want to search and replace in vim, the /find finds the pattern but :s%//g will not?
have a script that monitors software raid (if interested check it out https://dwaves.org/2019/09/06/linux-server-monitor-software-raid-mail-notification-on-failure/)
echo "=== smart status of all drives ==="| tee -a /scripts/monitor/raid_status_mail.log
# want to search and replace the /path/to/file.sh with $LOGFILE
# searching for the pattern works like charm
/\/scripts\/monitor\/raid_status_mail.log
# but replacing it won't
:s%/\/scripts\/monitor\/raid_status_mail\.log/\$LOGFILE/g
# what does one do wrong?
should replace /scripts/monitor/raid_status_mail.log with $LOGFILE
The substitution operation needs to be prefixed with %s and not the other way around as s%. So doing
%s/\/scripts\/monitor\/raid_status_mail\.log/\$LOGFILE/g
should work as expected. Or just the Vim's equivalent ex in command line mode as
printf '%s\n' "%s/\/scripts\/monitor\/raid_status_mail\.log/\$LOGFILE/g" w q | ex -s file
You inverted the beginning s%. Use %s instead.
Also, you use / as separation for the different fields, it works but makes the command less readable. You can replace the separation character by anything else. You could use : for example:
%s:/scripts/monitor/raid_status_mail.log:$LOGFILE:g
One last tip: install vim-over
This will highlight your searches in live while replacing something in vim.
Related
I'm using a shell & TCL script to login to a switch and get the output of certain commands and in some places I can see ^D coming up. I tried to use the dos2unix utility but still it didn't go away.
Eth1/37 NOM: xcvrAbsen routed auto auto --
^DEth1/38 NOM: xcvrAbsen routed auto auto --
Eth1/39 NOM: xcvrAbsen routed auto auto --
Eth101/1/45 eth 1000 NOM:NO_PATCHING CABLE
^DEth101/1/46 eth 1000 NOM:NO_PATCHING CABLE
Eth101/1/47 eth 1000 NOM:NO_PATCHING CABLE
How can this be eliminated, are there any standard tools like dos2unix which can get rid of such data?
What I'm trying to do is to compare two files which are from the same switch and the same command and the same output, but due to these ^D, Vimdiff shows it as different lines.
How to get this eliminated?
Command I'm using is something like this:
$cdir/ciscocmd -Y -u $operator -p $password -s $password -t $switch -r rfc_sa_commands | sed 's/^^D//' > $switch.$NOW
dos2unix removes carriage returns, no other control characters.
The tool to remove all occurrences of an arbitrary character is called tr.
tr -d '\004' <inputfile >outputfile
This assumes you have literal ctrl-D characters, not sequences of caret ^ and D. The tr utility cannot remove a specific sequence; it just processes individual characters. To remove a sequence, you'd need
sed 's/\^D//g' inputfile >outputfile
where the backslash is required because the caret alone has a special meaning in regular expressions (it matches beginning of line). Doubling it does not escape it; ^^ probably still just matches beginning of line, though it's not really well-defined, and could introduce apparently random behavior.
Even if the special character is visible as '^D', it may be NOT catchable like this.
Interesting readings, are:
https://en.wikipedia.org/wiki/ASCII#Character_groups
https://en.wikipedia.org/wiki/End-of-Transmission_character
I think a way to do it would be:
<your command>|sed -e 's/\x04//g'
Does it solve your issue?
I am not that good on linux shell script and I need little help.
I want to edit a file via script (finding the line and edit).
The Original line is:
# JVM_OPTS="$JVM_OPTS -Djava.rmi.server.hostname=< hostname >"
I want to uncomment and replaye hostname with 127.0.0.1
JVM_OPTS="$JVM_OPTS -Djava.rmi.server.hostname=127.0.0.1"
You can refer to the set command, change the filename with the name you are working at,
sed -i 's## JVM_OPTS="$JVM_OPTS -Djava.rmi.server.hostname=< hostname >"#JVM_OPTS="$JVM_OPTS -Djava.rmi.server.hostname=127.0.0.1"#' filename
Fine answers, but they don't do anything by way of TEACHING the gentleman how and why it works.
If you were using the mundane text editor, ed, you would use three commands after invoking the command "ed filename":
s/^# //
s/< hostname>/127.0.0.1/
w
So, you can use a pipe to submit those commands directly to ed, specifying "-" as its first argument so that it doesn't bother you by reporting character counts upon reading in and writing out the file:
( echo 's/^# //'; echo 's//127.0.0.1/'; echo w ) | ed - filename
You don't need to echo 'q' also because ed will automatically quit when it runs out of input or encounters "end of file" (you can simulate this on the keyboard by just hitting the CTRL-D key rather than actually typing q ).
Here's one way to do it:
sed -i -e 's/# \(JVM_OPTS=.*=\).*/\1127.0.0.1"/' path/to/file
That is, replace the line with the text captured within the group \(JVM_OPTS=.*=\), so everything from JVM_OPTS= until another = sign, and append 127.0.0.1" to the end.
If there might be other lines in the file starting with # JVM_OPTS=,
then you could make the pattern matching more strict, for example:
sed -i -e 's/# \(JVM_OPTS="$JVM_OPTS -Djava.rmi.server.hostname=\).*/\1127.0.0.1"/' path/to/file
I am trying to understand how
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
worked and what the pieces mean. The man page I read just confused me more and I tried the info sai Id but had no idea how to work it! I'm pretty new to Linux. Debian is my first distro but seemed like a rather logical place to start as it is a root of many others and has been around a while so probably is doing stuff well and fairly standardized. I am running Wheezy 64 bit as fyi if needed.
The sed command is a stream editor, reading its file (or STDIN) for input, applying commands to the input, and presenting the results (if any) to the output (STDOUT).
The general syntax for sed is
sed [OPTIONS] COMMAND FILE
In the shell command you gave:
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
the sed command is s/\^\[/\o33/g;s/\[1G\[/\[27G\[/' and /var/log/boot is the file.
The given sed command is actually two separate commands:
s/\^\[/\o33/g
s/\[1G\[/\[27G\[/
The intent of #1, the s (substitute) command, is to replace all occurrences of '^[' with an octal value of 033 (the ESC character). However, there is a mistake in this sed command. The proper bash syntax for an escaped octal code is \nnn, so the proper way for this sed command to have been written is:
s/\^\[/\033/g
Notice the trailing g after the replacement string? It means to perform a global replacement; without it, only the first occurrence would be changed.
The purpose of #2 is to replace all occurrences of the string \[1G\[ with \[27G\[. However, this command also has a mistake: a trailing g is needed to cause a global replacement. So, this second command needs to be written like this:
s/\[1G\[/\[27G\[/g
Finally, putting all this together, the two sed commands are applied across the contents of the /var/log/boot file, where the output has had all occurrences of ^[ converted into \033, and the strings \[1G\[ have been converted to \[27G\[.
I am experimenting with wildcards in bash and tried to list all the files that start with "xyz" but does not end with ".TXT" but getting incorrect results.
Here is the command that I tried:
$ ls -l xyz*[!\.TXT]
It is not listing the files with names "xyz" and "xyzTXT" that I have in my directory. However, it lists "xyz1", "xyz123".
It seems like adding [!\.TXT] after "xyz*" made the shell look for something that start with "xyz" and has at least one character after it.
Any ideas why it is happening and how to correct this command? I know it can be achieved using other commands but I am especially interested in knowing why it is failing and if it can done just using wildcards.
These commands will do what you want
shopt -s extglob
ls -l xyz!(*.TXT)
shopt -u extglob
The reason why your command doesn't work is beacause xyz*[!\.TXT] which is equivalent to xyz*[!\.TX] means xyz followed by any sequence of character (*) and finally a character in set {!,\,.,T,X} so matches 'xyzwhateveryouwant!' 'xyzwhateveryouwant\' 'xyzwhateveryouwant.' 'xyzwhateveryouwantT' 'xyzwhateveryouwantX'
EDIT: where whateveryouwant does not contain any of !\.TX
I don't think this is doable with only wildcards.
Your command isn't working because it means:
Match everything that has xyz followed by whatever you want and it must not end with sequent character: \, .,T and X. The second T doesn't count as far as what you have inside [] is read as a family of character and not as a string as you thought.
You don't either need to 'escape' . as long as it has no special meaning inside a wildcard.
At least, this is my knowledge of wildcards.
So far, I have been manually refactoring code by using the find-and-replace operation
%s:/stringiwanttoreplace/newstring/g
in vim.
But this is a slow and laborious process if I have stringiwanttoreplace in many files inside a specific directory.
My current/typical slow and laborious process involves a grep:-
grep -rn "stringiwanttoreplace" .
in my terminal to reveal all the locations/filenames where stringiwanttoreplace are; and now that I know which files contain stringiwanttoreplace, I will open each file one-by-one to perform the find-and-replace operation in each file.
Is there a more efficient workflow (in vim) to get this done?
CLARIFICATION: I would prefer a vim-based solution instead of a bash script/one-liner.
Here's the full sequence of commands that I would use:
/stringiwanttoreplace
:vimgrep /<c-r>// **
:Qargs
:argdo %s//newstring/g
:argdo update
In the first line, we search for the target pattern. That populates the last search pattern register (:help quote/), which means that we won't have to type it out in full again.
The :vimgrep command searches the entire project for the specified pattern. Type <c-r>/ as ctlr+r followed by / - this inserts the contents of the last search pattern register onto the command line. The first and last / symbols are delimiters for the search field. The trailing ** tells Vim to look inside every file and directory below the current directory.
At this point, the quickfix list will be populated with search matches from all matching files. :Qargs is a custom command, which populates the argument list with all of the files listed in the quickfix list. Here's the implementation:
command! -nargs=0 -bar Qargs execute 'args ' . QuickfixFilenames()
function! QuickfixFilenames()
" Building a hash ensures we get each buffer only once
let buffer_numbers = {}
for quickfix_item in getqflist()
let buffer_numbers[quickfix_item['bufnr']] = bufname(quickfix_item['bufnr'])
endfor
return join(values(buffer_numbers))
endfunction
Add that to your vimrc file.
Having run :Qargs, our argument list should now contain all of the files that include our target string. So we can run the substitution command with :argdo, to execute the command in each file. We can leave the search field of the substitution command blank, and it will automatically use the most recent search pattern. If you want, you could include the c flag when you run the substitution command, then you'll be prompted for confirmation.
Finally, the :argdo update command saves each file that was changed.
As #Peter Rincker pointed out, you should ensure that Vim's 'hidden' option is enabled, otherwise it will raise an error when you try to switch to another buffer before writing any changes to the active buffer.
Also, note that the last 3 commands can be executed in a single command line, by separating them with a pipe character.
:Qargs | argdo %s//replacement/gc | update
The :Qargs command is pinched from this answer (by me), which in turn was inspired by this answer by DrAl. A very similar solution was posted by #ib, which suggests to me that Vim should really implement something like :quickfixdo natively.
If you really want to do it in Vim you can follow the suggestions here.
You can call this from within Vim (:!find ...) but you don't need to:
find . -type f | xargs sed -i 's/stringiwanttoreplace/newstring/g'
Fine-tune the file selection with the dozens of parameters described in
man find
(e.g., replace only in HTML files: -name \*.html)
This solution will try to attempt the replacement in all files. You can filter that through grep before, but that is just doing twice the work for no gain.
By the way: sed uses almost the same syntax for regular expressions as Vim (stemming from the same history).
You could open all the files and type
:bufdo :s/stringiwanttoreplace/newstring/g
It performs the search/replace in all your buffers.
You don't need vim to do this, you can use command line tools. Using sed in a loop on the list of files to do this for you automatically. Something like this:
for each in `grep -l "stringiwanttoreplace" *` ;
do
cat $each | sed -e "s/stringiwanttoreplace/newstring/g" > $each
; done
vim7 has recursive grep built-in
:vimgrep /pattern/[j][g] file file1 file2 ... fileN
the result will be shown in a quickfix-window (:help quickfix)
to do the search recursively use the **-wildcard like
**/*.c to search through the current folder and recursively through all subdirectories.