I have a directory in which I will have some folders with date format (YYYYMMDD) as shown below -
david#machineX:/database/batch/snapshot$ ls -lt
drwxr-xr-x 2 app kyte 86016 Oct 25 05:19 20141023
drwxr-xr-x 2 app kyte 73728 Oct 18 00:21 20141016
drwxr-xr-x 2 app kyte 73728 Oct 9 22:23 20141009
drwxr-xr-x 2 app kyte 81920 Oct 4 03:11 20141002
Now I need to extract latest date folder from the /database/batch/snapshot directory and then construct the command in my shell script like this -
./file_checker --directory /database/batch/snapshot/20141023/ --regex ".*.data" > shardfile_20141023.log
Below is my shell script -
#!/bin/bash
./file_checker --directory /database/batch/snapshot/20141023/ --regex ".*.data" > shardfile_20141023.log
# now I need to grep shardfile_20141023.log after above command is executed
How do I find the latest date folder and construct above command in a shell script?
Look, this is one of approaches, just grep only folders that have 8 digits:
ls -t1 | grep -P -e "\d{8}" | head -1
Or
ls -t1 | grep -E -e "[0-9]{8}" | head -1
You could try the following in your script:
pushd /database/batch/snapshot
LATESTDATE=`ls -d * | sort -n | tail -1`
popd
./file_checker --directory /database/batch/snapshot/${LATESTDATE}/ --regex ".*.data" > shardfile_${LATESTDATE}.log
See BashFAQ#099 aka "How can I get the newest (or oldest) file from a directory?".
That being said, if you don't care for actual modification time and just want to find the most recent directory based on name you can use an array and globbing (note: the sort order with globbing is subject to LC_COLLATE):
$ find
.
./20141002
./20141009
./20141016
./20141023
$ foo=( * )
$ echo "${foo[${#foo[#]}-1]}"
20141023
Related
I want to fetch list of files from a server using SFTP one by one only if their size is less than 1 GB.
I am running the following command :
$sftp -oIdentityFile=/home/user/.ssh/id_rsa -oPort=22 user#hostname >list.txt <<EOF
cd upload/Example
ls -l iurygify*.zip
EOF
This results in:
$cat list.txt
sftp> cd upload/Example
sftp> ls -l iurygify*.zip
-rwxrwx--- 0 300096661 300026669 0 Mar 11 16:38 iurygify1.zip
-rwxrwx--- 0 300096661 300026669 0 Mar 11 16:38 iurygify2.zip
I could then use awk to calculate get the size and filename which I can save into logs for reference and then download only those files which meet the 1 GB criteria.
Is there any simpler approach to accomplish getting this file list and size? I want to avoid the junk entires of the prompt and commands in the list.txt and do not want to do this via expect command.
We are using SSH key authentication
You could place your sftp commands in a batch file and filter the output - no need for expect.
echo 'ls -l' > t
sftp -b t -oIdentityFile=/home/user/.ssh/id_rsa -oPort=22 user#hostname | grep -v 'sftp>' >list.txt
Or take it a step further and filter out the "right size" in the same step:
sftp -b t -oIdentityFile=/home/user/.ssh/id_rsa -oPort=22 user#hostname | awk '$1!~/sftp>/&&$5<1000000000' >list.txt
Maybe using lftp instead of sftp ?
$ lftp sftp://xxx > list.txt <<EOF
> open
> ls -l
> EOF
$ cat list.txt
drwxr-xr-x 10 ludo users 4096 May 24 2019 .
drwxr-xr-x 8 root root 4096 Dec 20 2018 ..
-rw------- 1 ludo users 36653 Mar 31 19:28 .bash_history
-rw-r--r-- 1 ludo users 220 Mar 21 2014 .bash_logout
-rw-r--r-- 1 ludo users 362 Aug 16 2018 .bash_profile
...
How do I calculate the sum total size of multiple files located in different directories?
I have a text file containing the full path and name of the files.
I figure a simple script using while read line and du -h might do the trick...
Example of text file (new2.txt) containing list of files to sum:
/mount/st4000/media/A/amediafile.ext
/mount/st4000/media/B/amediafile.ext
/mount/st4000/media/C/amediafile.ext
/mount/st4000/media/D/amediafile.ext
/mount/st4000/media/E/amediafile.ext
/mount/st4000/media/F/amediafile.ext
/mount/st4000/media/G/amediafile.ext
/mount/st4000/media/H/amediafile.ext
/mount/st4000/media/I/amediafile.ext
/mount/st4000/media/J/amediafile.ext
/mount/st4000/media/K/amediafile.ext
Note: the folder structure is not necessarily consecutive as in A..K
Based on the suggestion from AndreaT, adapting it slightly, I tried
while read mediafile;do du -b "$mediafile"|cut -f -1>>subtotals.txt;done<new2.txt
subtotals.txt looks like
733402685
944869798
730564608
213768
13332480
366983168
6122559750
539944960
735039488
1755005744
733478912
To add all the subtotals
sum=0; while read num; do ((sum += num)); done < subtotals.txt; echo $sum
Assuming that file input is like this
/home/administrator/filesum/cliprdr.c
/home/administrator/filesum/cliprdr.h
/home/administrator/filesum/event.c
/home/administrator/filesum/event.h
/home/administrator/filesum/main.c
/home/administrator/filesum/main.h
/home/administrator/filesum/utils.c
/home/administrator/filesum/utils.h
and the result of command ls -l is
-rw-r--r-- 1 administrator administrator 13452 Oct 4 17:56 cliprdr.c
-rw-r--r-- 1 administrator administrator 1240 Oct 4 17:56 cliprdr.h
-rw-r--r-- 1 administrator administrator 8141 Oct 4 17:56 event.c
-rw-r--r-- 1 administrator administrator 2164 Oct 4 17:56 event.h
-rw-r--r-- 1 administrator administrator 32403 Oct 4 17:56 main.c
-rw-r--r-- 1 administrator administrator 1074 Oct 4 17:56 main.h
-rw-r--r-- 1 administrator administrator 5452 Oct 4 17:56 utils.c
-rw-r--r-- 1 administrator administrator 1017 Oct 4 17:56 utils.h
the simplest command to run is:
cat filelist.txt | du -cb | tail -1 | cut -f -1
with following output (in bytes)
69370
Keep in mind that du prints actual disk usage rounded up to a multiple of (usually) 4kb instead of logical file size.
For small files this approximation may not be acceptable.
To sum one directory, you will have to do a while, and export the result to the parent shell.
I used an echo an the subsequent eval :
eval ' let sum=0$(
ls -l | tail -n +2 |\
while read perms link user uid size date day hour name ; do
echo -n "+$size" ;
done
)'
It produces a line, directly evaluated, which looks like
let sum=0+205+1201+1201+1530+128+99
You just have to reproduce twice this command on both folders.
The du command doesn't have a -b option on the unix systems I have available. And there are other ways to get file size.
Assuming you like the idea of a while loop in bash, the following might work:
#!/bin/bash
case "$(uname -s)" in
Linux) stat_opt=(-c '%s') ;;
*BSD|Darwin) stat_opt=(-f '%z') ;;
*) printf 'ERROR: I don'\''t know how to run on %s\n' "$(uname -s)" ;;
esac
declare -i total=0
declare -i count=0
declare filename
while read filename; do
[[ -f "$filename" ]] || continue
(( total+=$(stat "${stat_opt[#]}" "$filename") ))
(( count++ ))
done
printf 'Total: %d bytes in %d files.\n' "$total" "$count"
This would take your list of files as stdin. You can run it in BSD unix or in Linux -- the options to the stat command (which is not internal to bash) are the bit that are platform specific.
I am looking to create a bash script to query a remote server for the last time every instance of a file was modified. Each home directory has a version of this file.
For example, both owner and owner1 have a copy of foo.txt in their home directories on a remote box accessible via ssh.
/home/owner/
-rw-r--r-- 1 owner owner 3368 Jul 29 2014 foo.txt
/home/owner1/
-rw-r--r-- 1 owner1 owner1 3368 Jul 28 2014 foo.txt
I would like to output this information to a file that would look like:
User: owner Last Modified: Jul 29 2014
User: owner1 Last Modified: Jul 28 2014
You really ought to at least show that you attempted to write the script youself. Anyway, it's only a one-liner, so why quibble:
ssh remote-box 'ls -l /home/*/foo.txt'
It's not precisely the format you suggested, but it has all the information you asked for.
echo owner: `ssh owner#remote-box "date -r foo.txt"`>output.txt
echo owner1: `ssh owner1#remote-box "date -r foo.txt"`>>output.txt
The following function will print the data you're looking for:
remote_mod() {
ssh $1 ls -l $2 | awk '{ print "User: "$3" Last Modified: "$6" "$7" "$8 }'
}
This prints something like:
$ remote_mod yourmachine '~/.bashrc'
User: root Last Modified: Jun 2 15:01
You can then do this in a loop if you want to run the command against multiple remote files:
for d in owner owner1
do
remote_mod yourmachine /home/$d/foo.txt
done
The stat command will give you even more information, but it's in a more verbose format.
Here you go, maybe not exactly the output you want but I'm sure you will be able to modify the script to suit your needs. Make sure the user you ssh with have read access to the home directories.
ssh HOSTNAME "find /home/ -maxdepth 2 -name foo.txt | xargs -l -I{} bash -c '{
DIR=\$(dirname {});
LAST=\$(stat -c %y {});
echo "Dir:\${DIR} Last Modified :\${LAST}"
}'"
If the owner of the file is the "user" you want to be printed, you can simplify with :
ssh HOSTNAME "find /home/ -maxdepth 2 -name foo.txt | xargs -l -I{} bash -c '{
stat -c \"User: %U Last Modified : %y\" {};
}'"
I back up my files using rsync. Right after a sync, I ran it expecting to see nothing, but instead it looked like it was skipping directories. I've (obviously) changed names, but I believe I've still captured all the information I could. What's happening here?
$ ls -l /source/backup/myfiles
drwxr-xr-x 2 me me 4096 2010-10-03 14:00 foo
drwxr-xr-x 2 me me 4096 2011-08-03 23:49 bar
drwxr-xr-x 2 me me 4096 2011-08-18 18:58 baz
$ ls -l /destination/backup/myfiles
drwxr-xr-x 2 me me 4096 2010-10-03 14:00 foo
drwxr-xr-x 2 me me 4096 2011-08-03 23:49 bar
drwxr-xr-x 2 me me 4096 2011-08-18 18:58 baz
$ file /source/backup/myfiles/foo
/source/backup/myfiles/foo/: directory
Then I sync (expecting no changes):
$ rsync -rtvp /source/backup /destination
sending incremental file list
backup/myfiles
skipping non-regular file "backup/myfiles/foo"
skipping non-regular file "backup/myfiles/bar"
And here's the weird part:
$ echo 'hi' > /source/backup/myfiles/foo/test
$ rsync -rtvp /source/backup /destination
sending incremental file list
backup/myfiles
backup/myfiles/foo
backup/myfiles/foo/test
skipping non-regular file "backup/myfiles/foo"
skipping non-regular file "backup/myfiles/bar"
So it worked:
$ ls -l /source/backup/myfiles/foo
-rw-r--r-- 1 me me 3126091 2010-06-15 22:22 IMGP1856.JPG
-rw-r--r-- 1 me me 3473038 2010-06-15 22:30 P1010615.JPG
-rw-r--r-- 1 me me 3 2011-08-24 13:53 test
$ ls -l /destination/backup/myfiles/foo
-rw-r--r-- 1 me me 3126091 2010-06-15 22:22 IMGP1856.JPG
-rw-r--r-- 1 me me 3473038 2010-06-15 22:30 P1010615.JPG
-rw-r--r-- 1 me me 3 2011-08-24 13:53 test
but still:
$ rsync -rtvp /source/backup /destination
sending incremental file list
backup/myfiles
skipping non-regular file "backup/myfiles/foo"
skipping non-regular file "backup/myfiles/bar"
Other notes:
My actual directories "foo" and "bar" do have spaces, but no other strange characters. Other directories have spaces and have no problem. I 'stat'-ed and saw no differences between the directories that don't rsync and the ones that do.
If you need more information, just ask.
Are you absolutely sure those individual files are not symbolic links?
Rsync has a few useful flags such as -l which will "copy symlinks as symlinks". Adding -l to your command:
rsync -rtvpl /source/backup /destination
I believe symlinks are skipped by default because they can be a security risk. Check the man page or --help for more info on this:
rsync --help | grep link
To verify these are symbolic links or pro-actively to find symbolic links you can use file or find:
$ file /path/to/file
/path/to/file: symbolic link to `/path/file`
$ find /path -type l
/path/to/file
Are you absolutely sure that it's not a symbolic link directory?
try a:
file /source/backup/myfiles/foo
to make sure it's a directory
Also, it could very well be a loopback mount
try
mount
and make sure that /source/backup/myfiles/foo is not listed.
You should try the below command, most probably it will work for you:
rsync -ravz /source/backup /destination
You can try the following, it will work
rsync -rtvp /source/backup /destination
I personally always use this syntax in my script and works a treat to backup the entire system (skipping sys/* & proc/* nfs4/*)
sudo rsync --delete --stats --exclude-from $EXCLUDE -rlptgoDv / $TARGET/ | tee -a $LOG
Here is my script run by root's cron daily:
#!/bin/bash
#
NFS="/nfs4"
HOSTNAME=`hostname`
TIMESTAMP=`date "+%Y%m%d_%H%M%S"`
EXCLUDE="/home/gcclinux/Backups/root-rsync.excludes"
TARGET="${NFS}/${HOSTNAME}/SYS"
LOGDIR="${NFS}/${HOSTNAME}/SYS-LOG"
CMD=`/usr/bin/stat -f -L -c %T ${NFS}`
## CHECK IF NFS IS MOUNTED...
if [[ ! $CMD == "nfs" ]];then
echo "NFS NOT MOUNTED"
exit 1
fi
## CHECK IF LOG DIRECTORY EXIST
if [ ! -d "$LOGDIR" ]; then
/bin/mkdir -p $LOGDIR
fi
## CREATE LOG HEADER
LOG=$LOGDIR/"rsync_result."$TIMESTAMP".txt"
echo "-------------------------------------------------------" | tee -a $LOG
echo `date` | tee -a $LOG
echo "" | tee -a $LOG
## START RUNNING BACKUP
/usr/bin/rsync --delete --stats --exclude-from $EXCLUDE -rlptgoDv / $TARGET/ | tee -a $LOG
In some cases just copy file to another location (like home) then try again
I am pretty stuck with that script.
#!/bin/bash
STARTDIR=$1
MNTDIR=/tmp/test/mnt
find $STARTDIR -type l |
while read file;
do
echo Found symlink file: $file
DIR=`sed 's|/\w*$||'`
MKDIR=${MNTDIR}${DIR}
mkdir -p $MKDIR
cp -L $file $MKDIR
done
I passing some directory to $1 parameter, this directory have three symbolic links. In while statement echoed only first match, after using sed I lost all other matches.
Look for output below:
[artyom#LBOX tmp]$ ls -lh /tmp/imp/
total 16K
lrwxrwxrwx 1 artyom adm 19 Aug 8 10:33 ok1 -> /tmp/imp/sym3/file1
lrwxrwxrwx 1 artyom adm 19 Aug 8 09:19 ok2 -> /tmp/imp/sym2/file2
lrwxrwxrwx 1 artyom adm 19 Aug 8 10:32 ok3 -> /tmp/imp/sym3/file3
[artyom#LBOX tmp]$ ./copy.sh /tmp/imp/
Found symlink file: /tmp/imp/ok1
[artyom#LBOX tmp]$
Can somebody help with that issue?
Thanks
You forgot to feed something to sed. Without explicit input, it reads nothing in this construction. I wouldn't use this approach anyway, but just use something like:
DIR=`dirname "$file"`