I know there is this option for unix's find command:
find -version
GNU find version 4.1
-newer file Compares the modification date of the found file with that of
the file given. This matches if someone has modified the found
file more recently than file.
Is there an option that will let me find files that are older than a certain file. I would like to delete all files from a directory for cleanup. So, an alternative where I would find all files older than N days would do the job too.
You can use a ! to negate the -newer operation like this:
find . \! -newer filename
If you want to find files that were last modified more then 7 days ago use:
find . -mtime +7
UPDATE:
To avoid matching on the file you are comparing against use the following:
find . \! -newer filename \! -samefile filename
UPDATE2 (several years later):
The following is more complicated, but does do a strictly older than match. It uses -exec and test -ot to test each file against the comparison file. The second -exec is only executed if the first one (the test) succeeds. Remove the echo to actually remove the files.
find . -type f -exec test '{}' -ot filename \; -a -exec echo rm -f '{}' +
You can just use negation:
find ... \! -newer <reference>
You might also try the -mtime/-atime/-ctime/-Btime family of options. I don't immediately remember how they work, but they might be useful in this situation.
Beware of deleting files from a find operation, especially one running as root; there are a whole bunch of ways an unprivileged, malicious process on the same system can trick it into deleting things you didn't want deleted. I strongly recommend you read the entire "Deleting Files" section of the GNU find manual.
If you only need files that are older than file "foo" and not foo itself, exclude the file by name using negation:
find . ! -newer foo ! -name foo
Please note, that the negation of newer means "older or same timestamp":
As you see in this example, the same file is also returned:
thomas#vm1112:/home/thomas/tmp/ touch test
thomas#vm1112:/home/thomas/tmp/ find ./ ! -newer test
./test
Unfortunately, find doesnt support this
! -newer doesnt mean older. It only means not newer, but it also matches files that have equal modification time. So I rather use
for f in path/files/etc/*; do
[ $f -ot reference_file ] && {
echo "$f is older"
# do something
}
done
find dir \! -newer fencefile -exec \
sh -c '
for f in "$#"; do
[ "$f" -ot fencefile ] && printf "%s\n" "$f"
done
' sh {} + \
;
Related
AIM: to find all JS|TS excluding *.spec.js files in a directory but replace the base path with ./
I have this command
find src/app/directives -name '*.[j|t]s' ! -name '*.spec.js' -exec printf "import \"%s\";\n" {} \;
which in said directory prints the marked JS files. However I want to replace the src/app with ./
I've tried playing with [[]] and this command but they don't work.
find src/app/components -name '*.[j|t]s' ! -name '*.spec.js' -exec printf "import \"%s\";\n" ${{}/src
/hi} \;
zsh: bad substitution
Given your "AIM", all you really need is:
find src/app/directives -type f -name "*.[jt]s" ! -name "*.spec.js" -printf "./%f\n"
The reason being is the '|' in your character-class isn't matching anything, but isn't hurting anything for that matter. Your second ! -name "*.spec.js" is fine. You don't need -exec and can simply use -printf "./%f\n" (where "%f" provides the filename only for the current file). You simply prepend the "./" as part of the -printf format-string.
Let me know if I misunderstood your AIM or if you have further questions.
Removing src/app/directives While Preserving Remaining Path
If you want to preserve the remainder of the path after src/app/directives (essentially just replacing it with '.'), you can use a short helper-script with the POSIX parameter expansion to trim src/app/directives from the front of the string replacing it with '.' using printf in the helper script. For example the helper could be:
#!/bin/zsh
printf ".%s" "${1#./src/app/directives}"
(note: the leading "./" being removed along with src/app/directives is prepended by find, the '.' added by the printf format-string will result in the returned filename being ./rest/of/path/to/filename)
Call the script whatever you like, helper.sh below. Make it executable chmod +x helper.sh.
The find call would then be:
find src/app/directives -type f -name "*.[jt]s" ! -name "*.spec.js" -exec path/to/helper.sh '{}' \;
Give that a go and let me know if it does what you are needing.
I'm trying to move file from one place to another directory...So my program will read Log_Deleter, use parameters given in each line to delete the file.
When I execute the file, it seems like it runs fine (no errors) but non of the files are moved... I'm not sure why it's not moving the file nor display any error...
Can someone please identify the error?
my attempt:
#!/bin/ksh
while read -r line ; do
v=$line
set -- $v
cd /
$(find "$1" -type f -name "$2" -mtime +"$3" -exec mv {} "$4" \;)
done < Log_Deleter.txt
Log_Deleter.txt
/usr/IBM/WebSphere/AppServer/profiles/AppSrvSIT1/logs/Server1 'SystemOut_*' 5 /backup/Abackuptest1
/usr/IBM/WebSphere/AppServer/profiles/AppSrvSIT1/logs/Server2 'SystemOut_*' 5 /backup/Abackuptest2
Thanks for your help!
Find is looking for files that have a literal ' in the name. You need to remove the single quotes from $2 before invoking find. Try:
#!/bin/ksh
while read -r path name mtime dest ; do
name=$( echo $name | tr -d "'" )
find "$path" -type f -name "$name" -mtime +"$mtime" -exec mv {} "$dest" \;
done < Log_Deleter.txt
The problem is that you are trying to match a file whose name actually has the single quotes in it.
Barring other problems, I think your script will probably work once you take the quotes out of Log_Deleter.txt.
The quotes are only meaningful when the shell is parsing command input. This is not what the read builtin does. And even when reading command input, once the quotes get into a variable they stay there forever unless reread at the shells CLI layer via eval.
The shell is not exactly a macro processor. It's a complicated hybrid that a little bit CLI, a little bit programming language, and a little bit macro processor.
And, speaking of eval, it's not necessary to wrap the find in an eval-like construct. Simplify your script to run find directly and you will find it easier to debug and understand.
I am trying to write a script which will move files older than 1 day to an archive directory. I used the following find command:
for filename in `find /file_path/*.* -type f -mtime +1`
This fails since my argument list is too big to be handled by find. I got the following error:
/usr/bin/find: arg list too long
Is it possible to use find in an IF-ELSE statement? Can someone provide some examples of using mtime other then in find.
Edit: To add the for loop of which the find is a part.
find /file_path -name '*.*' -mtime +1 -type f |
while read filename
do ...move operation...
done
That assumes your original code was acceptable in the way it handled spaces etc in file names,
and that there is no sensible way to do the move in the action of find. It also avoids problems with overlong argument lists.
Why not just use the -exec part of find?
If you just want to cp files, you could use
find /file_path -name "." -mtime +1 -type f | xargs -i mv {} /usr/local/archived
How to rename all the files in one directory to new name using the command mv. Directory have 1000s of files and requirement is to change the last character of each file name to some specific char. Example: files are
abc.txt
asdf.txt
zxc.txt
...
ab_.txt
asd.txt
it should change to
ab_.txt
asd_.txt
zx_.txt
...
ab_.txt
as_.txt
You have to watch out for name collisions but this should work okay:
for i in *.txt ; do
j=$(echo "$i" | sed 's/..txt$/_.txt/')
echo mv \"$i\" \"$j\"
#mv "$i" "$j"
done
after you uncomment the mv (I left it commented so you could see what it does safely). The quotes are for handling files with spaces (evil, vile things in my opinion :-).
If all files end in ".txt", you can use mmv (Multiple Move) for that:
mmv "*[a-z].txt" "#1_.txt"
Plus: mmv will tell you when this generates a collision (in your example: abc.txt becomes ab_.txt which already exists) before any file is renamed.
Note that you must quote the file names, else the shell will expand the list before mmv sees it (but mmv will usually catch this mistake, too).
If your files all have a .txt suffix, I suggest the following script:
for i in *.txt
do
r=`basename $i .txt | sed 's/.$//'`
mv $i ${r}_.txt
done
Is it a definite requirement that you use the mv command?
The perl rename utility was written for this sort of thing. It's standard for debian-based linux distributions, but according to this page it can be added really easily to any other.
If it's already there (or if you install it) you can do:
rename -v 's/.\.txt$/_\.txt/' *.txt
The page included above has some basic info on regex and things if it's needed.
Find should be more efficient than for file in *.txt, which expands all of your 1000 files into a long list of command line parameters. Example (updated to use bash replacement approach):
find . \( -type d ! -name . -prune \) -o \( -name "*.txt" \) | while read file
do
mv $file ${file%%?.txt}_.txt
done
I'm not sure if this will work with thousands of files, but in bash:
for i in *.txt; do
j=`echo $i |sed 's/.\.txt/_.txt/'`
mv $i $j
done
You can use bash's ${parameter%%word} operator thusly:
for FILE in *.txt; do
mv $FILE ${FILE%%?.txt}_.txt
done
Let's say I have two directory structures:
/var/www/site1.prod
/var/www/site1.test
I want to use find(1) to see the files that are newer in /var/www/site1.test than their counterparts in /var/www/site1.prod.
Can I do this with find(1), and if so, how?
You also could use rsync -n
rsync -av -n /var/www/site1.test /var/www/site1.prod
should do it.
Using find,
cd /var/www/site1.test
find . -type f -print | while read file ; do
[ "$file" -nt /var/www/site1.prod/"$file" ] && echo "File '$file' changed"
done
This will work with filenames containing blanks, as well as work for a large volume of files.
To distinguish between modified and missing files, as per Eddie's comments,
cd /var/www/site1.test
find . -type f -print | while read file ; do
[ "$file" -nt /var/www/site1.prod/"$file" ] && reason=changed
[ \! -e /var/www/site1.prod/"$file" ] && reason=created
[ -n "$reason" ] && echo "$file $reason"
done
I think you can't do it with find alone, but you can do something like
$ cd /var/www/site1.test
$ files=`find .`
$ for $f in $files; do
if [ $f -nt /var/www/site1.prod/$f ]; then echo "$f changed!"; fi;
done
If you look at the -fprintf option you should be able to create two finds that output into two files that list the files name and modification time. You should then be able to just diff the two files to see what has changed.
I understand that you specifically asked for newer, but I think it's always good to know all your options.
I would use diff -ur dir1 dir2.
Maybe the timestamp changed, maybe it didn't, but that doesn't necessarily tell you if the contents are the same. Diff will tell you if the contents changed. If you really don't want to see the contents use diff -rq to just see the files that changed.