get the first value separated, not both of them...
You can use lookarounds (?<=..) and (?=..):
(?<=Ø)\d+(?=X)
or in Java style:
(?<=Ø)\\d+(?=X)
A second way is to use a capture group:
Ø(\d+)X
or
Ø(\\d+)X
Then you can extract the content of the group.
The regex engines I know parse \n as a newline. \d is used for numbers.
The following regex gives you the first number between a Ø and a X in a capture group:
^.*?Ø(\d+)X.*
Edit live on Debuggex
This Regex will do it for you, (\d+?)X, and here is a Rubular to prove it. See, you want to group digits together, but make it non-greedy, ending the evaluation on X.
Try this one:
\d+(?=\D)
Should find first number wich has some not a number ahead
With normal regular expressions, I would say:
Ø(\d+)X
This finds the Ø character, followed by one or more numbers, followed by an X. Also, the numbers will be stored in the first capture group. Capture groups differ from one regex implementation to another, but this would typically be denoted by \1. Capture group zero, \0, is usually the matched string itself. In this version, \d denotes digits 0-9, but if your regex engine uses \n for that purpose, use:
Ø(\n+)X
I am implementing the following problem in ruby.
Here's the pattern that I want :
1234, 1324, 1432, 1423, 2341 and so on
i.e. the digits in the four digit number should be between [1-4] and should also be non-repetitive.
to make you understand in a simple manner I take a two digit pattern
and the solution should be :
12, 21
i.e. the digits should be either 1 or 2 and should be non-repetitive.
To make sure that they are non-repetitive I want to use $1 for the condition for my second digit but its not working.
Please help me out and thanks in advance.
You can use this (see on rubular.com):
^(?=[1-4]{4}$)(?!.*(.).*\1).*$
The first assertion ensures that it's ^[1-4]{4}$, the second assertion is a negative lookahead that ensures that you can't match .*(.).*\1, i.e. a repeated character. The first assertion is "cheaper", so you want to do that first.
References
regular-expressions.info/Lookarounds and Backreferences
Related questions
How does the regular expression (?<=#)[^#]+(?=#) work?
Just for a giggle, here's another option:
^(?:1()|2()|3()|4()){4}\1\2\3\4$
As each unique character is consumed, the capturing group following it captures an empty string. The backreferences also try to match empty strings, so if one of them doesn't succeed, it can only mean the associated group didn't participate in the match. And that will only happen if string contains at least one duplicate.
This behavior of empty capturing groups and backreferences is not officially supported in any regex flavor, so caveat emptor. But it works in most of them, including Ruby.
I think this solution is a bit simpler
^(?:([1-4])(?!.*\1)){4}$
See it here on Rubular
^ # matches the start of the string
(?: # open a non capturing group
([1-4]) # The characters that are allowed the found char is captured in group 1
(?!.*\1) # That character is matched only if it does not occur once more
){4} # Defines the amount of characters
$
(?!.*\1) is a lookahead assertion, to ensure the character is not repeated.
^ and $ are anchors to match the start and the end of the string.
While the previous answers solve the problem, they aren't as generic as they could be, and don't allow for repetitions in the initial string. For example, {a,a,b,b,c,c}. After asking a similar question on Perl Monks, the following solution was given by Eily:
^(?:(?!\1)a()|(?!\2)a()|(?!\3)b()|(?!\4)b()|(?!\5)c()|(?!\6)c()){6}$
Similarly, this works for longer "symbols" in a string, and for variable length symbols too.
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 5 years ago.
I'm attempting something which I feel should be fairly obvious to me but it's not. I'm trying to match a string which does NOT contain a specific sequence of characters. I've tried using [^ab], [^(ab)], etc. to match strings containing no 'a's or 'b's, or only 'a's or only 'b's or 'ba' but not match on 'ab'. The examples I gave won't match 'ab' it's true but they also won't match 'a' alone and I need them to. Is there some simple way to do this?
Using a character class such as [^ab] will match a single character that is not within the set of characters. (With the ^ being the negating part).
To match a string which does not contain the multi-character sequence ab, you want to use a negative lookahead:
^(?:(?!ab).)+$
And the above expression disected in regex comment mode is:
(?x) # enable regex comment mode
^ # match start of line/string
(?: # begin non-capturing group
(?! # begin negative lookahead
ab # literal text sequence ab
) # end negative lookahead
. # any single character
) # end non-capturing group
+ # repeat previous match one or more times
$ # match end of line/string
Use negative lookahead:
^(?!.*ab).*$
UPDATE: In the comments below, I stated that this approach is slower than the one given in Peter's answer. I've run some tests since then, and found that it's really slightly faster. However, the reason to prefer this technique over the other is not speed, but simplicity.
The other technique, described here as a tempered greedy token, is suitable for more complex problems, like matching delimited text where the delimiters consist of multiple characters (like HTML, as Luke commented below). For the problem described in the question, it's overkill.
For anyone who's interested, I tested with a large chunk of Lorem Ipsum text, counting the number of lines that don't contain the word "quo". These are the regexes I used:
(?m)^(?!.*\bquo\b).+$
(?m)^(?:(?!\bquo\b).)+$
Whether I search for matches in the whole text, or break it up into lines and match them individually, the anchored lookahead consistently outperforms the floating one.
Yes its called negative lookahead. It goes like this - (?!regex here). So abc(?!def) will match abc not followed by def. So it'll match abce, abc, abck, etc.
Similarly there is positive lookahead - (?=regex here). So abc(?=def) will match abc followed by def.
There are also negative and positive lookbehind - (?<!regex here) and (?<=regex here) respectively
One point to note is that the negative lookahead is zero-width. That is, it does not count as having taken any space.
So it may look like a(?=b)c will match "abc" but it won't. It will match 'a', then the positive lookahead with 'b' but it won't move forward into the string. Then it will try to match the 'c' with 'b' which won't work. Similarly ^a(?=b)b$ will match 'ab' and not 'abb' because the lookarounds are zero-width (in most regex implementations).
More information on this page
abc(?!def) will match abc not followed
by def. So it'll match abce, abc,
abck, etc. what if I want neither def
nor xyz will it be abc(?!(def)(xyz))
???
I had the same question and found a solution:
abc(?:(?!def))(?:(?!xyz))
These non-counting groups are combined by "AND", so it this should do the trick. Hope it helps.
Using a regex as you described is the simple way (as far as I am aware). If you want a range you could use [^a-f].
Simplest way is to pull the negation out of the regular expression entirely:
if (!userName.matches("^([Ss]ys)?admin$")) { ... }
Just search for "ab" in the string then negate the result:
!/ab/.test("bamboo"); // true
!/ab/.test("baobab"); // false
It seems easier and should be faster too.
In this case I might just simply avoid regular expressions altogether and go with something like:
if (StringToTest.IndexOf("ab") < 0)
//do stuff
This is likely also going to be much faster (a quick test vs regexes above showed this method to take about 25% of the time of the regex method). In general, if I know the exact string I'm looking for, I've found regexes are overkill. Since you know you don't want "ab", it's a simple matter to test if the string contains that string, without using regex.
The regex [^ab] will match for example 'ab ab ab ab' but not 'ab', because it will match on the string ' a' or 'b '.
What language/scenario do you have? Can you subtract results from the original set, and just match ab?
If you are using GNU grep, and are parsing input, use the '-v' flag to invert your results, returning all non-matches. Other regex tools also have a 'return nonmatch' function, too.
If I understand correctly, you want everything except for those items which contain 'ab' anywhere.