I need to repoint a mount from /IsilonAsperaVOD/data to /Isilon/asperavod/data. Using vim I do:
:s/IsilonAsperaVOD/Isilon\\/asperavod/g
which returns the error. I can type out the path or copy/past with same effect...? how do I do this
You don't need :global (as in your own answer), prefixing a :% range for the entire file will suffice (without that, the :s only applies to the current line).
The problem with your original attempt is that you've escaped twice; there needs to be a single backslash in front of the /:
:%s/IsilonAsperaVOD/Isilon\/asperavod/g
But, it's better to avoid the escaping altogether by using a different separator character:
:%s#IsilonAsperaVOD#Isilon/asperavod#g
this worked
:g/IsilonAsperaVOD/s//IsilonDenver\/asperavod/g
so, single escape for directory slash, and g (global) and then :s, instead of :s (replace) even with %
Related
I have a bunch of t_ prefixed fonction names in a header file and i would like to remove this prefix from some of them.
I have ended by typing :
:s/\(\s+\)t_\([^(]+\)(/\1\2(/c
But vim complains with Pattern not found
What is wrong in my pattern ?
You forgot to put a backslash before + to give it its quantifier meaning. You can also probably simplify it using \< to match the start of a word instead of capturing spaces.
I suggest breaking the problem down, first the search:
/\v\s+\zst_\ze.*\(
Now the substitution:
:s///\c
We can reuse a search pattern simply typing an empty search on the substitution
OBS: I suppose your functions have () at the end of the line, so .*().
Another thing; \zs and \ze to match only t_, for more see: :h \zs.
If you are using neovim you can see what happens befor hitting enter, just put these lines on your init.vim:
if has("nvim")
set inccommand=nosplit
endif
I am little new to Vim world. I am trying to substitute *=, ~=(actually [special char]=) in to [whatever is symbol]=(adding space both sides). Here is my substitute command:
:%s/[~,\*]=/ = /g
the problem in this case is that I am not able to add respective special symbol before the equal sign. Can you help me...
This is a classic capture and replace use case. Capture the symbol part by enclosing it in \(...\), and then reference it in the replacement part via \1. You'll find more details at :help s/\1 (or :help :substitute in general):
:%s/\([~,\*]\)=/ \1= /g
Alternatively, you can start the match only on the = with \zs. This asserts that the symbol part is there, but as it isn't included in the match, you don't need to reference it:
:%s/[~,\*]\zs=/ = /g
The same trick can be applied with \ze at the end. As you can see, this often results in shorter commands.
This is probably the simplest answer to your question:
:%s/[~,\*]=/ & /
An& in the replace segment means 'entire match'.
In vi (from cygwin), when I do searching:
:%s/something
It just replaces the something with empty string like
:%s/something// .
I've googled for a while but nothing really mentions this. Is there anything I should add to the .vimrc or .exrc to make this work?
Thanks!
In vi and vim, when you search for a pattern, you can search it again by simply typing /. It is understood that the previous pattern has to be used when no pattern is specified for searching.
(Though, you can press n for finding next occurence)
Same way, when you give a source (pattern) and leave the replacement in substitute command, it assumes that the replacement is empty and hence the given pattern is replaced with no characters (in other words, the pattern is removed)
In your case, you should understand that % stand for whole file(buffer) and s for substitute. To search, you can simply use /, followed by a pattern. To substitute , you will use :s. You need not confuse searching and substituting. Hence, no need for such settings in ~/.exrc. Also, remember that / is enough to search the whole buffer and % isnt necessary with /. / searches the entire buffer implicitly.
You may also want to look at :g/Pattern/. Learn more about it by searching :help global or :help :g in command line.
The format of a substitution in vim is as follows:
:[range]s[ubstitute]/{pattern}/{string}/[flags] [count]
In your case you have omitted the string from the substitution command and here what vim documentation stated about it:
If the {string} is omitted the substitute is done as if it's empty.
Thus the matched pattern is deleted. The separator after {pattern}
can also be left out then. Example: >
:%s/TESTING This deletes "TESTING" from all lines, but only one per line.
For compatibility with Vi these two exceptions are allowed:
"/{string}/" and "\?{string}?" do the same as "//{string}/r".
"\&{string}&" does the same as "//{string}/".
E146
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. Example: >
:s+/+//+
In other words :%s/something and :%s;something or :%s,something have all the same behavior because the / ; and , in the last examples are considered only as SIMPLE SEPARATOR
I am trying to run this command in vi
:s/href="\//href="http:\/\/website.com\/folder\/subfolder\//g
but got this error E486: Pattern not found: href="\/
What am i doing wrong?
That error means pretty much what it says. vi didn't find any pattern href="/ (ignoring escapes) in your file.
Sometimes it's easier to use something besides / for the search delimiter if your search has a lot of slashes, so you don't need to escape them all. Try replacing the / delimiter with # instead, like this:
s#href="/#href="http://website.com/folder/subfolder/#g
Then maybe you can more easily see what's wrong with your pattern:
becouse there are many '/' chars, try use another delimiter, ex ',':
:s,some/pattern/with/slashes,new/string,g
On another note. That substition worked for me. Just copied and pasted. Are you on the same line that you are trying to perform the substitution on? the 'g' is meant globally on the line you are on. If you need to perform the search and replace on the file the use :%s/
I always wanted to know, how you can substitute within given parameters.
If you have a line like this:
123,Hello,World,(I am, here), unknown
and you wnat to replace World with Foobar then this is an easy task: :%s/World/Foobar/
Now I wonder how I can get rid of a , which is wihtin the ( ).
Theoretically I just have to find the first occurance of ( then substitute the , with a blank until ).
Try lookahead and lookbehind assertions:
%s/([^)]*\zs,\ze.*)//
(\zs and \ze tell where pattern starts and end)
or
%s/\(([^)]*\)\#<=,\(.*)\)\#=//
The first one is more readable, the second one uses \( ... \) groupings with parentheses inside groups which makes it look like obfuscated, and \#<= which apart from being a nice ASCII-art duck is the lookbehind operator, and \#= that is the lookahead operator.
References: :help pattern (more detail at :help /\#=, :help /\ze, etc.)
You use the GUI and want to try those commands? Copy them into the clipboard and run :#+ inside Gvim.
Modifying slightly the answer of #Tom can give you a quite good and "a bit" more readable result :
%s/\(.*\)(\(.*\),\(.*\))\(.*\)/\1(\2\3)\4/
That way you will have : in \1 will store what is at the left outside of the parenthesis, \4 what is at the right outside of the parenthesis and \2 and \3 what is inside the parenthesis, respectively on the left (\2) and on the right (\3).
With that you can easily swap your elements if your file is organised as column.
You can also select the text you want to change (either with visual or visual-block modes) and enter the : to start the replace command. vi will automatically start the command with :'<,'> which applies the command to the selected area.
Replacing a , can be done with:
:'<,'>s/,/ /g
For your example, this is the same thing as suggested by #ubuntuguy
%s/\(.*\)(\(.*\),\(.*\)/\1(\2\3
This will do the exact replacement you want.
Yet another approach, based on the fact that actually you want to substitute only the first occurrence of , inside the parenthesis:
:%s#\((.\{-}\),#\1 #
Explanation:
:%s for substitution in the whole file (don't use % if you want to work only with the current line)
we can use # or : as a delimiter to make the command more readable
in (.\{-} we ask to find any symbol (dot) after the left parenthesis and the rest stands for 0 or more occurrence (as few as possible) of this symbol. This expression is put inside \(...\) to be able to refer to this group as \1 in the future.