I need to grab the 2 character letter and numbers after that:
What I want is:
AB 12 CD-12345-67 -> CD-12345
AB12 CD 12345-67 -> CD 12345
AB-12CD12345-6 -> CD12345
ABC1234556 -> no match, as I look for 2 character letter and numbers after that.
ABC-1234556 -> no match, as I look for 2 character letter and numbers after that.
A1-BC-12D345-56 -> no match, after 2 characters letter, numbers must come
I used this regex
[A-Z]{2}[ |\-]?\d+
Which grabs CD-12345 and AB 12, in the first example. I just need CD-12345. Also it grabs BC1234556,BC-1234556, BC-12 in the last three example which i don't want.
Sometimes, space,no space or - character placed between numbers and letters block.
Thank you very much.
based on what you posted
^.*(?<![A-Z])([A-Z]{2}[- ]?\d++)(?![A-Z])
Demo
Related
I am facing some issues forming a regex that matches at least n times a given pattern within m characters of the input string.
For example imagine that my input string is:
00000001100000001110111100000000000000000000000000000000000000000000000000110000000111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100
I want to detect all cases where an 1 appears at least 7 times (not necessarily consecutively) in the input string, but within a window of up to 20 characters.
So far I have built this expression:
(1[^1]*?){7,}
which detects all cases where an 1 appears at least 7 times in the input string, but this now matches both the:
11000000011101111
and the
1100000001110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011
parts whereas I want only the first one to be kept, as it is within a substring composed of less than 20 characters.
It tried to combine the aforementioned regex with:
(?=(^[01]{0,20}))
to also match only parts of the string containing either an '1' or a '0' of length up to 20 characters but when I do that it stops working.
Does anyone have an idea gow to accomplish this?
I have put this example in regex101 as a quick reference.
Thank you very much!
This is not something that can be done with regex without listing out every possible string. You would need to iterate over the string instead.
You could also iterate over the matches. Example in Python:
import re
matches = re.finditer(r'(?=((1[^1]*?){7}))', string)
matches = [match.group(1) for match in matches if len(match.group(1)) <= 20]
The next Python snippet is an attempt to get the desired sequences using only the regular expression.
import re
r = r'''
(?mx)
( # the 1st capturing group will contain the desired sequence
1 # this sequence should begin with 1
(?=(?:[01]{6,19}) # let's see that there are enough 0s and 1s in a line
(.*$)) # the 2nd capturing group will contain all characters to the end of a line
(?:0*1){6}) # there must be six more 1s in the sequence
(?=.{0,13} # complement the 1st capturing group to 20 characters
\2) # the rest of a line should be 2nd capturing group
'''
s = '''
0000000
101010101010111111100000000000001
00000001100000001110111100000000000000000000000000000000000000000000000000110000000111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100
1111111
111111
'''
print([m.group(1) for m in re.finditer(r, s)])
Output:
['1010101010101', '11111100000000000001', '110000000111011', '1111111']
You can find an exhaustive explanation of this regular expression on RegEx101.
I have a list of different items. Some of them have 8-10 digits in front of the name, some others have these 8-10 digits behind the name and some others again don't have these numbers in the name.
I have two expressions that I use to remove these digits, but I can not manage to combine them with | (or). They work each for themselves, but if I use the first expression first, then the second expression, I don't get the result I want to have.
I use these to expressions for now:
(?<=[\d]{8,10}) (.*)
.*?(?=[\d]{8,10})
But if I use them both (first one and then the other), then some of the lines become totally empty.
How can I combine these to to do what I want, or if it's better, write a new expression that does what I want to do :)
List is like this:
12345678 Book
12345678 Book
Book 12345678
Book 12345678
Cabinet 120x30x145
Want this result:
Book
Book
Book
Book
Cabinet 120x30x145
Why not just use the following.
Check if there are 8 numbers in the beginning of the string, or at the end of it and remove them.
(^\d{8,10}\s*|\s*\d{8,10}$)
It gives the wanted behaviour
Instead of only matching everything but a number containing
8-10 digits + adjacent spaces, use a regex to substitute
such a number (also + adjacent spaces) with an empty string.
To match, use the following regex:
*\d{8,10} *
That is:
* - a space and an asterix - a sequence of spaces (may be empty),
\d{8,10} - a sequence of 8 to 10 digits,
* - another sequence of spaces (may be empty).
The replacement string is (as I said) empty. Of course, you should use
g (global) option.
Note that you can not use \s instead of the space, as \s matches also
CR and LF and we don't want this.
For a working example see https://regex101.com/r/1hsGzT/1
You need to use \b meta sequence boundary:
/\b[0-9\s]{8,10}\b/g;
var str = `12345678 Book
12345678 Book
Book 12345678
Book 12345678
Cabinet 120x30x145`;
var rgx = /\b[0-9\s]{8,10}\b/g;
var res = str.replace(rgx, `\n`)
console.log(res);
I have the following string:
"......(some chars) aaa bbb ###8/13/2018 ......(some chars)"
The ### in the string represent some random characters. ###'s length is unknown and it could be None (just "aaa bbb 8/13/2018").
My goal is to find the date from the string (8/13/2018) and the starting index of ###.
I currently used the following code:
m = re.search(r'\s.*?([0-9]{1,}/[0-9]{1,}/[0-9]{2,})', str)
m.groups()[0] ## The date
m.start() ## index of ###
But the regex is matching bbb ###8/13/2018 instead of ###8/13/2018
I also tried change the regex to:
r'\s(?!\s).*?[0-9]{1,}/[0-9]{1,}/[0-9]{2,}'
r'\s(?!\s)*?[0-9]{1,}/[0-9]{1,}/[0-9]{2,}'
But neither of them works.
I will be appreciated for any help or comments. Thank you.
I tend to believe you are looking for:
#*(?:\d{1,2}/){2}\d{2,4} or even \S*(?:\d{1,2}/){2}\d{2,4}
This is simply saying:
\S* start with 0 or more non-space charaters.
(?:\d{1,2}/){2} find two groups of \d{1,2}/ but do not capture them. ie not capturing: (?:..).this will match the month and date part 8/13/. \d{1,2} means atleast one digit and atmost two digits
\d{2,4} match the year .Atleast 2 digits and atmost 4 digits
Using a part of your regex, I think you mean something like this
r'\S*([0-9]+/[0-9]+/[0-9]{2,})'
https://regex101.com/r/dxF4sT/1
To find the starting index, it would be where the match was found.
Note that \S will find all consecutive non-whitespace.
You can change this to other things like [#a-zA-Z] etc..., just add it to the class.
I need a RegEx pattern that will return the first N words using a custom word boundary that is the normal RegEx white space (\s) plus punctuation like .,;:!?-*_
EDIT #1: Thanks for all your comments.
To be clear:
I'd like to set the characters that would be the word delimiters
Lets call this the "Delimiter Set", or strDelimiters
strDelimiters = ".,;:!?-*_"
nNumWordsToFind = 5
A word is defined as any contiguous text that does NOT contain any character in strDelimiters
The RegEx word boundary is any contiguous text that contains one or more of the characters in strDelimiters
I'd like to build the RegEx pattern to get/return the first nNumWordsToFind using the strDelimiters.
EDIT #2: Sat, Aug 8, 2015 at 12:49 AM US CT
#maraca definitely answered my question as originally stated.
But what I actually need is to return the number of words ≤ nNumWordsToFind.
So if the source text has only 3 words, but my RegEx asks for 4 words, I need it to return the 3 words. The answer provided by maraca fails if nNumWordsToFind > number of actual words in the source text.
For example:
one,two;three-four_five.six:seven eight nine! ten
It would see this as 10 words.
If I want the first 5 words, it would return:
one,two;three-four_five.
I have this pattern using the normal \s whitespace, which works, but NOT exactly what I need:
([\w]+\s+){<NumWordsOut>}
where <NumWordsOut> is the number of words to return.
I have also found this word boundary pattern, but I don't know how to use it:
a "real word boundary" that detects the edge between an ASCII letter
and a non-letter.
(?i)(?<=^|[^a-z])(?=[a-z])|(?<=[a-z])(?=$|[^a-z])
However, I would want my words to allow numbers as well.
IAC, I have not been able how to use the above custom word boundary pattern to return the first N words of my text.
BTW, I will be using this in a Keyboard Maestro macro.
Can anyone help?
TIA.
All you have to do is to adapt your pattern ([\w]+\s+){<NumWordsOut>} to, including some special cases:
^[\s.,;:!?*_-]*([^\s.,;:!?*_-]+([\s.,;:!?*_-]+|$)){<NumWordsOut>}
1. 2. 3. 4. 5.
Match any amount of delimiters before the first word
Match a word (= at least one non-delimiter)
The word has to be followed by at least one delimiter
Or it can be at the end of the string (in case no delimiter follows at the end)
Repeat 2. to 4. <NumWordsOut> times
Note how I changed the order of the -, it has to be at the start or end, otherwise it needs to be escaped: \-.
Thanks to #maraca for providing the complete answer to my question.
I just wanted to post the Keyboard Maestro macro that I have built using #maraca's RegEx pattern for anyone interested in the complete solution.
See KM Forum Macro: Get a Max of N Words in String Using RegEx
In the tester this works ... but not in PostgreSQL.
My data is like this -- usually a series of letters, followed by 2 numbers and a POSSIBLE '-' or 'space' with only ONE letter following. I am trying to isolate the 2 numbers and the Possible '-" or 'space' AND the ONE letter with my regex:
For ex:
AJ 50-R Busboys ## should return 50-R
APPLES 30 F ## should return 30 F
FOOBAR 30 Apple ## should return 30
Regex's (that have worked in the tester, but not in PostgreSQL) that I've tried:
substring(REF from '([0-9]+)-?([:space:])?([A-Za-z])?')
&
substring(REF from '([0-9]+)-?([A-Za-z])?')
So far everything tests out in the tester...but not the PostgreSQL. I just keep getting the numbers returns -- AND NOTHING AFTER IT.
What I am getting now(for ex):
AJ 50-R Busboys ## returns as "50" NOT as "50-R"
Your looking for: substring(REF from '([0-9]+(-| )([A-Za-z]\y)?)')
In SQLFiddle. Your primary problem is that substring returns the first or outermost matching group (ie., pattern surrounded with ()), which is why you get 50 for your '50-R'. If you were to surround the entire pattern with (), this would give you '50-R'. However, the pattern you have fails to return what you want on the other strings, even after accounting for this issue, so I had to modify the entire regex.
This matches your description and examples.
Your description is slightly ambiguous. Leading letters are followed by a space and then two digits in your examples, as opposed to your description.
SELECT t, substring(t, '^[[:alpha:] ]+(\d\d(:?[\s-]?[[:alpha:]]\M)?)')
FROM (
VALUES
('AJ 50-R Busboys') -- should return: 50-R
,('APPLES 30 F') -- should return: 30 F
,('FOOBAR 30 Apple') -- should return: 30
,('FOOBAR 30x Apple') -- should return: 30x
,('sadfgag30 D 66 X foo') -- should return: 30 D - not: 66 X
) r(t);
->SQLfiddle
Explanation
^ .. start of string (last row could fail without anchoring to start and global flag 'g'). Also: faster.
[[:alpha:] ]+ .. one or more letters or spaces (like in your examples).
( .. capturing parenthesis
\d\d .. two digits
(:? .. non-capturing parenthesis
[\s-]? .. '-' or 'white space' (character class), 0 or 1 times
[[:alpha:]] .. 1 letter
\M .. followed by end of word (can be end of string, too)
)? .. the pattern in non-capturing parentheses 0 or 1 times
Letters as defined by the character class alpha according to the current locale! The poor man's substitute [a-zA-Z] only works for basic ASCII letters and fails for anything more. Consider this simple demo:
SELECT substring('oö','[[:alpha:]]*')
,substring('oö','[a-zA-Z]*');
More about character classes in Postgres regular expressions in the manual.
It's because of the parentheses.
I've looked everywhere in the documentation and found an interesting sentence on this page:
[...] if the pattern contains any parentheses, the portion of the text that matched the first parenthesized subexpression (the one whose left parenthesis comes first) is returned.
I took your first expression:
([0-9]+)-?([:space:])?([A-Za-z])?
and wrapped it in parentheses:
(([0-9]+)-?([:space:])?([A-Za-z])?)
and it works fine (see SQLFiddle).
Update:
Also, because you're looking for - or space, you could rewrite your middle expression to [-|\s]? (thanks Matthew for pointing that out), which leads to the following possible REGEX:
(([0-9]+)[-|\s]?([A-Za-z])?)
(SQLFiddle)
Update 2:
While my answer provides the explanation as to why the result represented a partial match of your expression, the expression I presented above fails your third test case.
You should use the regex provided by Matthew in his answer.