I have a list of files:
XX_1
XX_2
XX_3
whose numbers keep incrementing every single time I run the program.
I want to find the latest file using a linux command. I tried:
find . -maxdepth 1 -name "*XX_*" -print
but this gives me all the files with XX_. I just want XX_3 and need to save the output that I get using this command to a variable so that I can copy the file. How do I do that?
I tried:
var=$(ssh pi#192.168.0.101 ls -1 FlightLog* | sort -t_ -k2 -nr | head -1)
ssh pi#192.168.0.101 sftp "$var"
And I got the following error:
/Users/ykathur2/bin/GetFile.sh: line 3: var: command not found
ssh: Could not resolve hostname flightlog_88.dat: Name or service not known
Couldn't read packet: Connection reset by peer
Please help!
How about this
$ ls -1 XX*
XX_1
XX_2
XX_3
$ ls -1 XX* | sort -t_ -k2 -nr | head -1
XX_3
Related
As I am very new to UNIX and struggling with two commands to get the desired output.
What will be the command to print the name of the largest file in size in
/usr/include directory.
I tried but not giving the filename whose size is largest :
find /usr/include -type f -exec ls -s {} \; | sort -n | tail -n 5
command to print user ids on a Linux environment which can’t be
used as login ids. [hint: consider /etc/passwd file]
I tried: cat /etc/passwd
Please let me know if these are the correct commands.
ls -lS /usr/include | head -2 | grep -v total
Flag S is to sort files by size. Head then grabs the first 2 lines of the long output and grep removes the summary line.
awk -F: '{ if($3<100) print $1;}' /etc/passwd
UID is the third field in /etc/passwd. Generally UIDs 0-99 are reserved for predefined system accounts. So this command prints all usernames with UID less than 100.
I am new to linux. I have a folder with many files in it and i need to get the latest file depending on the file name. Example: I have 3 files RAT_20190111.txt RAT_20190212.txt RAT_20190321.txt . I need a linux command to move the latest file here RAT20190321.txt to a specific directory.
If file pattern remains the same then you can try below command :
mv $(ls RAT*|sort -r|head -1) /path/to/directory/
As pointed out by #wwn, there is no need to use sort, Since the files are lexicographically sortable ls should do the job already of sorting them so the command will become :
mv $(ls RAT*|tail -1) /path/to/directory
The following command works.
ls | grep -v '/$' |sort | tail -n 1 | xargs -d '\n' -r mv -- /path/to/directory
The command first splits output of ls with newline. Then sorts it, takes the last file and then it moves this to the required directory.
Hope it helps.
Use the below command
cp ls |tail -n 1 /data...
I am trying to get the second file, when file contents sorted in reverse (desc order) and copy it to my local directory using scp
Here's what I got:
scp -r uname#host:./backups/dir1/$(ls -r | head -2| tail -1) /tmp/data_sync/dir1/
I still seem to copy all the files when I run this script. What am I missing? TIA.
The $(...) is being interpreted locally. If you want the commands to run on the remote, you'll need to use ssh and have the remote side use scp to copy files to your local system.
Since parsing ls's output has a number of problems, I'll use find to accomplish the same thing as ls, telling it to use NUL between each filename rather than newline. sort sorts that list of filenames, and sed -n 2p prints the second element of the sorted list of filenames. xargs runs the scp command, inserting the filename as the first argument.
ssh uname#host "find ./backups/dir1/ -mindepth 1 -maxdepth 1 -name '[^.]*' -print0 | \
sort -r -z | sed -z -n 2p | \
xargs -0 -I {} scp {} yourlocalhost:/tmp/data_sync/dir1/"
If I got your question, your command is ok with just one specification:
you first ran scp -r which recursively scps your files which have theri content sorted in reverse order.
Try without -r:
scp uname#host:./backups/dir1/$(ls -r | head -2 | tail -1) /tmp/data_sync/dir1/
The basic syntax for scp is:
scp username#source:/location/to/file username#destination:/where/to/put
Don't forget that -rrecursively copy entire directories. More, note that scp follows symbolic links encountered in the tree traversal.
I'm trying to write a script which help to follows the logs of my application.
The logs of my application are written to "var/log/MyLogs/" with the following pattern:
runningNumber_XXX.txt , for example:
0_XXX.txt
37_xxx.txt
99_xxx.txt
101_xxx.txt
103_xxx.txt
I'm trying to write a bash script (without a success for now) which will print last 20 rows of the last log file (the last log file is the file with has the biggest prefix number).
I know I need to go over the files in the folder (for file in /var/log/MyLogs/*) and check which file name has the biggest prefix, and after it print the last 20 rows from the selected file.
please help me....
Thanks...
find /var/log/MyLogs -iname '*_xxx.txt' | sort -n | tail -1 | xargs tail -20
Get correct files
Sort numerically
Get last log file
Get last 20 rows
tail -20 $(ls -1 /var/log/MyLogs/*_*.txt | sort -n -t _ -k 1 -r | head -1)
ls -1 [0-9]*_XXX.txt | sort -rn | head -1 | xargs tail -20
Usually is the bad practice using ls in shell scripts, but if you can ensure than the logfiles doesn't contains spaces and other strange characters, you can use a simple:
tail -20 $(ls -t1 /var/log/[0-9]*_XXX.txt | head -1)
The:
ls -t sorts the files my modification time newest comes first
head the the 1st
tail print the last lines
AGAIN, this is usually a bad practice, you can use it only when you knows what you're doing.
Found several posts like this one to tell how to find the latest file inside of a folder.
My question is one step forward, how to find the second latest file inside the same folder? The purpose is that I am looking for a way to diff the latest log with a previous log so as to know what have been changed. The log was generated in a daily basis.
Building on the linked solutions, you can just make tail keep the last two files, and then pass the result through head to keep the first one of those:
ls -Art | tail -n 2 | head -n 1
To do diff of the last (lately modified) two files:
ls -t | head -n 2 | xargs diff
Here's a stat-based solution (tested on linux)
for x in ./*;
do
if [[ -f "$x" ]]; then
stat --printf="%n %Y\n" "$x"; fi;
done |
sort -k2,2 -n -r |
sed -n '2{p;q}'
ls -dt {{your file pattern}} | head -n 2 | tail -n 1
Will provide second latest file in the pattern you search.
Here's the command returns you latest second file in the folder
ls -lt | tail -n 1 | head -n 2
enjoy...!