The question and answers here cover in detail how the following vim command collapses a series of empty lines into a single line:
:g/^$/,/./-j
However, I want to do the same but also treat lines with onlywhite space in them as blank. The following command is what I tried but it doesn't work:
:g/^\s*$/,/./-j
As far as I can tell, that should find the lines that are empty and have only whitespace on them, but not all lines are being collapsed.
You're halfway there.
Remember that the initial command consisted of a search part and an action part. The search part :g/^$/ found all empty lines and the action part ,/./-j was executed for each (well, each that hadn't already been deleted by a previous j).
The modification you made to the search part of the string is correct in that it will now find lines that are either empty or contain only whitespace.
However, it's the action that you're executing after that that's causing you grief. The original action to be executed on the found line was ,/./-j which basically means execute a join j over the range from this line to the one before the next 'real' character. More detail on how this works can be found in the question you linked to.
The first 'real' character that it finds in your case actually includes whitespace so, while the search bit will find whitespace lines and act on them, the range of the join in the action will not be what you want.
What you need to specify for the end of the range in the action is the line previous to the next one that has something other than whitespace (rather than just a line with any 'real' character). A line with a non-whitespace character is simply one that matches the regex \S (the backslash with uppercase S denotes a non-whitespace character).
So, in the end, what you're looking for is:
:g/^\s*$/,/\S/-j
Having said that, keep in mind that the line that remains behind is (I think) the first from the range. So, it's not necessarily empty, it may contain white-space.
If you wish to ensure all whitespace-only lines are made empty, just execute:
:g/^\s*$/s/.*//
after the collapsing command above. Or, you can combine both into a single command using | as an action separator:
:g/^\s*$/,/\S/-j|s/.*//
Related
I have a file which is structured like this:
Line
foo Änderbar: PM baz
Line
Line
foo Änderbar: OM baz
Line
Line
foo Änderbar: ++ baz
Line
Line
foo Änderbar: -- baz
Line
So the file consists of "blocks" which are separated by a newline (I have converted the file to Unix line endings). Each block can have an arbitrary number of lines. Each line of a block contains at least one character which is not a newline, and is finished by a newline character. The lines which separate the blocks consist of exactly one newline character.
In each block, there is exactly one line in the following format:
at least one character which is not newline, followed by
the literal string 'Änderbar: ', followed by
exactly one of the literal strings '++', '--', 'OM', 'PM', followed by
at least one character which is not newline, followed by
the line-terminating newline character
There is always at least one other non-empty line in the same block above this special line and one other non-empty line below this special line.
I need an effective method to find (and thereby select) all blocks where the literal after Änderbar: is -- (find / select one block after another, each one after hitting Find Next again, i.e. not selecting all of those blocks at the same time).
Normally, I have fun solving such problems with Notepad++. However, in that case, it seems that I either get more and more stupid as I get older, or that there is a bug in Notepad++'s regex handling engine.
Notepad++ uses BOOST (and supports PCRE expressions via BOOST). Since this is in wide use, I consider that problem important enough to post it here, just in case that BOOST really is the reason for the misbehavior.
Having said this: I loaded that file into Notepad++, fired up the Search and Replace dialog, ticked . matches newline, ticked Regular Expression and entered the following regex in the Find What: textbox:
\n([^\n]+\n)+[^\n]+(Änderbar\:\ --[^\n]+\n)([^\n]+\n)+
I was quite surprised that this made Notepad++ behave weirdly: When the cursor was placed in the empty line immediately before a block with Änderbar: --, hitting Find Next found / selected that block as expected. But when the cursor was at another place, hitting Find Next made Notepad++ find / select the whole rest of the file, i.e. all blocks below the cursor position.
I then have tested if it would find the blocks having ++ after Änderbar:, i.e. I changed my regex to
\n([^\n]+\n)+[^\n]+(Änderbar\:\ \+\+[^\n]+\n)([^\n]+\n)+
Guess what: This was working reliably in each situation. The same is true for the last both:
\n([^\n]+\n)+[^\n]+(Änderbar\:\ PM[^\n]+\n)([^\n]+\n)+
\n([^\n]+\n)+[^\n]+(Änderbar\:\ OM[^\n]+\n)([^\n]+\n)+
So Notepad++ / PCRE seems to have a problem with the correct interpretation of - under certain circumstances, or I have a subtle bug in my regex which only triggers when I am searching for -- (instead of ++, OM or PM) at the respective place.
Please note that I already have tried to leave away the \ in front of the space character (which actually could only make the situation worse, but I've tried just in case) and that I also have tried to use \-\- instead of -- (although the latter should be fine). That did not alter the (mis-)behavior in any way.
So what is the problem here? Is there a bug in my regex, or is there a bug in Notepad++?
UPDATE
I have stripped down the actual file in question and have uploaded it to https://pastebin.com/w62E57U5. To reproduce the problem, please do the following:
Download the file from the link above and save it somewhere on your HDD (do not copy the text directly into Notepad++).
Load the file into Notepad++. The cursor now is in the topmost line, and nothing is selected.
This is essential: Click Edit -> EOL Conversion -> Unix (LF).
Verify that the cursor is still in the topmost line (which is empty) and that nothing is selected.
Open the Find dialog and choose the settings and enter the search string as described above.
Click "Find Next".
Note that now the complete text is found / selected.
Keeping the Find window open, delete the third line of the file (it reads "Funktionspaket(e): ML"). Do not just empty that line, but really delete it so that no empty line remains between the line before and the line after.
Again, place the cursor in the topmost line (which is still empty) and make sure nothing is selected.
Click "Find Next".
Note that the regular expression now works as expected.
Obviously, somebody is trying to make a fool of me, right?
I think the key is: you need to begin your regex with ^ (beginning of line).
Your original regex becomes:
^\n([^\n]+\n)+[^\n]+(Änderbar\:\ --[^\n]+\n)([^\n]+\n)+
But you can simplify it with:
^\R(?:.+\R)+.+Änderbar: --.+\R(?:.+(?:\R|\z))+
Note: tick . matches newline
Where:
\R matches any kind of linebreak, no needs to change the EOL.
\z matches the end of file, if you don't use it, you can't match the last line of the file if there're no linebreak.
(?:...) is a non capture group, much more efficient (if you don't need to capture, of course)
Both works fine with your 2 sample files.
It's not a bug. You're just forgetting something very important - with Windows line endings, your lines have a \r before the \n, so the \n([^\n]+\n)+ part of your RegEx will also match your blank lines which is why clicking "Find Next" matches everything from the cursor position instead of from the start of the block.
Go to Edit > EOL Conversion > Unix (LF) and you'll see that it works now. If you want to support Windows and Unix line endings you'll have to change every [^\n] to [^\r\n] and every \n to \r?\n.
I'm trying to removing a specific line from many files I'm working on with Notepad++.
Upon searching, I found Notepad++ Remove line with specific word in multiple files within a directory but somehow the regex provided (^.*(?:YOURSTRINGHERE).*\r\n$) from the answers doesn't work for me (screenshot: https://cdn.discordapp.com/attachments/311547963883388938/407737068475908096/unknown.png).
I read on some other questions/answers that certain regex doesn't work in newer/older Notepad++ versions. I was using Notepad++ 5.x.x then updated to the latest 7.5.4, but neither worked with the regex provided in the question above.
At the moment I can work around it by replacing that line with nothing, twice (because there are only 2 variants that I need to remove from those files) but that leaves an empty line at the end of the files. So I have to do another step further to remove that empty line.
I'm hoping someone can offer helps that allow me to remove that line and leave no empty line/space behind.
The regex you attempt to use will only match your line, if it is followed by an empty line and Windows linebreaks (CR LF) are used. This is due to \r\n$ which matches a linebreak sequence followed by the end of the line.
Instead you might want to use
^.*(?:YOURSTRINGHERE).*\R?
To match the line containing your string and optionally a following line break sequence to remove the line instead of emptying it out. This will leave you with a trailing newline, if your word is contained in the last line of a file. You can use
(\R)?.*(?:YOURSTRINGHERE).*(?(1)|\R)
To avoid this. It uses a conditional to either match the previous linebreak, or the following if there is none.
I had spent a while trying to narrow down a way of retrieving only web links from a few thousand lines that ended with either jpg or png.
If I use
%s/\(http.*\(jpg\|png\)\)\=\(.*\|\_s\)/\1/g|%s/\n\=
I can grab links just fine. The some thousands of lines are removed and replaced by only matching links. But if I remove the first \=, like here
%s/\(http.*\(jpg\|png\)\)\(.*\|\_s\)/\1/g|%s/\n\=
nothing in the file is changed or removed, and all the text is highlighted as a match.
If I remove it from the end of the pattern string, it concatenates every match onto a single line. I understand the basic reason for why this happens (being used by itself). That said, I am lost as to why it does not happen the same way when used in this specific case. (Meaning, the links do not get piled onto one line.)
My questions are:
Why do the links remain unchanged in the first example rather than replace the entire file or be removed entirely?
Why does specifying \n as an optional element not remove the nulls when the meaning of \= is "match 0 OR 1"?
Starting from the end of your regexp, with
%s/\n\=
You're substituting in every line 0 or 1 \n with //, hence and since you're not using the g flag, in any line that begins with anything but a \n, there'll be a match of the 0 part and nothing will be substituted with nothing: i.e. the line remains the same. (Led zeppelin quote)
It's equivalent to:
:%s/^\n
If you remove the \=, the first \n actually found in every line will be removed, that's why empty lines and the newlines at the end of your non empty lines get removed.
Now, here:
%s/\(http.*\(jpg\|png\)\)\=\(.*\|\_s\)/\1/g
The \= makes so that any string with 0 or 1 \(http.*\(jpg\|png\)\) patterns followed by anything (since you have \(.*\|\_s\)), will be replaced by the first saved pattern.
Basically, you're matching your whole file and preventing only this pattern: \(http.*\(jpg\|png\)\) from being removed.
When you remove \=, the 0 part of the match drops, and only in the lines that actually have the \(http.*\(jpg\|png\)\) pattern there will be a substitution of the matched pattern with itself from http up to jpg/png with anything after that being removed.
On a side note, if you save a pattern but don't use it in the substitution string, you're losing that pattern anyway.
If you actually only want to keep the http..jpg/png lines and remove the others, you can use the g! or v command:
:v/http.*jpg\|png/d
deletes all the lines that don't have the matched pattern.
I have a C++ source file containing many functions.
I want to find the beginning of every function quickly.
How can I form an expression for )newline{newline?
The newline symbol can be either one of the following:
\n
\r
\n\r
\r\n
Presumably, the same symbol is used all across the file, so instead of a single expression for all options combined, I need a single expression for each option.
I assume that a regular-expression can be used, but I'm not sure how.
Thanks
Barak, before we look at individual options, for all options, this will do it:
\)[\r\n]+{[\r\n]+
The [\r\n] is a character class that allows either of \r or \n. It is quantified with a + which means we are looking for one or more of these characters.
You said you want individual options, so this can be turned to:
\)\r\n{\r\n
\)\r{\r
\)\n{\n
\)\n\r{\n\r (this sequence of newlines is quite surprising)
If you simply want to use the regex search in VS to find the beginning of each function then this should work for you:
\)\r?\n\s*{\r?\n
Although that assumes the { is always on the next line with no white space before the line break.
This would be less strict where white space is concerned, but still expect the { to be on the next line and to be followed by a line break:
\)\s*\r?\n\s*{\s*\r?\n
And this would basically just look for the 2 brackets even if they're on the same line:
\)\s*\r?\n?\s*{
And if you expect there could be several line breaks between the 2 brackets:
\)\s*(\r?\n\s*)*{
Last example should find anything that could resemble the beginning of a method. But not sure how strict you want your search to be.
I have a word list in alphabetical order.
It is ranked as a column.
I do not use any programming languages.
The list in notepad format.
I need to match every similar words and take them on same line.
I use regex but I can't achieve correct results.
First list is like:
accept
accepted
accepts
accepting
calculate
calculated
calculates
calculating
fix
fixed
A list I want:
accept accepted accepts accepting
calculate calculated calculates calculating
fix fixed
This seems to work, but you will have to do Replace All multiple times:
Find (^(.+?)\s*?.*?)\R\2 and replace with \1\t\2. . matches newline should be disabled.
How it works:
It finds some characters at the start of line ^(.+?), then any linebreak \R, and those same characters again \2.
\s*?.*? is used to skip unnecessary characters after multiple Replace All. \s*? skips the first whitespace, and .*? any remaining chars on the line.
Match is replaced with \1\t\2, where \1 is anything matched in (^(.+?)\s*?.*?), and \2 is anything matched with (.+?). \t is used to insert tab character to replace linebreak.
How it breaks:
Note that this will not work well with different words with similar prefix, like:
hand
hands
handle
handles
This will be hand hands handle handles after 2 replaces.
I can imagine doing this programatically with limited success (take first word which comes as a root and if derived word with this root follows, place it on the same line, else take the word as a new root and put it to new line). This will still fail at irregular words where root is not the same for all forms.
Without programming there is a way only with (manual) preprocessing – if there are less than 4 forms for given word in the list, you insert blank line for each missing verb form, so there are always 4 lines for each word. Then you can use regex to get each such a quadruple into one line.