I'm terrible at regex and need to remove everything from a large portion of text except for a certain variable declaration that occurs numerous times, id like to remove everything except for instances of mc_gross=anyint.
Generally we'd need to use "negative lookarounds" to find everything but a specified string. But these are fairly inefficient (although that's probably of little concern to you in this instance), and lookaround is not supported by all regex engines (not sure about notepad++, and even then probably depends on the version you're using).
If you're interested in learning about that approach, refer to How to negate specific word in regex?
But regardless, since you are using notepad++, I'd recommend selecting your target, then inverting the selection.
This will select each instance, allowing for optional white space either side of the '=' sign.
mc_gross\s*=\s*\d+
The following answer over on super user explains how to use bookmarks in notepad++ to achieve the "inverse selection":
https://superuser.com/questions/290247/how-to-delete-all-line-except-lines-containing-a-word-i-need
Substitute the regex they're using over there, with the one above.
You could do a regular expression replace of ^.*\b(mc_gross\s*=\s*\d+)\b.*$ with \1. That will remove everything other than the wanted text on each line. Note that on lines where the wanted text occurs two or more times, only one occurrence will be retained. In the search the ^.*\b matches from start-of-line to a word boundary before the wanted text; the \b.*$ matches everything from a word boundary after the wanted text until end of line; the round brackets capture the wanted text for the replacement text. If text such as abcmc_gross=13def should be matched and retained as mc_gross=13 then delete the \bs from the search.
To remove unwanted lines do a regular expression search for ^mc_gross\s*=\s*\d+$ from the Mark tab, tick Bookmark line and click Mark all. Then use Menu => Search => Bookmark => Remove unmarked lines.
Find what: [\s\S]*?(mc_gross=\d+|\Z)
Replace with: \1
Position the cursor at the start of the text then Replace All.
Add word boundaries \b around mc_gross=\d+ if you think it's necessary.
Related
Doing a find and replace in VsCode on a large amount of files. I'm looking to replace all spaces after a set of quotes, but only on a specific line.
I can very easily find all spaces using \s+, but I don't understand how to capture only the spaces after a specific string(one specific line). I've tried positive look behinds, but I can only get it to match the first space, but I need to match all spaces on that line.
Example code:
variable = "01 - Testing this thing"
I need to find and replace all the spaces between the quotation marks with underscores, but I can't get any regex to match all the spaces between the quotes. I might want to replace the dash(-) as well, but the spaces are more important and I'm struggling to figure it out.
Here is a pretty good workflow.
Open a Search Editor (from the Command Palette or set a keybinding to it).
Use this regex (?<=variable = ")[^"]*.
That will find all matches in all files in your workspace or whatever folders you designate in the file to include filter. I suggest setting the context lines option to 0.
Ctrl+Shift+L to select all your matches. The matches are the 01 - Testing this thing part.
Now do a regular find in that search editor tab - with the Find in Selection option enabled.
Simply doing a find of and replaceAll with _ will make all those changes (in the Search Editor only).
To apply those changes to all the files with your initial search results, use the extension search-editor-apply-changes Apply Search Editor Changes... command.
Then you can check to see if the changes were as you expected and save all. It will open all affected files so you can inspect them.
Seems like a few steps but notice the first regex can be very simple. And then you are doing a simple find/replace in just those selections. Demo:
You search for a string that matches, it has A space between the quotes. Replace with what is before and after the space but the space is now an underscore. You have to apply this as often as the max number od spaces in a string. It can't be done in 1 regex search-replace.
In the Search Bar
Find Regex:
(variable = "[^" ]*) ([^"]*")
Replace:
$1_$2
Then apply Replace All (button) and Refresh (button) until no more searches found.
I use Notepad++,
i need to search and replace entire word that contain a specific keyword.
Ex: someting HELP.blablabla.blabla someting
i would like to search entire text for words that contain the keyword "HELP" untill the first space OR the first comma.
In this case: HELP.blablabla.blabla
thanks a lot
Go to the search panel, check the regex checkbox on the bottom and try: (HELP)([^ ,]*)
Note: There are a space character after the ^
This regex means: Search for the entire word HELP (HELP) followed by anything that it isn't an space or an comma [^ ,] the ^ inside the brackets is a denial
Edit:
You can use just HELP[^ ,]* the parenthesis is just to create capturing groups if you need to use the specific groups to replace later. As pointed by #alphabravo
You say search and replace an entire word but if it were that simple then I wonder why a regular search and replace isn't sufficient. So I'm reading between the lines and assuming you want to match on full lines of text.
I think I've used npp enough to get the syntax right. I don't remember any eccentricities that would apply. Is the comma/space optional?
^[^, ]*HELP[^, ]*[, ]
I'm kinda thinking this one might be good enough:
^[^, ]*HELP
I have a large text file which contains many timestamps. The timestamps look like this: 2013/11/14 06:52:38AM. I need to remove the last two characters (am/pm/AM/PM) from each of these. The problem is that a simple find and replace of "AM" may remove text from other parts of the file (which contains a lot of other text).
I have done a find using the regular expression (:\d\d[ap]m), which in the above example would track down the last bit of the timestamp: :38AM. I now need to replace this with :38, but I don't know how this is done (allowing for any combination of two digits after the colon).
Any help would be much appreciated.
EDIT: What I needed was to replace (:\d\d)[ap]m with \1
Make (:\d\d[ap]m) into (:\d\d)[ap]m and use $1 not \1
Go to Search > Replace menu (shortcut CTRL+H) and do the following:
Find what:
[0-9]{2}\K[AP]M
Replace:
[leave empty]
Select radio button "Regular Expression"
Then press Replace All
You can test it at regex101.
Note: the use of [0-9] is generally better than \d (read why), and avoiding to use a capture group $1 with the use of \K is considered better. It's definitely not important in your case, but it is good to know :)
I have some text like this:
dagGeneralCodes$_ctl1$_ctl0
Some text
dagGeneralCodes$_ctl2$_ctl0
Some text
dagGeneralCodes$_ctl3$_ctl0
Some text
dagGeneralCodes$_ctl4$_ctl0
Some text
I want to create a regular expression that extracts the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0 from the text above.
the result should be: dagGeneralCodes$_ctl4$_ctl0
Thanks in advance
Wael
This should do it:
.*(dagGeneralCodes\$_ctl\d\$_ctl0)
The .* at the front is greedy so initially it will grab the entire input string. It will then backtrack until it finds the last occurrence of the text you want.
Alternatively you can just find all the matches and keep the last one, which is what I'd suggest.
Also, specific advice will probably need to be given depending on what language you're doing this in. In Java, for example, you will need to use DOTALL mode to . matches newlines because ordinarily it doesn't. Other languages call this multiline mode. Javascript has a slightly different workaround for this and so on.
You can use:
[\d\D]*(dagGeneralCodes\$_ctl\d+\$_ctl0)
I'm using [\d\D] instead of . to make it match new-line as well. The * is used in a greedy way so that it will consume all but the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0.
I really like using this Regular Expression Cheatsheet; it's free, a single page, and printed, fits on my cube wall.
I have been fighting this problem with the help of a RegEx cheat sheet, trying to figure out how to do this, but I give up... I have this lengthy file open in Notepad++ and would like to remove all lines that do not start with a digit (0..9). I would use the Find/Replace functionality of N++. I am only mentioning this as I am not sure what Regex implementation is N++ using... Thank you
Example. From the following text:
1hello
foo
2world
bar
3!
I would like to extract
1hello
2world
3!
not:
1hello
2world
3!
by doing a find/replace on a regular expression.
You can clear up those line with ^[^0-9].* but it will leave blank lines.
Notepad++ use scintilla, and also using its regex engine to match those.
\r and \n are never matched because in
Scintilla, regular expression searches
are made line per line (stripped of
end-of-line chars).
http://www.scintilla.org/SciTERegEx.html
To clear up those blank lines, only way is choose extended mode, and replace \n\n to \n, If you are in windows mode change \r\n\r\n to \r\n
[^0-9] is a regular expression that matches pretty much anything, except digits. If you say ^[^0-9] you "anchor" it to the start of the line, in most regular expression systems. If you want to include the rest of the line, use ^[^0-9].+.
^[^\d].* marks a whole line whose first character is not a digit. Check if there are really no whitespaces in front of the digits. Otherwise you'd have to use a different expression.
UPDATE:
You will have to do ot in two steps. First empty the lines that do not start with a digit. Then remove the empty lines in extended mode.
One could also use the technique of bookmarking in Notepad++. I started benefiting from this feature (long time present but only more recently made somewhat more visible in the UI) not very long ago.
Simply bring up the find dialogue, type regex for lines not starting with digit ^\D.*$ and select Mark All. This will place blue circles, like marbles, in the left gutter - these are line bookmarks. Then just select from main menu Search -> Bookmark -> Remove bookmarked lines.
Bookmarks are cool, you could extract these lines by simply selecting to copy bookmarked lines, opening new document and pasting lines there. I sometimes use this technique when reviewing log files.
I'm not sure what you are asking. but the reg exp for finding the lines with a digit at the beginning would be
^\d.*
you can remove all the lines that match the above or alternatly keep all the lines that match this expression:
^[^\d].*