I use Notepad++, and I need a regular expression to match consecutive numbers only
Example:
verses 3-4
verses 11-12
verses 26-27
so the regex finds these matches, and not lines like: verses 3-9, or verses 26-32.. etc.
I use the \d+, but don't know how to increase the same one by just one.
Regular expressions match text, not numbers. Therefore, you can't do this with a regex alone (unless you spell out all possible combinations). You need a scripting language that converts the matched texts to integers and compares those.
For example, in Python:
for potentialmatch in re.finditer(r"(\d+)-(\d+)", mytext):
if int(potentialmatch.group(1)) + 1 == int(potentialmatch.group(2)):
# Match found
You cannot do that, short of using a regex that encompasses all such options.
You can generate one, though:
(1..99 | %{"$_-$($_+1)"}) -join '|'
in PowerShell yields a regex that will match everything from 1-2 to 99-100.
Related
I have data like this:
~10~682423~15~Test Data~10~68276127~15~More Data~10~6813~15~Also Data~
I'm trying to use Notepad++ to find and replace the values within tag 10 (682423, 68276127, 6813) with zeroes. I thought the syntax below would work, but it selects the first occurrence of the text I want and the rest of the line, instead of just the text I want (~10~682423~, for example). I also tried dozens of variations from searching online, but they also either did the same thing or wouldn't return any results.
~10~.*~
You can use: (?<=~10~)\d+(?=~) and replace with 0. This uses lookarounds to check that ~10~ precedes the digit sequence and the (?=~) ensures a ~ follows the digit sequence. If any character could be after the ~10~ field, use (?<=~10~)[^~]+(?=~).
The problem with ~10~.*~ is that the * is greedy, so it just slurps away matching any character and ~.
Use
\b10~\d+
Replace with 10~0. See proof. \b10~ will capture 10 as entire number (no match in 210 is allowed) and \d+ will match one or more digits.
Sample input:
___file___name___2000___ed2___1___2___3
DIFFERENT+FILENAME+(2000)+1+2+3+ed10
Desired output (eg, all letters and 4-digit numbers and literal 'ed' followed immediately by a digit of arbitrary length:
file name 2000 ed2
DIFFERENT FILENAME 2000 ed10
I am using:
[A-Za-z]+|[\d]{4}|ed\d+ which only returns:
file name 2000 ed
DIFFERENT FILENAME 2000 ed
I see that there is a related Q+A here:Regular Expression to match specific string followed by number?
eg using ed[0-9]* would match ed#, but unsure why it does not match in the above.
As written, your regex is correct. Remember, however, that regex tries to match its statements from left to right. Your ed\d+ is never going to match, because the ed was already consumed by your [A-Za-z] alternative. Reorder your regex and it'll work just fine:
ed\d+|[a-zA-Z]+|\d{4}
Demo
Nick's answer is right, but because in-order matching can be a less readable "gotcha", the best (order-insensitive) ways to do this kind of search are 1) with specified delimiters, and 2) by making each search term unique.
Jan's answer handles #1 well. But you would have to specify each specific delimiter, including its length (e.g. ___). It sounds like you may have some unusual delimiters, so this may not be ideal.
For #2, then, you can make each search term unique. (That is, you want the thing matching "file" and "name" to be distinct from the thing matching "2000", and both to be distinct from the thing matching "ed2".)
One way to do this is [A-Za-z]+(?![0-9a-zA-Z])|[\d]{4}|ed\d+. This is saying that for the first type of search term, you want an alphabet string which is followed by a non-alphanumeric character. This keeps it distinct from the third search term, which is an alphabet string followed by some digit(s). This also allows you to specify any range of delimiters inside of that negative lookbehind.
demo
You might very well use (just grab the first capturing group):
(?:^|___|[+(]) # delimiter before
([a-zA-Z0-9]{2,}) # the actual content
(?=$|___|[+)]) # delimiter afterwards
See a demo on regex101.com
I am re-phrasing my question to clear confusions!
I want to match if a string has certain letters for this I use the character class:
[ACD]
and it works perfectly!
but I want to match if the string has those letter(s) 2 or more times either repeated or 2 separate letters
For example:
[AKL] should match:
ABCVL
AAGHF
KKUI
AKL
But the above should not match the following:
ABCD
KHID
LOVE
because those are there but only once!
that's why I was trying to use:
[ACD]{2,}
But it's not working, probably it's not the right Regex.. can somebody a Regex guru can help me solve this puzzle?
Thanks
PS: I will use it on MYSQL - a differnt approach can also welcome! but I like to use regex for smarter and shorter query!
To ensure that a string contains at least two occurencies in a set of letters (lets say A K L as in your example), you can write something like this:
[AKL].*[AKL]
Since the MySQL regex engine is a DFA, there is no need to use a negated character class like [^AKL] in place of the dot to avoid backtracking, or a lazy quantifier that is not supported at all.
example:
SELECT 'KKUI' REGEXP '[AKL].*[AKL]';
will return 1
You can follow this link that speaks on the particular subject of the LIKE and the REGEXP features in MySQL.
If I understood you correctly, this is quite simple:
[A-Z].*?[A-Z]
This looks for your something in your set, [A-Z], and then lazily matches characters until it (potentially) comes across the set, [A-Z], again.
As #Enigmadan pointed out, a lazy match is not necessary here: [A-Z].*[A-Z]
The expression you are using searches for characters between 2 and unlimited times with these characters ACDFGHIJKMNOPQRSTUVWXZ.
However, your RegEx expression is excluding Y (UVWXZ])) therefore Z cannot be found since it is not surrounded by another character in your expression and the same principle applies to B ([ACD) also excluded in you RegEx expression. For example Z and A would match in an expression like ZABCDEFGHIJKLMNOPQRSTUVWXYZA
If those were not excluded on purpose probably better can be to use ranges like [A-Z]
If you want 2 or more of a match on [AKL], then you may use just [AKL] and may have match >= 2.
I am not good at SQL regex, but may be something like this?
check (dbo.RegexMatch( ['ABCVL'], '[AKL]' ) >= 2)
To put it in simple English, use [AKL] as your regex, and check the match on the string to be greater than 2. Here's how I would do in Java:
private boolean search2orMore(String string) {
Matcher matcher = Pattern.compile("[ACD]").matcher(string);
int counter = 0;
while (matcher.find())
{
counter++;
}
return (counter >= 2);
}
You can't use [ACD]{2,} because it always wants to match 2 or more of each characters and will fail if you have 2 or more matching single characters.
your question is not very clear, but here is my trial pattern
\b(\S*[AKL]\S*[AKL]\S*)\b
Demo
pretty sure this should work in any case
(?<l>[^AKL\n]*[AKL]+[^AKL\n]*[AKL]+[^AKL\n]*)[\n\r]
replace AKL for letters you need can be done very easily dynamicly tell me if you need it
Is this what you are looking for?
".*(.*[AKL].*){2,}.*" (without quotes)
It matches if there are at least two occurences of your charactes sorrounded by anything.
It is .NET regex, but should be same for anything else
Edit
Overall, MySQL regular expression support is pretty weak.
If you only need to match your capture group a minimum of two times, then you can simply use:
select * from ... where ... regexp('([ACD].*){2,}') #could be `2,` or just `2`
If you need to match your capture group more than two times, then just change the number:
select * from ... where ... regexp('([ACD].*){3}')
#This number should match the number of matches you need
If you needed a minimum of 7 matches and you were using your previous capture group [ACDF-KM-XZ]
e.g.
select * from ... where ... regexp('([ACDF-KM-XZ].*){7,}')
Response before edit:
Your regex is trying to find at least two characters from the set[ACDFGHIJKMNOPQRSTUVWXZ].
([ACDFGHIJKMNOPQRSTUVWXZ]){2,}
The reason A and Z are not being matched in your example string (ABCDEFGHIJKLMNOPQRSTUVWXYZ) is because you are looking for two or more characters that are together that match your set. A is a single character followed by a character that does not match your set. Thus, A is not matched.
Similarly, Z is a single character preceded by a character that does not match your set. Thus, Z is not matched.
The bolded characters below do not match your set
ABCDEFGHIJKLMNOPQRSTUVWXYZ
If you were to do a global search in the string, only the italicized characters would be matched:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Given a series of bits, e.g 10100011110010101 and 010001001010100.
How can I write a regular expression to reflect the patterns where no 1s are neighbors?
For example, 010001001010100 is a good one, but 10100011110010101 is not as there are 1s neighboring.
edit:
Sorry my above statement may be misleading. i don't want to check whether a sequence is non-1-neighboring. I want to use an regex to find all non-1-neighborings.
Assume I have a very lone series of bits, I want to write a regex to find out all sub sequences where no 1s are neighbors.
Simply "11" will return true if neighboring ones exist.
Regards
You can try with following regex:
^0*(10+)*1?$
The following regex will match any run of zeros where any embedded ones are not adjacent to another one. A leading or trailing one at beginning/end of string is accepted as well.
(^1)?0+(10+)*(1$)?
A test with your example strings yields:
bash$ grep -Eo '(^1)?0+(10+)*(1$)?' <<<10100011110010101
101000
0010101
bash$ grep -Eo '(^1)?0+(10+)*(1$)?' <<<010001001010100
010001001010100
Search for 11+, i.e. a 1 followed by at least one 1.
You can use this, if your regex flavor supports lookarounds:
(?<!1)1(?!1)
(?<!1): not preceded by 1
(?!1): not followed by 1
If your regex flavor doesn't support lookarounds, you can use a capturing group and 2 non capturing groups instead of:
(?:^|0)(1)(?:0|$)
(Note that the capturing group is usefull only if you want to catch the offset of the capture with an appropriate function)
What do I use to search for multiple words in a string? I would like the logical operation to be AND so that all the words are in the string somewhere. I have a bunch of nonsense paragraphs and one plain English paragraph, and I'd like to narrow it down by specifying a couple common words like, "the" and "and", but would like it match all words I specify.
Regular expressions support a "lookaround" condition that lets you search for a term within a string and then forget the location of the result; starting at the beginning of the string for the next search term. This will allow searching a string for a group of words in any order.
The regular expression for this is:
^(?=.*\bword1\b)(?=.*\bword2\b)(?=.*\bword3\b)
Where \b is a word boundary and the ?= is the lookaround modifier.
If you have a variable number of words you want to search for, you will need to build this regular expression string with a loop - just wrap each word in the lookaround syntax and append it to the expression.
AND as concatenation
^(?=.*?\b(?:word1)\b)(?=.*?\b(?:word2)\b)(?=.*?\b(?:word3)\b)
OR as alternation
^(?=.*?\b(?:word1|word2|word3)\b
^(?=.*?\b(?:word1)\b)|^(?=.*?\b(?:word2)\b)|^(?=.*?\b(?:word3)\b)
Maybe using a language recognition chart to recognize english would work. Some quick tests seem to work (this assumes paragraphs separated by newlines only).
The regexp will match one of any of those conditions... \bword\b is word separated by boundaries word\b is a word ending and just word will match it in any place of the paragraph to be matched.
my #paragraphs = split(/\n/,$text);
for my $p (#paragraphs) {
if ($p =~ m/\bthe\b|\band\b|\ban\b|\bin\b|\bon\b|\bthat\b|\bis\b|\bare\b|th|sh|ough|augh|ing\b|tion\b|ed\b|age\b|’s\b|’ve\b|n’t\b|’d\b/) {
print "Probable english\n$p\n";
}
}
Firstly I'm not certain what you're trying to return... the whole sentence? The words in between your two given words?
Something like:
\b(word1|word2)\b(\w+\b)*(word1|word2)\b(\w+\b)*\.
(where \b is the word boundary in your language)
would match a complete sentence that contained either of the two words or both..
You'd probably need to make it case insensitive so that if it appears at the start of the sentence it will still match
Assuming PCRE (Perl regexes), I am not sure that you can do it at all easily. The AND operation is concatenation of regexes, but you want to be able to permute the order in which the words appear without having to formally generate the permutation. For N words, when N = 2, it is bearable; with N = 3, it is barely OK; with N > 3, it is unlikely to be acceptable. So, the simple iterative solution - N regexes, one for each word, and iterate ensuring each is satisfied - looks like the best choice to me.