I have a text file and want to keep lines started with <Path>
, and delete all the other lines.
How can I do it?
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Sign up to join this communityThere is an easy way to achieve this. You need to perform 2 steps.
Go to Search menu > Find... > Select "Mark" Tab. Activate regular expressions. Search for ^<Path>
(^
is for line start). Don't forget to check "Bookmark lines" and Press "Mark All"
==> All Rows you want to keep got a Bookmark
Go to Menu "Search - Bookmark - Remove unmarked lines"
==> All lines that aren't Bookmarked are deleted.
This can actually be done in two steps as of 6.3. I think it can be done earlier than that as I had 5.9 when I first tried it.
Using stema's post as the basis of this answer. There's one less step now. Mark lines and remove unmarked lines. Done. Detailed instructions follow.
Search menu "Find". In the Find dialog, click the "Mark" tab. Activate regular expressions. Search for ^<Path>
(^
is for line start). Don't forget to check "Bookmark lines" and Press "Mark All"
==> All Rows you want to keep now have a Bookmark
Search Menu -> Bookmark -> Remove Unmarked Lines.
==> All NON Bookmarked lines are deleted.
^
in my search term, nor use regular expression. Hope this helps someone.
May 20, 2019 at 5:23
regex replace
(?!^.*test.*$)^.+
replace test with your requested text
replace
[\r\n]{2,}
with \r\n
Use ^(?!<Path>).*\r\n
to replace matches with empty string. Generalized version would be ^(?!.*?test).*\r\n
. This won't remove empty line at the end of the file. All other lines are removed, including multiple consecutive empty lines.
(?!)
is a negative look up. ^.*test.*$
selects the whole line that contains the requested text.
[\r\n]{2,}
matches any \r\n
that occurs more then once this is Windows New line. if you have Linux or another operating system you might need to mess with this. the second is to replace it with one return line.
\r\n
EOL mentioned in the post is what Windows uses and therefore may or may not be what you are looking for. Often on Linux environments it's just \n
, or in Mac environments just \r
, so if you pull a file from either, it won't be Windows style EOL. However, if you download via FileZilla and ASCII mode triggers, it may change them back into Windows EOL (like \n
into \r\n
). So if regex is not working, check the EOL style by going to "View > Show Symbols > Show End Of Line". CR = \r
. LF = \n
.
It seems to me that the easiest way is to just use the "Find All in Current Document" feature and then either copy the results into a new file or select all and replace in the current one.
This would find all lines containing your text and list them at the bottom. Just right click on the search result and copy / paste.
\tLine [\d]*:
. Still a great answer.
Better solution with regex replace:
(?!^.*SOMETEXT.*$)^.+\r?\n
And replace with nothing
Go to menu Search -> Find... -> Activate regular expressions. Search for "^Path" (^ is for line start).
Click on the "Find all in Current Document" button.
The "Find result" window will appear with all the lines the the pattern. Select copy/paste them to a new tab in Notepad++.
In this new tab, got to: menu Search -> Replace... -> Activate regular expressions.
In the "Find what:" field, use the pattern: "Line \d+: ". Leave the "Replace with:" field blank.
Click on the "Replace all" button.
Providing that you actually want to match <Path>
and not a file system path, you can try this from a command line using Perl:
perl -pe " if ($_ !~ /<Path>/) { s/$_// } " < in.txt > out.txt
It worked with Strawberry Perl on Windows, so adjust accordingly if the results are not what you expect.
It is clumsy, but copy it all to Excel, and then use =IF(LEFT(A1,6)="<Path>",A1,"")
and copy that formula all the way down. Then copy that back to Notepad++. It's not ideal, but it's pretty easy (if you have Excel). Warning: It will not work well with indented lines (Excel will shift the columns, etc.).
There is no easy way to do what you want with Notepad++. You'll need to either download a program to your computer or script something in VB (I assume you're on Windows).
You can do what you want one of two ways with sed
. The sed
utility is a favorite on *nix and can be found for Windows from the great people at GnuWin (http://gnuwin32.sourceforge.net/packages/sed.htm). You would download this program, and then run your command from the command prompt.
Delete all lines not containing :
sed -i '/^<PATH>/!d' file
Print all lines containing to a new file:
sed -n '/^<PATH>/p' file > newfile
I suggest you use print the lines you want to a new file. The reason for this is that you probably won't get the regex statement for <PATH>
the first time around. The sed
utility uses Regular Expression Basic Syntax (view the reference at http://www.regular-expressions.info/reference.html). If <PATH>
is something like a *nix path (/var/www
) then you'll need to escape the /
character for your regex to work.
Example: sed -n '/^\/var\/www/p' file > newfile
This will print out all lines that start with /var/www
. If I filed to escape the /
character, then the command would have thrown an error. You can escape a special character (such as /
) with the backslash character \
.
<PATH>
disappeared because you didn't format it as code, and that's an invalid html tag so it's not shown
Use Search->Replace and enter a regular expression like ^[^ ].*
and replace all with an empty string using Regular expression
. Next step is to find empty lines searching for \n\n
replacing with \n
using Extended
multiple times until 0 occurrences were found.
(use \r\n\r\n
and \r\n
depending on file format). If you have very many empty lines in a row, it is quickier to use \n\n\n\n\n\n\n
or even more \n
:s in the search string.