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.
Related
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
I need to search for multiple pattern of files and check their mtime and if it morethan 30 days then delete all the files. I am using the below command but it's deleting only one pattern of file and not all. Kindly let me know where is the mistake in my command.
find /root -type f \( -name "*.tgz" -o -name "*.bz2" \) -mtime +30 -print -exec rm '{}' +
Try escaping parentheses in the command and adding a wildcard character:
find /root -type f \( -name "*.tgz" -o -name "*.bz2" \) -mtime +30 -exec rm {} \+
I'm trying to do a find and replace function, finding files which match a criteria then find/replace text within them.
Find statement (works find and returns list of files):
find / -type f -name "*.properties" -o -name "*.xml" -not \( -path '/tmp/*' -o -path '/var/tmp/*' \)
Sed find/replace:
sed -i 's/find/replace/g' {} \;
Putting together:
find / -type f -name "*.properties" -o -name "*.xml" -not \( -path '/tmp/*' -o -path '/var/tmp/*' \) -exec sed -i 's/10\.32\.19\.156/10.32.19.165/g' {} \;
However this does not seem to work. Removing some 'find' parameters causes it to work, for example this works:
find / -type f -name "*.properties" -exec sed -i 's/10\.32\.19\.156/10.32.19.165/g' {} \;
How can I get sed to work with the extended 'find' parameters?
Currently these two 'find' statements return exactly the same result in a test folder with only 2 files:
find /var/tmp/ipreplace/ -type f -name "*.properties"
find /var/tmp/ipreplace/ -type f -name "*.properties" -o -name "*.xml" -not \( -path '/tmp/*' -o -path '/var/tmp/*' \)
I guess the use of -path parameter in your find command is wrong.
Try the following:
find / -not \( -path '/tmp' -prune \) -not \( -path '/var/tmp' -prune \) -type f -name "*.properties" -o -name "*.xml" -exec sed -i 's/10\.32\.19\.156/10.32.19.165/g' {} \;
Look at this post for reference
I've a bash command who should find and remove files:
/usr/bin/find /srv/www/vest2/produktion/ -type f \( -iname 'vest*.xml' -o -iname 'vest*.xls'-o -iname 'vwk*.xls' \) -ctime +90 -exec rm {} \ ';'
but it doesn't work.
Whats the problem?
Many thanks
Martin
Remove the single quotes around the final semi colon
Before:
/usr/bin/find /srv/www/vest2/produktion/ -type f \( -iname 'vest*.xml' -o -iname 'vest*.xls'-o -iname 'vwk*.xls' \) -ctime +90 -exec rm {} \ ';'
/usr/bin/find: missing argument to `-exec'
Now:
/usr/bin/find /srv/www/vest2/produktion/ -type f \( -iname 'vest*.xml' -o -iname 'vest*.xls'-o -iname 'vwk*.xls' \) -ctime +90 -exec rm {} \;
/usr/bin/find: `/srv/www/vest2/produktion/': No such file or directory
If that directory existed for me, it would have executed.
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" {} \;