In Word, I had to convert my footnotes to lines appearing at the end of each file to able to make changes in formatting. Some macro I found online was using braces and I ended up using also highlighting so I can see easily where my footnotes used to be. In this way, I have the following strings twice in my documents in the main text and also at the end of each document, sort of like makeshift endnotes.
=={1}==
.
.
.
=={99}==
I want to be able to match those instances in the text and convert them to proper markdown now. The problem is that the in-text format
[^1], [^2], etc.
will be different from what needs to come at the bottom with a semi-colon added:
[^1]:
etc.
So I'm guessing I'll have to live with replacing my old formatting with the new ones with semi-colons and deleting the semi-colons individually while I edit/clean up my text in the future. Without adding the semi-colon, it won't work.
My question is how to use the regex to match the two-digit strings with braces and equation marks.
This
==(\{d{1,2}\})==
did not work.
Also, as I am no pro, I would need the replacement as well. It probably will be
[^($1)]:
I reckon. Apparently, the equal mark doesn't have to be escaped.
Current format:
...some text...makeshift footnote in the format of
=={one- or two-digit number with no spaces in between}==
For example,
=={1}==
=={23}==
etc.
Desired result for all occurences recursively:
[^1]:
.
.
.
[^99]:
The markdown format is single square brackets with a caret and a number, also a semi-colon with the actual footnotes. Usually the number goes up to 42-45 maximum but it doesn't matter, the two digit regex is needed. As I said, the semi-colon will be needed in all instances.
Cheers
You have just some errors in your regex, you forget to escaped the d for digit, it should be \d and the capture group must not include the curly braces.
Use:
Ctrl+H
Find what: =={(\d{1,2})}==
Replace with: [^$1]:
TICK Wrap around
SELECT Regular expression
Replace all
Explanation:
=={ # literally
(\d{1,2}) # group 1, 1 or 2 digits
}== # literally
Screenshot (before):
Screenshot (after):
Related
This question already has an answer here:
Notepad++: add parentheses to timestamps
(1 answer)
Closed 1 year ago.
so i have a big list of items in excel. i copied them to Notepad++ because it has regex built in.
it could be AuAC21-XTS02L or BgUX20-C02S etc. basically i want to replace thses two with Au(AC21-XTS02)L and Bg(UX20-C02)S.
with the regular expression \D\D\d\d-(\D){1,3}\d\d i can perfectly find the part of the text that i want to enclose with parentheses but now i dont know how.
i tried using (\D\D\d\d-(\D){1,3}\d\d) as replacement but then i just receive something like Au(DDdd-D{1,3}dd)L.
any help would be appreciated.
You can store the whole matched string in a group and then replace that with ($1). Note that depending on your Notepad++ version you may need to use \ instead of $ to refer to a matching group (i.e. the replacement string would be (\1))
Take a look at this Regex101 snippet: https://regex101.com/r/0e1Wcc/1
It will convert a sample input like,
it could be AuAC21-XTS02L or BgUX20-C02S etc.
it could be AuAC21-XTS02L or BgUX20-C02S etc.
it could be AuAC21-XTS02L or BgUX20-C02S etc.
into
it could be Au(AC21-XTS02)L or Bg(UX20-C02)S etc.
it could be Au(AC21-XTS02)L or Bg(UX20-C02)S etc.
it could be Au(AC21-XTS02)L or Bg(UX20-C02)S etc.
You can take the full match $0 for pattern \D\D\d\d-\D{1,3}\d\d without a capture group because that is not needed, and use it in the replacement between parenthesis \($0\)
The output will be
Au(AC21-XTS02)L or Bg(UX20-C02)S
Note that \D matches any character except a digit, so it could also match a space or a newline.
Looking at the example strings, at bit more precise match (using the same replacement \($0\) could be:
[A-Z][a-z]\K[A-Z]{2}\d\d-[A-Z0-9]{1,3}\d\d(?=[A-Z])
Regex demo
I'm trying to replace all spaces in a log file with commas (to convert it to CSV format). However, some log entries have spaces that I don't want replaced. These entries are bounded by quotation marks. I looked at a couple of examples and came up with the following code, which seems to work in RegExr.com and regex101.com.
[\s](?=(?:"[^"]*"|[^"])*$)
However, when I do a find/replace with that expression, it runs correctly until it hits the first quotation with a space and then selects the entire contents of the file.
Sample log file entry:
date=2020-08-24 time=07:35:15 idseq=216296511061885345 itime="2020-08-24 07:35:15" euid=3 epid=4107 dsteuid=3 dstepid=101 type="utm" subtype="webfilter" level="notice" action="passthrough" msg="URL belongs to an allowed category in policy"
Desired result:
date=2020-08-24,time=07:35:15,idseq=216296511061885345,itime="2020-08-24 07:35:15",euid=3,epid=4107,dsteuid=3,dstepid=101,type="utm",subtype="webfilter",level="notice",action="passthrough",msg="URL belongs to an allowed category in policy"
RegExr result:
EDIT: After more testing, it appears that with a single line, the replace works. However, if you have more than one line, it replaces all lines with the replace character (in my case, the comma).
Ctrl+H
Find what: "[^"\r\n]+"(*SKIP)(*FAIL)|\h+
Replace with: ,
CHECK Wrap around
CHECK Regular expression
Replace all
Explanation:
"[^"\r\n]+" # everything between quotes
(*SKIP)(*FAIL) # kip and fail the match
| # OR
\h+ # 1 or more horizontal spaces
Screenshot (before):
Screenshot (after):
While lengthy, if you have a known list of values, you can simply use them as replacement keys
first value is skipped as it shouldn't be prefixed with ,
must capture and = around labels to be more sure, (though this does not guarantee it will not find substrings in the msg field)
's/ (time|idseq|itime|euid|epid|dsteuid|dstepid|type|subtype|level|action|msg)=/,$1='
Example in Python
import re
>>> source = '''date=2020-08-24 time=07:35:15 idseq=216296511061885345 itime="2020-08-24 07:35:15" euid=3 epid=4107 dsteuid=3 dstepid=101 type="utm" subtype="webfilter" level="notice" action="passthrough" msg="URL belongs to an allowed category in policy"'''
>>> regex = ''' (time|idseq|itime|euid|epid|dsteuid|dstepid|type|subtype|level|action|msg)='''
>>> print(re.sub(regex, r",\1=", source)) # raw string to prevent loss of 1
date=2020-08-24,time=07:35:15,idseq=216296511061885345,itime="2020-08-24 07:35:15",euid=3,epid=4107,dsteuid=3,dstepid=101,type="utm",subtype="webfilter",level="notice",action="passthrough",msg="URL belongs to an allowed category in policy"
You may find some values contain \" or similar, which can break even quite careful regular expressions!
Also note for a CSV you may wish to replace the field names entirely
I have a problem that I really hope that somebody could help me. So, I want to delete some parts of text from a notepad++ document using Regex. If there's another software that I can use to delete this part of text, let me know please, I am really really noob with regex
So, my document its like this:
1
00:00:00,859 --> 00:00:03,070
text over here
2
00:00:03,070 --> 00:00:09,589
text over here
3
00:00:09,589 --> 00:00:10,589
some numbers here
4
00:00:10,589 --> 00:00:12,709
Text over here
5
00:00:12,709 --> 00:00:18,610
More text with numbers here
What I want to learn is how can I delete the first 2 lines of numbers in all the document? So I could get only the text parts (the "text over here" parts)
I would really appreciate any kind of help!
My solution:
^[\s\S]{1,5}\d{1,3}:\d{1,3}:\d{1,3},\d{1,5}\s-->\s*?\d{1,3}:\d{1,3}:\d{1,3},\d{1,5}\s
This solution match both types: either all data in one line, or numbers in one line and data in the second.
Demo: https://regex101.com/r/nKD0DQ/1/
Simplest solution;
\d+(\r\n|\r|\n)\d{2}:\d{2}.*(\r\n|\r|\n)
Get line with some number \d+ with its line break (\r\n|\r|\n)
Also the next line that starts with two 2-digit numbers and a colon \d{2}:\d{2} with the rest .* and its line break. No need to match all since we already are in the correct line, since subtitle file is defined well with its predictable structure.
Put this as Find what: value in Search -> Replace.. in Notepad++, with Seach Mode: Regular Expression and with replace value (Replace with:) of empty space. Will get you the correct result, lines of expected text with empty line in between each.
to see it on action on regex101
Subtitles, for accuracy you can use this:
\d+(\r\n|\n|\r)(\d\d:){2}\d\d,\d{3}\s*-->\s*(\d\d:){2}\d\d,\d{3}(\r\n|\n|\r)
Check Regular Expression, Find what with this and Replace with empty would do.
Regxe Demo
srt subtitles are basically ordered. And it's better accurate than lose texts.
\d : a single digit.
+ : one or more of occurances of the afore character or group.
\r\n: carriage and return. (newline)
* : zero or more of occurances of the afore character or group.
| : Or, match either one.
{3}: Match afore character or group three times.
I'm going for a less specific regex:
^[0-9]*\n[0-9:,]*\s-->\s[0-9:,]*
Demo # regex101
I am working with government measures and am required to parse a string that contains variable information based on delimiters that come from issuing bodies associated with the fda.
I am trying to retrieve the delimiter and the value after the delimiter. I have searched for hours to find a regex solution to retrieve both the delimiter and the value that follows it and, though there seems to be posts that handle this, the code found in the post haven't worked.
One of the major issues in this task is that the delimiters often have repeated characters. For instance: delimiters are used such as "=", "=,", "/=". In this case I would need to tell the difference between "=" and "=,".
Is there a regex that would handle all of this?
Here is an example of the string :
=/A9999XYZ=>100T0479&,1Blah
Notice the delimiters are:
"=/"
"=>'
"&,1"
Any help would be appreciated.
You can use a regex like this
(=/|=>|&,1)|(\w+)
Working demo
The idea is that the first group contains the delimiters and the 2nd group the content. I assume the content can be word characters (a to z and digits with underscore). You have then to grab the content of every capturing group.
You need to capture both the delimiter and the value as group 1 and 2 respectively.
If your values are all alphanumeric, use this:
(&,1|\W+)(\w+)
See live demo.
If your values can contain non-alphanumeric characters, it get complicated:
(=/|=>|=,|=|&,1)((?:.(?!=/|=>|=,|=|&,1))+.)
See live demo.
Code the delimiters longest first, eg "=," before "=", otherwise the alternation, which matches left to right, will match "=" and the comma will become part of the value.
This uses a negative look ahead to stop matching past the next delimiter.
In Notepad++ 6.5.1 I need to replace certain patterns within quote pairs. I want to save the replace as part of a macro, so all replacements need to happen in one step.
For example, in the following string, replace all 'a' characters within quote pairs with a dash, while leaving characters outside the quote pairs untouched:
Input: aa"bbabaavv"kdjhas"bbabaavv"x
Desired result: aa"bb-b--vv"kdjhas"bb-b--vv"x
Note that the quotes are matched up pairwise, such that the 'a' in kdjhas is untouched.
So far I have tried searching for (?:"[^"a]*|\G)\Ka([^"a]*) and replacing with -$1, but that simply replaces all the a's, with the result --"bb-b--vv"kdjh-s"bb-b--vv"x. I'm attempting PCRE regex that will let me recursively replace the quote-delimited text.
Edit: Quote marks within a quoted string are escaped with an extra quote, e.g. "". However, assume I will have already replaced these in a previous pass with a special character. Therefore a regex solution to this problem will not have to deal with escaped quotes.
It is hard to tell if this is possible as you've only provided one line of input text.
But assuming that input follows this pattern:
BOL|any text|string with two groups of a's|any text|string with two groups of a's|any text|EOL
aa "bbabaavv" kdjhas "bbabaavv" x
I was able to create this regexp search string:
^(.+?\".+?)([a]+)(.+?)([a]+)(.*?\")(.+?\".+?)([a]+)(.+?)([a]+)(.*?\".*)$
With this replace string:
\1-\3-\5\6-\8-\A
and it turn your input string from this:
aa"bbabaavv"kdjhas"bbabaavv"x
into this:
aa"bb-b-vv"kdjhas"bb-b-vv"x
Now naturally the search an replace will fail if the input varies from that pattern described as the search is looking for those four groups of a's inside the two groups of quoted strings.
Also I tested that regexp using Zeus which can create a regexp with more than 9 groups.
As you can see the regexp requires 10 groups.
I'm not familar with Notpad++ so I don't know if it supports that many groups.
If your data have variable number of occurrences of quoted strings, then it is not possible to perform replacements only via regex at least in its form offered by Notepad++.
To replace using regex, you would need to perform regex find in existing regex match. As far as I know such a functionality is not available in Notepad++ regexes.
Self-answer
I may have been reaching for the stars in trying to get Notepad++ to do this regex replace, but I think I found a workaround.
The actual task I was attempting involved creating a SQL Server VALUES list from an Excel spreadsheet, where I was copying and pasting selected cells into Notepad++. The delimiters are \t and \r\n. But, cells can have linefeeds too, which are delimited by ". So, I was going to replace these linefeeds with <br> (or something like it), so that
"line1
line2"
would become "line1<br>line2", before processing the actual end-of-row line feeds.
Having such parsing work reliably, especially when more than two lines were in a single cell, may have been too much to ask of Notepad++'s regex capability.
So I came up with a workaround that seems to be working:) Basically it starts with selecting a blank "dummy" column to the right of my column selection (which I can insert if I'm partially selecting from the middle). This will leave a trailing \t at the end of each row, which effectively sets these EOL's apart from ones that might exist with a text cell, freeing me from having to parse line feeds from a "..." field.
So I compiled a macro from the following steps, which seems to be working well:
replace ' with ''
replace \t\r\n with '\)\r\n, \('
replace \t with ', '
replace "" with ''
replace " with <blank>
replace ^ with \(' (cleanup - first row only)
replace ^, \('$ with <blank> (cleanup - last row only)
Example transformation:
from
line1 line 2
"line3
line3b
line3c" line 4
to
('line1', 'line 2')
, ('line3
line3b
line3c', 'line 4')
which can now be easily modified into a SELECT statement:
SELECT *
FROM (VALUES('line1', 'line 2')
, ('line3
line3b
line3c', 'line 4')
) t(a,b)