The objective is to find and list anything with "messages" and/or "error.log" etc.. in the beginning then list both "messages.1..99" and "error.log.1..99" using regular expressions.
This command works for however, it would require me to make many -or searches, but to simplify, I would like to have multiple in a set within the search. Like for instance:
# find /var/log -maxdepth 1 -type f -size +1M -name [messages|error.log|secure.log|kern.log...]?[0-9]|[0-9][0-9] ! -iname "*.gz"
not
# find /var/log -maxdepth 1 -type f -size +1M -name "messages?[0-9]" -o -name "messages?[0-9][0-9]"
How might I perform this command with regular expressions?
# find /var/log -maxdepth 1 -type f -size +1M -name "[messages,error.log,kern,secure]?[0-9]" ! -iname "*.gz"
My attempt with regex doesn't print anything in standard out:
# find /var/log -maxdepth 1 -type f -size +1M -regex -name "[messages,error,kern,secure]?[0-9]" ! -iname "*.gz"
Try this:
find /var/log -maxdepth 1 -type f -size +1M -type f -regextype egrep -regex '.*(messages|error|kern|secure)\.[0-9]+.*' -not -name \*gz
Related
I am doing some practice on find command but I don't get the expected result when I attempt to use -execoption of it. The command I wrote just works without -exec option as the following:
$ find ~ \( -type f -not -perm 0600 \) -or \( -type d -name 'D*' \)
/home/baki/.bashrc
/home/baki/.bash_logout
/home/baki/.cache/motd.legal-displayed
/home/baki/.config/wslu/baseexec
/home/baki/.config/wslu/oemcp
/home/baki/.gitconfig
/home/baki/.landscape/sysinfo.log
/home/baki/.motd_shown
/home/baki/.profile
/home/baki/.ssh/known_hosts
/home/baki/.sudo_as_admin_successful
/home/baki/ssh_start
/home/baki/token
However, when I add the -exec option to the end of the command, it doesn't give any output:
find ~ \( -type f -not -perm 0600 \) -or \( -type d -name 'D*' \) -exec ls -l '{}' ';'
I have searched about it but I couldn't find a piece of useful information that can solve my problem.
Is my command wrong or is it about something else?
Thank you for your help.
The default -and operation has higher precedence than -or. Use extra parentheses:
find ~ \( \( -type f -not -perm 0600 \) -or \( -type d -name 'D*' \) \) -exec ls -l '{}' ';'
You can probably omit the inner parentheses in this case.
First one :
We found several files and we have to copy that to kat4 and here is code, but it doesn't seem to work corectly
find /home/imk-prac/ -type f -size -13c -name '*\?plik\?*' -exec cp {} /home/inf-19/aduda/\*kat1\*/\*kat2\*/\*kat4\*/ \; 2> /dev/null
'cp' I assume that it is copy, but I don't know what 'exec' and '{}' do.
Second one:
find /home/imk-prac/ \( -type f -size -13c -name '*\?plik\?*' \) -o\( -type d -name '\[Kolo1\]*' \)2> /dev/null
Generally,I understand this line (except for '2' and '-o') , but I want to add looking for files which were modificated in less that 30 days and here is what I wanted to combine with upper command :
find /home/imk-prac/ -type f -mtime -30 -exec ls -l {} \; > /dev/null
As a result I wrote it down as:
find /home/imk-prac/ \( -type f -size -13c -name '*\?plik\?' -mtime -30 -exec ls -l{}\) -o \( -type d -name '\[Kolo1\]*' \) 2> /dev/null
but it doesn't work
Moreover, I wanted to add looking for files with speciefied quantity of symbols and I found this command:
grep -Po '(^|\s)\S{64}(\s|$)' file
But I have no idea how to combine all of those 3 upper commands.
I will be grateful for any help, thank you for your time!
Im trying to find all files modified during de last 24 hours in /var/www/vhost directory.
That is working ok with the find command, then, I want to filter the list because i don't want jpg files, jpeg files and so on.
Now i have this and it's working ok:
find /var/www/vhosts/ -ctime 0 -type f | grep -ve ".jpg$" | grep -ve ".jpeg"
I guess (and know) there's a better solution to my problem.
Any help?
change your find command itself to
find /var/www/vhosts/ -not \( -name "*.jpeg" -o -name "*.jpg" \) -ctime 0 -type f
You can do it all with one find command:
find /var/www/vhosts -ctime 0 -type f \! -iname \*.jpg \! -iname \*.jpeg
Use -regex and ! (negation):
find $DIR -regextype posix-extended ! -regex '.*\.(gif|jpg|pdf|png)$'
In the following command i want to search only only the directories which are non hidden how can i do this using the following command .Iwant to ignore hidden directories while searching the log file
find /home/tom/project/ -name '.log.txt'
ls /home/tom/project/
dir1
dir2
.backup
.snapshot/
.ignore/
Try
find /home/tom/project -type d -name '.*' -prune -o -name .log.txt -print
This will find all files but ignore those that start with a dot so hidden files.
find /home/tom/project/ -type f \( -iname ".log.txt" ! -iname ".*" \)
EDIT:
If the above those not work, this should do the trick. It has a better regex.
find /home/tom/project/ \( ! -regex '.*/\..*' \) -type f -name ".log.txt"
EDIT2:
The following will exclude hidden folders but will search for the hidden files that have the requested pattern:
find /home/tom/project/ \( ! -regex '.*/\..*/..*' \) -type f -name ".log.txt"
EDIT3:
The grep solution :) if this doesn't work i'm lost :)
find /home/tom/project/ \( ! -regex '.*/\..*/..*' \) -exec grep -l ".log.txt" {} \;
EDIT4:
Have you tried the simple solutions?
find /home/tom/project/ -type f -name ".log.txt"
OR
find /home/tom/project/ -type f -name "*" -exec grep -l ".log.txt" {} \;
In my home dir , i have folders like .cpan .cpcpan and they are also showing.
How i hide them . i am using
find /home -mindepth 1 -maxdepth 1 -type d -printf "%f\n"
Prune them.
find /home -mindepth 1 -maxdepth 1 -type d \( -name '.*' -prune -o -printf "%f\n" \)