I am at the beginning of learning Regex, and I use every opportunity to understand how it's working. Currently I am trying to extract dates from a text file (which is in fact a vnt-file type from my mobile phone). It looks like following:
BEGIN:VNOTE
VERSION:1.1
BODY;ENCODING=QUOTED-PRINTABLE;CHARSET=UTF-8:18.07.=0A14.08.=0A15.09.=0A15.10.=
=0A13.11.=0A13.12.=0A12.01.=0A03.02. Grippe=0A06.03.=0A04.04.2015=0A0=
5.05.2015=0A03.06.2015=0A03.07.2015=0A02.08.2015=0A30.08.2015=0A28.09=
17.11.2017=0A
DCREATED:20171118T095601
X-IRMC-LUID:150
END:VNOTE
I want to extract all dates, so that the final list is like that:
18.07.
14.08.
15.09.
15.10.
and so on. If the date has also a year, it should also be displayed.
I almost found out how to detect the dates by the following regex:
.+(\d\d\.\d\d\.(2015|2016|2017)?).+
But it only detect very few of the dates. The result is this:
BEGIN:VNOTE
VERSION:1.1
15.10.
04.04.2015
30.08.2015
24.01.2016
DCREATED:20171118T075601
X-IRMC-LUID:150
END:VNOTE
Then I tried to add a question mark which makes the .+ not greedy, as far as I read in tutorials. Then the regex looks like:
.+?(\d\d\.\d\d\.(2015|2016|2017)?).+?
But the result is still not what I am looking for:
BEGIN:VNOTE
VERSION:1.1
21.03.20.04.18.05.18.06.18.07.14.08.15.09.15.10.
13.11.13.12.12.01.03.02.06.03.04.04.20150A0=
03.06.201503.07.201502.08.201530.08.20150A28.09=
28.10.201525.11.201528.12.201524.01.20160A
DCREATED:20171118T075601
X-IRMC-LUID:150
END:VNOTE
For someone who is familiar with regex I am pretty sure this is very easy to solve, but I don't get it. It's very confusing when you are new to regex. I tried to find a hint in some tutorials or stackoverflow posts, but all I found is this: Notepad++ how to extract only the text field which is needed?
But it doesn't work for me. I assume it might have something to do with the fact that my text file is not one single line.
I have my example on regex101 too.
I would be very thankful if maybe someone can give me a hint what else I can try.
Edit: I would like to detect the dates with the regex and as a result have a list with only the dates (maybe it is called substitute?)
Edit 2: Sorry for not mentioning it earlier: I just want to use the regex in e.g. Notepad++ or an online regex test website. Just to get the result of the dates and save the result in a new txt-file. I don't want to use the regex in an programming language. My apologies for not being precisely before.
Edit 3: The result should be a list with the dates, and each date in a new line:
I want to extract all dates, so that the final list is like that:
18.07.
14.08.
15.09.
15.10.
I suggest this pattern:
(?:.*?|\G)(\d\d\.\d\d\.(?:\d{4})?)
This makes use of the \G flag that, in this case, allows for multiple matches from the very start of the match without letting any single unmatched character in the text, thus allowing the removal of all but what's wanted.
If you want to remove the extra matches as well, add |.* at the end:
(?:.*?|\G)(\d\d\.\d\d\.(?:\d{4})?)|.*
regex101 demo
In N++, make sure the options underlined are selected, and that the cursor is at the beginning. In the picture below, I replaced then undid the replacement, only to show that matches were identified (16 replacements).
You can try using the following pattern:
\d{2}\.\d{2}\.(?:\d{4})?
This will match day.month dates of the form 18.07., but it also allows such a date to be followed by a four digit year, e.g. 18.07.2017. While it would be nice to make the pattern more restrictive, to avoid false fire matches, I do not see anything obvious which can be added to the above pattern. Follow the demo link below to see the pattern in action.
Demo
Related
As you know, Google links can be pretty unwieldy:
https://www.google.com/search?q=some+search+here&source=hp&newwindow=1&ei=A_23ssOllsUx&oq=some+se....
I have MANY Google links saved that I would like to clean up to make them look like so:
https://www.google.com/search?q=some+search+here
The only issue is that I cannot figure out the correct regex pattern for Vim to do this.
I figure it must be something like this:
:%s/&source=[^&].*//
:%s/&source=[^&].*[^&]//
:%s/&source=.*[^&]//
But none of these are working; they start at &source, and replace until the end of the line.
Also, the search?q=some+search+here can appear anywhere after the .com/, so I cannot rely on it being in the same place every time.
So, what is the correct Vim regex pattern to use in order to clean up these links?
Your example can easily be dealt with by using a very simple pattern:
:%s/&.*
because you want to keep everything that comes before the second parameter, which is marked by the first & in the string.
But, if the q parameter can be anywhere in the query string, as in:
https://www.google.com/search?source=hp&newwindow=1&q=some+search+here&ei=A_23ssOllsUx&oq=some+se....
then no amount of capturing or whatnot will be enough to cover every possible case with a single pattern, let alone a readable one. At this point, scripting is really the only reasonable approach, preferably with a language that understands URLs.
--- EDIT ---
Hmm, scratch that. The following seems to work across the board:
:%s#^\(https://www.google.com/search?\)\(.*\)\(q=.\{-}\)&.*#\1\3
We use # as separator because of the many / in a typical URL.
We capture a first group, up to and including the ? that marks the beginning of the query string.
We match whatever comes between the ? and the first occurrence of q= without capturing it.
We capture a second group, the q parameter, up to and excluding the next &.
We replace the whole thing with the first capture group followed by the second capture group.
I'm trying to see if its possible to extend an existing arbitrary regex by prepending or appending another regex to match within matches.
Take the following example:
The original regex is cat|car|bat so matching output is
cat
car
bat
I want to add to this regex and output only matches that start with 'ca',
cat
car
I specifically don't want to interpret a whole regex, which could be quite a long operation and then change its internal content to match produce the output as in:
^ca[tr]
or run the original regex and then the second one over the results. I'm taking the original regex as an argument in python but want to 'prefilter' the matches by adding the additional code.
This is probably a slight abuse of regex, but I'm still interested if it's possible. I have tried what I know of subgroups and the following examples but they're not giving me what I need.
Things I've tried:
^ca(cat|car|bat)
(?<=ca(cat|car|bat))
(?<=^ca(cat|car|bat))
It may not be possible but I'm interested in what any regex gurus think. I'm also interested if there is some way of doing this positionally if the length of the initial output is known.
A slightly more realistic example of the inital query might be [a-z]{4} but if I create (?<=^ca([a-z]{4})) it matches against 6 letter strings starting with ca, not 4 letter.
Thanks for any solutions and/or opinions on it.
EDIT: See solution including #Nick's contribution below. The tool I was testing this with (exrex) seems to have a slight bug that, following the examples given, would create matches 6 characters long.
You were not far off with what you tried, only you don't need a lookbehind, but rather a lookahead assertion, and a parenthesis was misplaced. The right thing is: Put the original pattern in parentheses, and prepend (?=ca):
(?=ca)(cat|car|bat)
(?=ca)([a-z]{4})
In the second example (without | alternative), the parentheses around the original pattern wouldn't be required.
Ok, thanks to #Armali I've come to the conclusion that (?=ca)(^[a-z]{4}$) works (see https://regexr.com/3f4vo). However, I'm trying this with the great exrex tool to attempt to produce matching strings, and it's producing matches that are 6 characters long rather than 4. This may be a limitation of exrex rather than the regex, which seems to work in other cases.
See #Nick's comment.
I've also raised an issue on the exrex GitHub for this.
I have a list of the following numbers and want a Regular expression that matches when a number is not in the list.
0,1,2,3,4,9,11,12,13,14,15,16,18,19,250
I have written the following REGEX statement.
^(?!.*(0|1|2|3|4|9|11|12|13|14|15|16|18|19|250)).*$
The problem is that it correctly gives a match for 5,6,7,8 etc but not for 17 or 251 for example.
I have been testing this on the online REGEX simulators.
This should resolve your issue..
^(?!\D*(0|1|2|3|4|9|11|12|13|14|15|16|18|19|250)\b).*$
In your earlier regex you were basically saying eliminate all numbers that start with 0/1/2/3/4/9!
So your original regex would actually match 54/623/71/88 but not the others. Also the 11-19 and 250 in the list were rendered useless.
Although as others have I would also recommend you to not use regex for this, as I believe it is an overkill and a maintenance nightmare!
Also an extra note "Variable length look arounds are very inefficient too" vs regular checks.
I would do \b\d+\b to get each number in the string and check if they are in your list. It would be way faster.
You can use the discard technique by matching what you do not want and capturing what you really want.
You can use a regex like this:
\b(?:[0-49]|1[1-689]|250)\b|(\d+)
Here you can check a working demo where in blue you have the matches (what you don't want) and in green the content you want. Then you have to grab the content from the capturing group
Working demo
Not sure what regex engine you are using, but here I created a sample using java:
https://ideone.com/B7kLe0
I am trying to extract data from semi structured text, it is an email composed of tab delimited tables. Users have entered time stamp on top of each table and within the table they list security identifiers that I am looking for.
The goal is to extract correctsecurity and the time stamp on top of the table that correctsecurity is located at.
For example...
10:00 AM
not it
not it
9:00 AM
not it
correctsecurity
..is supposed to return 9:00 AM correctsecurity. However my current regex is returning 10:00 AM correctsecurity, meaning right item, but not the right time.
Here is my regex so far:
((1[0-2]|[0-9]):[0-5][0-9](\s?(AM|PM))?)(?:(.*\n)+)(correctsecurity)
Note that the last part correctsecurity is being created dynamically based on other criteria so even if I were to provide the actual item in this question it would do little help(because it is one of many), for simplicity sake please assume that correctsecurity is exactly the item I am looking for.
Lastly I am doing this in VBA so maybe solving this whole problem is easier without using a long regex, so feel free proposed non regex solutions.
To solve the main problem simply change the central section of your regex to not accept empty lines:
.*\n -> .+\n
Then add a newline anchor \n before the central section to avoid the skip of the AM|PM section:
So your regex will be:
((1[0-2]|[0-9]):[0-5][0-9](\s?(AM|PM))?)\n(?:(.+\n)+)(correctsecurity)
^ ^
Changes --------------------------------|------|
Optional Optimization
You may remove many unneeded groups and add a generic multi-os regex for the newline (?:\r\n?|\n):
((?:1[0-2]|[0-9]):[0-5][0-9](?: [AP]M)?)(?:\r\n?|\n)(?:[^\r\n]+(?:\r\n?|\n))+(correctsecurity)
You can solve this with negative lookahead :
a((?!a).)*correctsecurity
Where a is the pattern that you want to start the match at and don't want to encouter in the middle of the match.
Applied to your specif needs :
\d*:\d* [AP]M((?!\d*:\d* [AP]M).)*correctsecurity
Don't forget to let the dot match line breaks.
I assume VBA uses thee VBScript regex dialect which requires the following modification:
\d*:\d* [AP]M((?!\d*:\d* [AP]M)[\s\S])*correctsecurity
Hopefully someone can help me out. Been all over google now.
I'm doing some zone-ocr of documents, and want to extract some text with regex. It is always like this:
"Til: Name Name Name org.nr 12323123".
I want to extract the name-part, it can be 1-4 names, but "Til:" and "org.nr" is always before and after.
Anyone?
If you can't use capturing groups (check your documentation) you can try this:
(?<=Til:).*?(?=org\.nr)
This solution is using look behind and lookahead assertions, but those are not supported from every regex flavour. If they are working, this regex will return only the part you want, because the parts in the assertions are not matched, it checks only if the patterns in the assertions are there.
Use the pattern:
Til:(.*)org\.nr
Then take the second group to get the content between the parenthesis.