I have taken this line of code from a book to format html code into a string-concatenated form for use in javascript in vim. I can't seem to understand what the numbers '1' and '2 represent and what the question marks are for at the end of the regexes. I'm used to seeing substitutions like %s/foo/bar/g, so the absence of forward slashes confuses me a bit. Summarized, I don't understand the '1' and '2', the question marks after the dollar sign and before the carriage return, and why the forward slashes are not used.
vmap <silent> ;q :s?^\(\s*\)\(.*\)\s*$? \1 + '\2'?<CR>
Forward slashes are typically used as the separator, but the substitute command uses the first character after the 's' as the separator, allowing it to be changed to anything. It seems the author thought that all the slashes might be confusing, so changed it to a '?'. This is how the command would appear with the more traditional forward slashes:
:s/^\(\s*\)\(.*\)\s*$/ \1 + '\2'/<CR>
So the above would mean, search forward, finding lines starting with any amount of whitespace, followed by any number of any characters, and any amount of whitespace, and then substitute it for <space><the first whitespace><space>+<space>'<the other characters>'. I think it was intending to strip any trailing whitespace, but in my testing it doesn't do that, because the .* will match everything to the end of the line.
As it was said in other answers, the / delimiter can be replaced by some other char : when many slashes are used in the command, it may be more clear; see a question about it, https://stackoverflow.com/a/36568901/3271687.
\1, \2, \n... match the nth sub-expression used in the pattern. A sub-expression is defined with \( and \). So in your example:
:s?^\(\s*\)\(.*\)\s*$? \1 + '\2'?
\s* --> note that this part can't be reached, it's useless
, \1 is replaced by the spaces found in \(\s*\), and \2 is replaced by all the chars (the whole rest of the line) found in \(.*\).
Related
For instance, if I wanted to a find and replace with strings containing backward or forward slashes, how would this be accomplished in vim?
Examples
Find & Replace is: :%s/foo/bar/g
what if I wanted to find all occurrences of <dog/> and replace it with <cat\>
Same way you escape characters most anywhere else in linuxy programs, with a backslash:
:%s/<dog\/>/<cat\\>
But note that you can select a different delimiter instead:
:%s#<doc/>#<cat\\>#
This saves you all typing all those time-consuming, confusing backslashes in patterns with a ton of slashes.
From the documentation:
Instead of the / which surrounds the pattern and replacement string, you
can use any other single-byte character, but not an alphanumeric character,
\, " or |. This is useful if you want to include a / in the search
pattern or replacement string.
%s:<dog/>:<cat>
You can replace the / delimiters if they become annoying for certain patterns.
Quote them with a backslash. Also, it often helps to use another delimiter besides slash.
:%s#<dog/>#<cat\\>#
or if you have to use slash as the substitute command delimiter
:%s/<dog\/>/<cat\\>/
I was looking for something similar, to search for register values containing the / character (to record a macro). The solution was to search using the ? token instead of the /.
The syntax is:
:%s/<dog\/>/<cat\\>/g
backslash slash backslash star
/(<- the prompt)\/\*
so after you type it looks like
/\/\*
I want to replace a line, that represents a part of mathematical equation:
f(x,z,time,temp)=-(2.0)/(exp(128*((x-2.5*time)*(x-2.5*time)+(z-0.2)*(z-0.2))))+(
with a new one similar to the above. Both new and old lines are saved in bash variables.
Main problem is that mathematical equation is full with special characters that do not allow proper search and replace in bash mode, even when I used as delimiter special character that is not used in equation.
I used
sed -n "s|$OLD|$NEW|g" restart.k
and
sed -i "s|$OLD|$NEW|g" restart.k
but all times I get wrong results.
Any idea to solve this?
There is only * in your pattern here that is special for sed, so escape it and do replacement as usual:
sed "s:$(sed 's:[*]:\\&:g' <<<"$old"):$new:" infile
if there are more special characters in your real sample, then you will need to add them inside bracket []; there are some exceptions like:
if ^ character: it can be place anywhere in [] but not first character, because ^ character at first negates the characters within its bracket expression.
if ] character: it should be the first character, because this character is also used to end the bracket expression.
if - character: it should be the first or last character, because this character is also can be used for defining the range of characters too.
I wanted to replace underscores with hyphens in all places where the character('_') is preceded and following by uppercase letters e.g. QWQW_IOIO, OP_FD_GF_JK, TRT_JKJ, etc. The replacement is needed throughout one document.
I tried to replace this in vim using:
:%s/[A-Z]_[A-Z]/[A-Z]-[A-Z]/g
But that resulted in QWQW_IOIO with QWQ[A-Z]-[A-Z]OIO :(
I tried using a sed command:
sed -i '/[A-Z]_[A-Z]/ s/_/-/g' ./file_name
This resulted in replacement over the whole line. e.g.
QWQW_IOIO variable may contain '_' or '-' line was replaced by
QWQW-IOIO variable may contain '-' or '-'
You had the right idea with your first vim approach. But you need to use a capturing group to remember what character was found in the [A-Z] section. Those are nicely explained here and under :h /\1. As a side note, I would recommend using \u instead of [A-Z], since it is both shorter and faster. That means the solution you want is:
:%s/\(\u\)_\(\u\)/\1-\2/g
Or, if you would like to use the magic setting to make it more readable:
:%s/\v(\u)_(\u)/\1-\2/g
Another option would be to limit the part of the search that gets replaced with the \zs and \ze atoms:
:%s/\u\zs_\ze\u/-/g
This is the shortest solution I'm aware of.
This should do what you want, assuming GNU sed.
sed -i -r -e 's/([A-Z]+)_([A-Z]+)/\1-\2/g' ./file_name
Explanation:
-r flag enables extended regex
[A-Z]+ is "one or more uppercase letters"
() groups a pattern together and creates a numbered memorized match
\1, \2 put those memorized matches in the replacement.
So basically this finds a chunk of uppercase letters followed by an underscore, followed by another chunk of uppercase letters, memorizes only the letter chunks as 2 groups,
([A-Z]+)_([A-Z]+)
Then it replays those groups, but with a hyphen in between instead of an underscore.
\1-\2
The g flag at the end says to do this even if the pattern shows up multiple times on one line.
Note that this falls apart a little in this case:
QWQW_IOIO_ABAB
Because it matches the first time, but not the second; the second part won't match because IOIO was consumed by the first match. So that would result in
QWQW-IOIO_ABAB
This version drops the + so it only matches one uppercase letter, and won't break in the same way:
sed -i -r -e 's/([A-Z])_([A-Z])/\1-\2/g'
It still has a small flaw, if you have a string like this:
A_B_C
Same issue as before, just one letter now instead of multiple.
I am a beginner at Vim and I've been reading about substitution but I haven't found an answer to this question.
Let's say I have some numbers in a file like so:
1
2
3
And I want to get:
(1)
(2)
(3)
I think the command should resemble something like :s:\d\+:........ Also, what's the difference between :s/foo/bar and :s:foo:bar ?
Thanks
Here is an alternative, slightly less verbose, solution:
:%s/^\d\+/(&)
Explanation:
^ anchors the pattern to the beginning of the line
\d is the atom that covers 0123456789
\+ matches one or more of the preceding item
& is a shorthand for \0, the whole match
Let me address those in reverse.
First: there's no difference between :s/foo/bar and :s:foo:bar; whatever delimiter you use after the s, vim will expect you to use from then on. This can be nice if you have a substitution involving lots of slashes, for instance.
For the first: to do this to the first number on the current line (assuming no commas, decimal places, etc), you could do
:s:\(\d\+\):(\1)
The \(...\) doesn't change what is matched - rather, it tells vim to remember whatever matched what is inside, and store it. The first \(...\) is stored in \1, the second in \2, etc. So, when you do the replacement, you can reference \1 to get the number back.
If you want to change ALL numbers on the current line, change it to
:s:\(\d\+\):(\1):g
If you want to change ALL numbers on ALL lines, change it to
:%s:\(\d\+\):(\1):g
You can do what you want with:
:%s/\([0-9]\)/(\1)/
%s means global search and replace, that is do the search/replace for every line in the file. the \( \) defines a group, which in turn is referenced by \1. So the above search and replace, finds all lines with a single digit ([0-9]), and replaces it with the matched digit surrounded by parentheses.
I have a line in a source file: [12 13 15]. In vim, I type:
:%s/\([0-90-9]\) /\0, /g
wanting to add a coma after 12 and 13. It works, but not quite, as it inserts an extraspace [12 , 13 , 15].
How can I achieve the desired effect?
Use \1 in the replacement expression, not \0.
\1 is the text captured by the first \(...\). If there were any more pairs of escaped parens in your pattern, \2 would match the text capture between the pair starting at the second \(, \3 at the third \(, and so on.
\0 is the entire text matched by the whole pattern, whether in parentheses or not. In your case this includes the space at the end of your pattern.
Also note that [0-90-9] is the same as [0-9]: each [...] collection matches just one character. It happens to work anyway, because in your data ‘a digit followed by a space’ matches in the same places as ‘2 digits followed by a space’. (If you actually needed to only insert commas after 2 digits, you could write [0-9][0-9].)
"I have a line in a source file:..."
then you type :%s/... this will do the substitution on all lines, if it matched. or that is the single line in your file?
If it is the single line, you don't have to group, or [0-9], just :%s/ \+/,/g will do the job.
The fine answers already point interesting solutions, but here's another one,
making use of the \zs, which marks the start of the match. In this pattern:
/[0-9]\zs /
The searched text is /[0-9] /, but only the space counts as a match. Note
that you can use the class \d to simplify the digit character class, so the
following command shall work for your needs:
:s/\d\d\zs /, /g ; matches only the space, replace by `, '
You said you have multiple lines and these changes are only to certain lines.
You can either visually select the lines to be changed or use the :global
command, which searches for lines matching a pattern and applies a command to
them. Now you'd need to build an expression to match the lines to be changed
in a less precise as possible way. If the lines that begins with optional
spaces, a [ and two digits are the only lines to be matched and no other
ones, then this would work for you:
:g/\s*[\d\d/s/\d\d\zs /, /g
Check the help for pattern.txt for \ze and similar and
:global.
Homework: use the help to understand \zs and see how this works:
:s/\d\d\zs\ze /,/g