I am trying to tail the latest file in a directory that matches a name.
Below example doesn't work
tail -f | ls -t /var/log/impala/impalad.demo.local.impala.log.INFO.* | head -1
tail: warning: following standard input indefinitely is ineffective
/var/log/impala/impalad.demo.local.impala.log.INFO.20180322-104843.43442
What is the best way to tail impalad.demo.local.impala.log.INFO.* that has latest time on it?
Use
tail -f $(ls -t /var/log/impala/impalad.demo.local.impala.log.INFO.* | head -1)
instead. tail expects a file to read from, which you obtain with ls -t /var/log/impala/impalad.demo.local.impala.log.INFO.* | head -1. Piping follows an input | output pattern, though, so your current scheme won't get you far. Piping the filename into tail won't work either because tail doesn't expect filenames from standard input. To pass a filename to tail you must pass it as an argument.
Related
I need to run a random script in a folder. I can use 'ls -1 /home/sepinto/EML-Samples/scripts/regular | sort -R | head -1' to get a random file name but how can I execute it?
Thanks
I can use 'ls -1 /home/sepinto/EML-Samples/scripts/regular | sort -R | head -1' to get a random file name but how can I execute it?
First, change your ls command to list the files with complete path, then just execute them either directly with backticks or storing them in a variable.
SCRIPTNAME=`ls -1 /home/sepinto/EML-Samples/scripts/regular/*.sh | sort -R | head -1`
echo "Executing $SCRIPTNAME"
"$SCRIPTNAME"
The above assumes that all your files have a .sh ending, change the ls argument if this is not the case.
It depends on the shell you are using.
I will assume your shell is bash or sh and the scripts are already executable.
Then all you need to do is enclose your command line above in backticks.
Everything you type between backticks is evaluated (executed) by the shell and then replaced by the command's output.
`ls -1 /home/sepinto/EML-Samples/scripts/regular | sort -R | head -1`
Edit:
Turns out that the path is stripped if ls -1 is piped into another command.
A solution is to store the path in a variable so you can use it twice:
d="/home/sepinto/EML-Samples/scripts/regular"; $d/`ls -1 $d/ | sort -R | head -1`
I'm currently making a shell program and I want to display the total amount of bytes in a specific file using the pipe command. I know that the pipe command takes whatever is on the left side and gives it to the right as input. (Assuming you are in the directory the file is in)
I know that the command (wc -c) displays the number of bytes in a file but I'm not sure how to pipe it. What I've tried was:
ls fileName.sh | wc -c
wc takes the filename as argument, not as input. Try this:
wc -c fileName.sh
The wc program takes multiple arguments. You can do this to apply it to all entries in the current working directory:
wc -c $(ls)
Another approach is to use xargs to convert input to arguments:
ls | xargs wc -c
You may need to use a more complex line if you have spaces in your filenames. ls can output a single file per line, and xargs can be told to split only on \n:
ls -1 | xargs -d '\n' wc -c
If you prefer to use find instead of ls (a more powerful tool), the -print0 option for find plays along with the -0 option to xargs.
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.
I want to store the result of a command to a variable in my shell script. I cant seem to get it to work. I want the most recently dated file in the directory.
PRODUCT= 'ls -t /some/dir/file* | head -1 | xargs -n1 basename'
it wont work
you have two options, either $ or backsticks`.
1) x=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
or
2) x=`ls -t /some/dir/file* | head -1 | xargs -n1 basename`
echo $x
Edit: removing unnecessary bracket for (2).
The problem that you're having is that the command needs to be surrounded by back-ticks rather than single quotes. This is known as 'Command Substitution'.
Bash allows you to use $() for command substitution, but this is not available in all shells. I don't know if it's available in KSH; if it is, it's probably not available in all versions.
If the $() syntax is available in your version of ksh, you should definitely use it; it's easier to read (back ticks are too easy to confuse with single quotes); back-ticks are also hard to nest.
This only addresses one of the problems with your command, however: ls returns directories as well as files, so if the most recent thing modified in the specified directory is a sub-directory, that is what you will see.
If you only want to see files, I suggest using some version of the following (I'm using Bash, which supports default variables, you'll probably have to play around with the syntax of $1)
lastfile ()
{
find ${1:-.} -maxdepth 1 -type f -printf "%T+ %p\n" | sort -n | tail -1 | sed 's/[^[:space:]]\+ //'
}
This runs find on the directory, and only pulls files from that directory. It formats all of the files like this:
2012-08-29+16:21:40.0000000000 ./.sqlite_history
2013-01-14+08:52:14.0000000000 ./.davmail.properties
2012-04-04+16:16:40.0000000000 ./.DS_Store
2010-04-21+15:49:00.0000000000 ./.joe_state
2008-09-05+17:15:28.0000000000 ./.hplip.conf
2012-01-31+13:12:28.0000000000 ./.oneclick
sorts the list, takes the last line, and chops off everything before the first space.
You want $() (preferred) or backticks (``) (older style), rather than single quotes:
PRODUCT=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
or
PRODUCT=`ls -t /some/dir/file* | head -1 | xargs -n1 basename`
You need both quotes to ensure you keep the name even if it contains spaces, and also in case you later want more than 1 file, and "$(..)" to run commands in background
I believe you also need the '-1' option to ls, otherwise you could have several names per lines (you only keep 1 line, but it could be several files)
PRODUCT="$(ls -1t /some/dir/file* | head -1 | xargs -n1 basename)"
Please do not put space around the "=" variable assignments (as I saw on other solutions here) , as it's not very compatible as well.
I would do something like:
Your version corrected:
PRODUCT=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
Or simpler:
PRODUCT=$(cd /some/dir && ls -1t file* | head -1)
change to the directory
list one filename per line and sort by time/date
grab the first line
I am using ls -l -t to get a list of files in a directory ordered by time.
I would like to limit the search result to the top 2 files in the list.
Is this possible?
I've tried with grep and I struggled.
You can pipe it into head:
ls -l -t | head -3
Will give you top 3 lines (2 files and the total).
This will just give you the first 2 lines of files, skipping the size line:
ls -l -t | tail -n +2 | head -2
tail strips the first line, then head outputs the next 2 lines.
To avoid dealing with the top output line you can reverse the sort and get the last two lines
ls -ltr | tail -2
This is pretty safe, but depending what you'll do with those two file entries after you find them, you should read Parsing ls on the problems with using ls to get files and file information.
Or you could try just this
ls -1 -t | head -2
The -1 switch skips the title line.
You can use the head command to grab only the first two lines of output:
ls -l -t | head -2
You have to pipe through head.
ls -l -t | head -n 3
will output the two first results.
Try this:
ls -td -- * | head -n 2