In a line like this:
something "one!", "and two", "and three", "and more"
How should I use the command ci" to replace "and three"?
I have tried some made up combos with no luck.
First navigate to the desired substring:
/three
or in the line:
f";;;;
(which moves to the next " 5 times)
or
f,;
Then, ci" zaps the string leaving the cursor for insert inside the empty quotes
Move into substring:
2fa
Then replace and three:
ci"
Related
I have a file which is as following
!J INCé0001438823
#1 A LIFESAFER HOLDINGS, INC.é0001509607
#1 ARIZONA DISCOUNT PROPERTIES LLCé0001457512
#1 PAINTBALL CORPé0001433777
$ LLCé0001427189
$AVY, INC.é0001655250
& S MEDIA GROUP LLCé0001447162
I just want to keep the last 10 characters of each line so that it becomes as following:-
0001438823
0001509607
0001457512
0001433777
0001427189
0001655250
:%s/.*\(.\{10\}\)/\1
: ex-commaned
% entire file
s/ substitute
.* anything (greedy)
. followed by any character
\{10\} exactly 10 of them
\( \) put them in a match group
/ replace with
\1 said match group
I would treat this as a shell script problem. Enter the following in vim:
:%! rev|cut -c1-10|rev
The :%! will pipe the entire buffer through the following filter, and then the filter comes straight from here.
for a single line you could use:
$9hd0
$ go to end of line
9h go 9 characters left
d0 delete to beginning of line
Assuming the é character appears only once in a line, and only before your target ten digits, then this would seem to work:
:% s/^.*é//
: command
% all lines
s/ / substitute (i.e., search-and-replace) the stuff between / and /
^ search from beginning of line,
. including any character (wildcard),
* any number of the preceding character,
é finding "é";
// replace with the stuff between / and / (i.e., nothing)
Note that you can type the é character by using ctrl-k e' (control-k, then e, then apostrophe, without spaces). On my system at least, this works in insert mode and when typing the "substitute" command. (To see the list of characters you can invoke with the ctrl-k "digraph" feature, use :dig or :digraph.
I'd like to replace double quotes " characters which come in pairs. Let me explain what I mean.
"Some sentence"
Here double quotes should be replaced because they come in pair.
"Some sentence
Here should not be replaced - there is no matching pair for the first quote character.
I'd like to replace first quote character with „.
❯ echo „ |hexdump -C
00000000 e2 80 9e 0a
And the second quote character with ”
❯ echo ” |hexdump -C
00000000 e2 80 9d 0a
Summing it up, the following:
Hi, "how
are you"
Should be the following after being replacement is made.
Hi, „how
are you”
I've come up with the following code, but it fails to work:
'sed -r s/(\")(.+)(\")/\1\xe2\x80\x9e\3\xe2\x80\x9d/g'
" hi " gives "„"”.
EDIT
As requested in the comments, here comes a sample from a file to be modified. Important note: the file is structured - perhaps it may help. The file is always a srt file, i.e. movie subtitle format.
104
00:10:25,332 --> 00:10:27,876
Kobieta mówi do drugiej:
"Widzisz to, co ja?"
105
00:10:28,001 --> 00:10:30,904
A tamta: "No to co?
Każdy wygląda tak samo."
Your expression doesn't work because you have three capturing groups: The three sets of (). You are putting the 1st (the first quote) and the 3rd (the last quote) in the output and ignoring the 2nd, which is the part you want to keep.
There's no reason to capture the quotes, since you don't want to inject them into the output. Only the bit in the middle needs to be captured.
There is also a flaw, the (.*) will itself match against a string containing a quote. So /"(.*)"/ would match the entire sequence "one"two", with the capture, (.*), matching one"two. Use [^"]* to match a sequence of non-quote characters.
Fixing this, and treating the entire text file as one line with -z, which only works if there are no nul characters in the text file, it appears this works:
sed -zE 's/"([^"]+)"/„\1“/g'
sed -rn ':a;s/"([^"]*)"/„\1”/g;/"/!{p;b;};$p;N;ba'
It substitutes all "xx" with „xx”. If the result contains no more " it is printed and we restart with next line. Else we concatenate the next line and we restart. The $p is just here to print the last lines if they contain a dangling ".
I need to find all pairs of strings that have the same pattern.
For example:
another string, that is not interesting
la-di-da-di __pattern__ -di-la-di-la
la-di-da-da-di-la __pattern__ -la-da-li-la
and yet another usual string
So I want to delete strings with __pattern__ inside.
I don't know how to do it just with builtin commands and now I have the function, that doesn't work properly:
function! DelDup(pattern)
echom a:pattern
redir => l:count
execute "normal! :%s/a:pattern//n\<cr>"
redir END
echo l:count
endfunction
Here I try to run ":%s/a:pattern//n" to find the count of occurrences of pattern in the text.
And at the same time I try to put it into the variable "l:count".
Then I tried to echo the count I got, but nothing happens when I try to do it.
So the last my problem in function writing is that I can't write the command execution result to variable.
If you have another solution -- please describe it to me.
Update:
Excuse me for bad description. I want to delete only strings, that has pattern-twins in text.
I'm not sure if I understand your question correctly, but I'm assuming you want to remove all lines where there are at least 2 matches. If that's the case you can use the following command:
:g/\(__pattern__.*\)\{2,}/d
How this works is that it deletes all the lines where there is a match (:g/../d).
The pattern is made up of a group (\(..\)) which needs to be matched at least 2 times (\{2,}). And the pattern has a .* at the end so it matches everything between the matches of the pattern.
There are many ways to count occurrences of a pattern, and I'm quite sure there exist a Q/A on the subject. Let's do it yet another way and chain with the next step. (Yes this is completely obfuscated, but it permits to obtain programmatically the information without the need to parse the localized result of :substitute after redirection.)
" declare a list that contain all matches
let matches = []
" replace each occurrence of the "pattern" with:
" the result of the expression "\=" that can be
" interpreted as the last ([-1]) element of the
" list "matches" returned by the function (add)
" that adds the current match (submatch(0)) to the
" list
:%s/thepattern/\=add(matches, submatch(0))[-1]/gn
" The big caveat of this command is that it modifies
" the current buffer.
" We need something like the following to leave it unmodified:
:g/thepattern/call substitute(getline('.'), 'thepattern', '\=add(counter, submatch(0))[-1]', 'g')
" Note however that this flavour won't work with multi-lines patterns
" Now you can test the number of matches or do anything fancy with it
if len(matches) > 1
" replaces matches with nothing
:%s/thepattern//g
endif
Only if you want to define this as a function you'll need to play with:
exe 'normal :%s/'.escape(a:pattern, '/\').'/replacement..../flags....'
I want a register to contain backslashes, but when I try they get removed. Example (for"first_name last_name" → "last_name, first name" in a file with all lines according to regex pattern:/^[^ \n]+ [^ \n]+$/gm):
Setting register 1 with let:
:let #1="%s/\(.*\) \(.*\)/\2, \1/g"
Displaying register 1 (inserted at the current line):
:put 1
The insertion:
"%s/(.*) (.*)/^B, ^A/g"
I want that register to contain backslashes so that it can be executes as a command in command mode.
Fix (escape the escapes):
:let #1="%s/\\(.*\\) \\(.*\\)/\\2, \\1/g"
Say I have this line of code:
$query = "SELECT * FROM table";
Is there a command in vi/vim which can instantly delete everything between quotes and position the cursor between them so I can start typing?
Use ci", which means: change what inside the double quotes.
You can also manipulate other text objects in a similar way, e.g.:
ci' - change inside the single quotes
ciw - change inside a word
ci( - change inside parentheses
dit - delete inside an HTML tag, etc.
More about different vim text objects here.
You can select between quotes and then delete (d), change (c) etc. using
vi"
Similarly, you can substitute braces, brackets, XML elements etc. thus:
vi(
vi{
vit
or to simply change/delete, do the corresponding di", ci" etc. Substituting a for i will encompassing the surrounding elements (so you mark or change the brackets and contents, for example)
I've made a plugin vim-textobj-quotes: https://github.com/beloglazov/vim-textobj-quotes
It provides text objects for the closest pairs of quotes of any type and supports quotes spanning multiple lines. Using only iq or aq it allows you to operate on the content of single ('), double ("), or back (`) quotes that currently surround the cursor, are in front of the cursor, or behind (in that order of preference). In other words, it jumps forward or backwards when needed to reach the quotes.
It's easier to understand by looking at examples (the cursor is shown with |):
Before: foo '1, |2, 3' bar; after pressing diq: foo '|' bar
Before: foo| '1, 2, 3' bar; after pressing diq: foo '|' bar
Before: foo '1, 2, 3' |bar; after pressing diq: foo '|' bar
Before: foo '1, |2, 3' bar; after pressing daq: foo | bar
Before: foo| '1, 2, 3' bar; after pressing daq: foo | bar
Before: foo '1, 2, 3' |bar; after pressing daq: foo | bar
The examples above are given for single quotes, the plugin works exactly the same way for double (") and back (`) quotes.
You can also use any other operators: ciq, diq, yiq, viq, etc.
Please have a look at the github page linked above for more details.
An addition to Brian's answer, you can also p(paste) and y(yank) the new value, so if you want to replace the value inside quotes with another value, you could do yi" on the selection that you want to copy, vi" to select the area that you want to replace and then just p to properly replace the value.
From already inside the quotes you can do
di"
Read it as delete inside "
The chosen answer is suitable ONLY for ViM but NOT for vi. The question is inaccurate as well because the author did not mention what is initial position of the cursor. If we assume that the cursor is inside the double quotes then for vi the answer will be:
T"ct"
Where:
T" - move back just after the " character
c - change command
t" - provide end position for c command, where it should stop erasing characters, in other words the range to change