I'm having a bunch of comma separated CSV files.
I would like to replace exact one value which is between the third and fourth comma. I would love to do this with Notepad++ 'Find in Files' and Replace functionality which could use RegEx.
Each line in the files look like this:
03/11/2016,07:44:09,327575757,1,5434543,...
The value I would like to replace in each line is always the number 1 to another one.
It can't be a simple regex for e.g. ,1, as this could be somewhere else in the line, so it must be the one after the third and before the fourth comma...
Could anyone help me with the RegEx?
Thanks in advance!
Two more rows as example:
01/25/2016,15:22:55,276575950,1,103116561,10.111.0.111,ngd.itemversions,0.401,0.058,W10,0.052,143783065,,...
01/25/2016,15:23:07,276581704,1,126731239,10.111.0.111,ll.browse,7.133,1.589,W272,3.191,113273232,,...
You can use
^(?:[^,\n]*,){2}[^,\n]*\K,1,
Replace with any value you need.
The pattern explanation:
^ - start of a line
(?:[^,\n]*,){2} - 2 sequences of
[^,\n]* - zero or more characters other than , and \n (matched with the negated character class [^,\n]) followed with
, - a literal comma
[^,\n]* - zero or more characters other than , and \n
\K - an operator that forces the regex engine to discard the whole text matched so far with the regex pattern
,1, - what we get in the match.
Note that \n inside the negated character classes will prevent overflowing to the next lines in the document.
You can replace value between third and fourth comma using following regex.
Regex: ([^,]+,[^,]+,[^,]+),([^,]+)
Replacement to do: Replace with \1,value. I used XX for demo.
Regex101 Demo
Notepad++ Demo
Related
I have a document with multiple information. What I want is to build a Notepad++ Regex replace function, that finds the following lines in the document and replaces the blank spaces between the "" with an underline (_).
Example:
The line is:
&LOG Part: "NAME TEST.zip"
The result should be:
&LOG Part: "NAME_TEST.zip"
The perfect solution would be that the regex finds the &LOG Part: "NAME TEST.zip" lines and replaces the blank space with an underline.
What I have tried for now is this expression to find the text between the " ":
\"[^"]*\"
It should do it, but I don't know which expression to use to replace the blank spaces with an underline.
Anyone could help with a solution?
Thanks!
The \"[^"]*\" will only match whole substrings from " up to another closest " without matching individual spaces you want to replace.
Since Notepad++ does not support infinite width lookbehind, the only possible solution is using the \G - based regex to set the boundaries and use multiple matching (this one will replace consecutive spaces with 1 _):
(?:"|(?!^)\G)\K([^ "]*) +(?=[^"]*")
Or (if each space should be replaced with an underscore):
(?:"|(?!^)\G)\K([^ "]*) (?=[^"]*")
And replace with $1_. If you need to restrict to replacing inside &LOG Part only, just add it to the beginning:
(?:&LOG Part:\s*"|(?!^)\G)\K([^ "]*) (?=[^"]*")
A human-readable explanation of the regex:
(?:"|(?!^)\G)\K - Find a ", or, with each subsequent successful match, the end of the previous successful match position, and omit all the text in the buffer (thanks to \K)
([^ "]*) - (Group 1, accessed with$1from the replacement pattern) 0+ characters other than a space and"`
+ - one or more literal spaces (replace with \h to match all horizontal whitespace, or \s to match any whitespace)
(?=[^"]*") - check if there is a double quote ahead of the current position
I've got a CSV file with lines like:
57,13,"Bob, Bill and Susan",Student,Club,Funded,64,3200^M
I need them to look like
57,13,Bob-Bill-and-Susan,Student,Club,Funded,64,3200
I'm using vim regexes. I've broken it down into 4 steps:
Remove ^M and insert newlines:
:%s:<ctrl-V><ctrl-M>:\r:g`
Replace all with -:
:%s: :\-:g
Remove commas between quotes: Need help here.
Remove quotes:
:%s:\"\([^"]*\)\":\1:g
How do I remove commas between quotes, without removing all commas in the file?
Something like this?
:%s:\("\w\+\),\(\w\+"\):\1 \2:g
My preferred solution to this problem (removing commas inside quoted regions) is to use replacements with an expression instead of trying to get this done in one regex.
To do this you need to prepend you replacement with \= to get the replacement treated as a vim expression. From here you can extract just the parts between quotes and then manipulate the the matched part separately. This requires having two short regexes instead of one complicated one.
:%s/".\{-}"/\=substitute(submatch(0), ',', '' , 'g')/g
So ".\{-}" matches anything in quotes (non greedy) and substitute(submatch(0), ',', '' , 'g') takes what was matched and removes all of the commas and its return value is used as the actual replacement.
The relevant help page is :help sub-replace-special.
As for the other parts of your question. Step 1 is essentially trying to remove all carriage returns since the file format is actually the dos file format. You can remove them with the dos2unix program.
In Step 2 escaping the - in the replacement is unnecessary. So the command is just
:%s/ /-/g
In Step 4, you have an overly complicated regex if all you want to do is remove quotes. Since all you need to do is match quotes and remove them
:%s/"//g
:%s:\("\w*\)\(,\)\(.*"\):\1\3:g
example: "this is , an, example"
\("\w*\) match start of " every letter following qoutes group \1 for back reference
\(,\) capture comma group \2 for back reference
(.*"\) match every other character upto the second qoute ->group 3 for backreference
:\1\3: only include groups without comma, discard group 2 from returned string which is \2
:%s:\("\w*\)\(,\)\(.*"\):\1\3:g removes commas
I have a file abc.txt which has data like this when I opened up in notepad++
10.114.128.196, 10.149.53.72, 40.169.74.47
Is there any way I can make it like this using regular expressions in notepad++?
10.114.128.196,abc
10.149.53.72,abc
40.169.74.47,abc
Search for
(\d{1,3}(\.\d{1,3}){3})[,\s]*
and replace with
$1,abc\n
(\d{1,3}(\.\d{1,3}){3}) matches 1 to 3 digits followed by 3 more such groups starting with a ".". Because of the round brackets around the found pattern is stored in capturing group 1, you can reuse this matched text in the replacement by inserting $1.
[,\s]* matches zero or more commas and whitespace characters.
Global replace ", " with ",abc\n"?
On the field search put: ((\d+\.?){4}(.)( ?))
On the field replace put: $1abc\r\n
The last line will not have a comma , so I think it is ok to have just this one to fix ;)
In Notepad++, I'd like to replace only the first and second comma (","), by ":".
Example :
blue,black,red -> blue:black:red (2 first commas replaced)
blue,black,red,yellow -> blue:black:red,yellow (third comma still here)
Thanks!
I believe you can do this by replacing this regex:
^([^,]*),([^,]*),(.*)$
With this:
$1:$2:$3
For compatibility with cases where there are less than 2 commas, use these:
^(([^,]*),)?(([^,]*),)?(.*)$
$2:$4:$5
Something along this line,
^([^,]*),([^,]*),(.*)$
And replace with
$1:$2:$3
Or \1:\2:\3
Just two capturing groups is enough.
Regex:
^([^,]*),([^,]*),
Replacement string:
$1:$2:
DEMO
Explanation:
^ Asserts that we are at the start.
([^,]*) Captures any character not of , zero or more times and stored it into a group.(ie, group 1)
, Matches a literal , symbol.
([^,]*) Captures any character not of , zero or more times and stored it into a group.(ie, group 2)
, Matches a literal , symbol.
Well you can try to capture the parts in groups and then replace them as follows:
/^([^,]*),([^,]*),(.*)$/$1:$2:$3
How does it work: each line is matched such that the first part contains all data before the first comma, the second part in between the two commas and the third part all other characters (including commas).
This is simply replaced by joining the groups with colons.
A no-brainer; virtually "GREP 1-0-1". Not really an effort.
Just find
^([^,]+),([^,]+),
and replace with
\1:\2:
Click on the menu item: Search > Replace
In the dialog box that appears, set the following values...
Find what: ^([^,]+),([^,]+),
Replace with: $1:$2:
Search Mode: Regular expression
I have a large file with content inside every bracket. This is not at the beginning of the line.
1. Atmos-phere (7800)
2. Atmospheric composition (90100)
3.Air quality (10110)
4. Atmospheric chemistry and composition (889s120)
5.Atmospheric particulates (10678130)
I need to do the following
Replace the entire content, get rid of line numbers
1.Atmosphere (10000) to plain Atmosphere
Delete the line numbers as well
1.Atmosphere (10000) to plain Atmosphere
make it a hyperlink
1.Atmosphere (10000) to plain linky study
[I added/Edit] Extract the words into a new file, where we get a simple list of key words. Can you also please explain the numbers in replace the \1\2, and escape on some characters
Each set of key words is a new line
Atmospheric
Atmospheric composition
Air quality
Each set is a on one line separated by one space and commas
Atmospheric, Atmospheric composition, Air quality
I tried find with regex like so, \(*\) it finds the brackets, but dont know how to replace this, and where to put the replace, and what variable holds the replacement value.
Here is mine exression for notepad ([0-9(). ]*)(.*)(\s\()(.*)
You need split your search in groups
([0-9. ]*) numbers, spaces and dots combination in 0 or more times
(.*) everything till next expression
(\s\() space and opening parenthesis
(.*) everything else
In replace box - for practicing if you place
\1\2\3\4 this do nothing :) just print all groups from above from 1.1 to 1.4
\2 this way you get only 1.2 group
new_thing\2new_thing adds your text before and after group
<a href=blah.com/\2.html>linky study</a> so now your text is added - spaces between words can be problematic when creating link - so another expression need to be made to replace all spaces in link to i.e. _
If you need add backslash as text (or other special sign used by regex) it must be escaped so you put \\ for backslash or \$ for dolar sign
Want more tune - <a href=blah.com/\2.html>\2</a> add again 1.2 group - or use whichever you want
On the screenshot you can see how I use it (I had found and replaced one line)
Ok and then we have case 4.2 with colon at the end so simply add colon after extracted section:
change replace from \2 to \2,
Now you need join it so simplest way is to Edit->Line Operations->Join Lines
but if you want to be real pro switch to Extended mode (just above Regular expression mode in Replace window) and Find \r\n and replace with space.
Removing line endings can differ in some cases but this is another story - for now I assume that you using windows since Notepad++ is windows tool and line endings are in windows style :)
The following regex should do the job: \d+\.\s*(.*?)\s*\(.*?\).
And the replacement: <a href=example.com\\\1.htm>\1</a>.
Explanation:
\d+ : Match a digit 0 or more times.
\. : Match a dot.
\s* : Match spaces 0 or more times.
(.*?) : Group and match everything until ( found.
\s* : Match spaces 0 or more times.
\(.*?\) : Match parenthesis and what's between it.
The replacement part is simple since \1 is referring to the matching group.
Online demo.
Try replacing ^\d+\.(.*) \(\w+\)$ with <a href=blah.com\\\1.htm>linky study</a>.
The ^\d+. removes the leading number and dot. The (.*) collects the words. Then there is a single space. The \(\w+\)$ matches the final number in brackets.
Update for the added Q4.
Regular expressions capture things written between round brackets ( and ). Brackets that are to be found in the text being searched must be escaped as \( and \). In the replacement expression the \1 and \2 etc are replaced by the corresponding capture expression. So a search expression such as Z(\d+)X([aeiou]+)Y might match Z29XeieiY then the replacement expression P\2Q\1R would insert PeieiQ29R. In the search at the top of this answer there is one capture, the (.) captures or collects the words and then the \1 inserts the captured words into the replacement text.