In Vim, I want to do a search and replace that includes:
[0-9]*
And in the replace part, I want whatever was in that vague range to be carried over and used.
For example let's say I have this:
getNumber(42).roundToTenth();
getNumber(43).roundToTenth();
getNumber(44).roundToTenth();
getNumber(45).roundToTenth();
getNumber(46).roundToTenth();
getNumber(47).roundToTenth();
and want to do a search and replace to change it to
DontGetNumber(42).roundDownTenth();
DontGetNumber(43).roundDownTenth();
DontGetNumber(44).roundDownTenth();
DontGetNumber(45).roundDownTenth();
DontGetNumber(46).roundDownTenth();
DontGetNumber(47).roundDownTenth();
How do I do that?
I think maybe you mean [0-9].* (number followed by zero or more characters). You can use capturing parentheses and use the backreference in the substitution.
s/[0-9]\(.*\)/\1/
The \1 will be whatever was captured. Change the replacement expression as needed.
s/getNumber(\([0-9]*\)).*/DontGetNumber(\1).roundDownTenth();
You really want to use search/replace for this?
Personnally, I'd rather do something like :%g/Number/norm! ^~IDont^[fT2sDown^[
use a record with q[register] and fill the part after norm! with c-r [register]
Related
I'm writing a shell script to modify a file and I have a line something like this in it:
sed s/here \(.*\n\)/gone \1/g
Unfortunately, the search seems to match the longest string (i.e., it goes all the way to the last \n -- thus giving me just one replacement) but I want it to match only up to the first \n it finds (so I can get replacements on every line).
Is this possible?
Thanks for your help!
Looks like you want the feature called non-greedy (or lazy) match. Unfortunately sed does not provide such feature. To emulate it you need to search for anything except separator match until separator match. Like this:
s/here \([^\n]*\n\)/gone \1/g
When i need to replace a string with a new string in vim.
First i would use search-mode to check that the search pattern is correct.
/search pattern
Then use the 's' command to do substitution.
:%s/search pattern/new string/
The search pattern need to type twice. If it is too complex, it would be boring.
Is there a method to avoid this?
You can simply omit the pattern in the substitution command, e.g.
:%s//new string/
This is documented in :help last-pattern (emphasis mine):
The last used pattern and offset are remembered. They can be used to
repeat the search, possibly in another direction or with another
count. Note that two patterns are remembered: One for 'normal' search
commands and one for the substitute command ":s". Each time an empty
pattern is given, the previously used pattern is used.
Also (in addition to Marco Baldelli's correct answer), the last search pattern is stored in the special register /. You can insert this in the command-line via Ctrl + R, followed by /. (This also works in insert mode, and also with other registers.) It's helpful when you want to tweak your search pattern before substituting.
I have several functions that start with get_ in my code:
get_num(...) , get_str(...)
I want to change them to get_*_struct(...).
Can I somehow match the get_* regex and then replace according to the pattern so that:
get_num(...) becomes get_num_struct(...),
get_str(...) becomes get_str_struct(...)
Can you also explain some logic behind it, because the theoretical regex aren't like the ones used in UNIX (or vi, are they different?) and I'm always struggling to figure them out.
This has to be done in the vi editor as this is main work tool.
Thanks!
To transform get_num(...) to get_num_struct(...), you need to capture the correct text in the input. And, you can't put the parentheses in the regular expression because you may need to match pointers to functions too, as in &get_distance, and uses in comments. However, and this depends partially on the fact that you are using vim and partially on how you need to keep the entire input together, I have checked that this works:
%s/get_\w\+/&_struct/g
On every line, find every expression starting with get_ and continuing with at least one letter, number, or underscore, and replace it with the entire matched string followed by _struct.
Darn it; I shouldn't answer these things on spec. Note that other regex engines might use \& instead of &. This depends on having magic set, which is default in vim.
For an alternate way to do it:
%s/get_\(\w*\)(/get_\1_struct(/g
What this does:
\w matches to any "word character"; \w* matches 0 or more word characters.
\(...\) tells vim to remember whatever matches .... So, \(w*\) means "match any number of word characters, and remember what you matched. You can then access it in the replacement with \1 (or \2 for the second, etc.)
So, the overall pattern get_\(\w*\)( looks for get_, followed by any number of word chars, followed by (.
The replacement then just does exactly what you want.
(Sorry if that was too verbose - not sure how comfortable you are with vim regex.)
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.
I'm trying to search and replace $data['user'] for $data['sessionUser'].
However, no matter what search string I use, I always get a "pattern not found" as the result of it.
So, what would be the correct search string? Do I need to escape any of these characters?
:%s/$data['user']/$data['sessionUser']/g
:%s/\$data\[\'user\'\]/$data['sessionUser']/g
I did not test this, but I guess it should work.
Here's a list of all special search characters you need to escape in Vim: `^$.*[~)+/
There's nothing wrong with with the answers given, but you can do this:
:%s/$data\['\zsuser\ze']/sessionUser/g
\zs and \ze can be used to delimit the part of the match that is affected by the replacement.
You don't need to escape the $ since it's the at the start of the pattern and can't match an EOL here. And you don't need to escape the ] since it doesn't have a matching starting [. However there's certainly no harm in escaping these characters if you can't remember all the rules. See :help pattern.txt for the full details, but don't try to digest it all in one go!
If you want to get fancy, you can do:
:%s/$data\['\zsuser\ze']/session\u&/g
& refers to the entire matched text (delimited by \zs and \ze if present), so it becomes 'user' in this case. The \u when used in a replacement string makes the next character upper-case. I hope this helps.
Search and replace in vim is almost identical to sed, so use the same escapes as you would with that:
:%s/\$data\['user'\]/$data['session']/g
Note that you only really need to escape special characters in the search part (the part between the first set of //s). The only character you need to escape in the replace part is the escape character \ itself (which you're not using here).
The [ char has a meaning in regex. It stands for character ranges. The $ char has a meaning too. It stands for end-line anchor. So you have to escape a lot of things. I suggest you to try a little plugin like this or this one and use a visual search.