i have the next code:
files=$(lftp -u mgtwrk35,Unix11! sftp://illin634<<EOF
cd some_dir
ls | tail -1
EOF)
why does the code above work,
but if i add to the ls function -tr like:
ls -tr | tail -1
it doesn't work and gives me the next message:
ls: invalid option -- t
ls: invalid option -- r
Thanks in advance for the answer
The ls in lftp is not the real thing. It's also named "ls" but does not support all the parameters you can find in the standard ls.
find lets you recursively list files.
cls --sort=date lets you sort by modification date.
I don't know how to combine the two. There's a work-around mentioned by Nicolas Noble, which is to use awk to post-process the result of find.
Related
I'm currently moving from Windows administrator to Linux. I'm trying to monitor a folder when there will be new log files created everyday.
For example:
log-02-17-2023
log-02-18-2023
How can we figureout the latest file using the modified time of the file in that folder and after finding the latest log file I would want it to parse that file and search for the keyword error.
I kind of have something written in perl, but its still in progress. Is it possible to execute the *.sh file and it should show me the parsed output of that keyword error and the line number and echo/print the complete line containing the error as an output from the latest file. Is it possible?
If I do this I can get it to display the file names that has this keyword error, but I would also want it to display the keywork error line as well as it should pick only the latest logfile.
$ find /var/monitor/ -type f -name "*" -exec grep -l "error" {} \+ 2>/dev/null
Output:
/var/monitor/logfile
/var/monitor/logfile-16022023
Is it possible to get any simple sh script doing this process? I know this is not a code donating forums but it would be greatly appreciable for any quick help and I promise to mark the answer immediately. Thanks in advance.
Try this:
ls -dlst /var/monitor/* | awk '{print $10}' | xargs grep -s "error" | head -1
To put it in a shell.sh:
#!/bin/bash
cmd="ls -dlst /var/monitor/* | awk '{print $10}' | xargs grep -s "error" | head -1"
bash -c $cmd
I am trying to get the most recent .CSV or .csv file name among other comma separated value files where the extension name is case insensitive.
I am achieving this with the following command, provided by someone else without any explanation:
ls -t ~(i:*.CSV) | head -1
or
ls -t -- ~(i:*.CSV) | head -1
I have two questions:
What is the use of ~ and -- in this case? Does -- helps here?
How can I get a blank response when there is no .csv or .CSV file in
the folder? At the moment I get:
/bin/ls: cannot access ~(i:*.CSV): No such file or directory
I know I can test the exit code of the last command, but I was wondering maybe there is a --silent option or something.
Many thanks for your time.
PS: I made my research online quite thorough and I was unable to find an answer.
The ~ is just a literal character; the intent would appear to be to match filenames starting with ~ and ending with .csv, with i: being a flag to make the match case-insensitive. However, I don't know of any shell that supports that particular syntax. The closest thing I am aware of would be zsh's globbing flags:
setopt extended_glob # Allow globbing flags
ls ~(#i)*.csv
Here, (#i) indicates that anything after it should be matched without regard to case.
Update: as #baptistemm points out, ~(i:...) is syntax defined by ksh.
The -- is a conventional argument, supported by many commands, to mean that any arguments that follow are not options, but should be treated literally. For example, ls -l would mean ls should use the -l option to modify its output, while ls -- -l means ls should try to list a file named -l.
~(i:*.CSV) is to tell to shell (this is only supported apparently in ksh93) the enclosed text after : must be treated as insensitive, so in this example that could all these possibilites.
*.csv or
*.Csv or
*.cSv or
*.csV or
*.CSv or
*.CSV
Note this could have been written ls -t *.[CcSsVv] in bash.
To silent errors I suggest you to look for in this site for "standard error /dev/null" that will help.
I tried running commands like what you have in both bash and zsh and neither worked, so I can't help you out with that, but if you want to discard the error, you can add 2>/dev/null to the end of the ls command, so your command would look like the following:
ls -t ~(i:*.CSV) 2>/dev/null | head -1
This will redirect anything written to STDERR to /dev/null (i.e. throw it out), which, in your case, would be /bin/ls: cannot access ~(i:*.CSV): No such file or directory.
I was working my way through a primer on Shell (Bash) Scripting and had the following doubt :
Why does not the following command print the contents of cp's directory : which cp | ls -l
Does not piping by definition mean that we pass the output of one command to another i.e. redirect the output ?
Can someone help me out ? I am a newbie ..
The output of which is being piped to the standard input of ls. However, ls doesn't take anything on standard input. You want it (I presume) to be passed as a parameter. There are a couple of ways of doing that:
which cp | xargs ls -l
or
ls -l `which cp`
or
ls -l $(which cp)
In the first example the xargs command takes the standard output of the previous previous command and makes each line a parameter to the command whose name immediately follows xargs. So, for instance
find / | xargs ls -l
will do an ls -l on each file in the filesystem (there are some issues with this with peculiarly named files but that's beyond the scope of this answer).
The remaining two are broadly equivalent and use the shell to do this, expanding the output from which into the command line for cp.
It would be,
$ ls -l $(which cp)
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
OR
$ which cp | xargs ls -l
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
To pass the output of one command as parameter of another command, you need to use xargs along with the pipe symbol.
From man xargs
xargs - build and execute command lines from standard input.xargs reads items
from the standard input, delimited by blanks (which can be protected
with double or single quotes or a backslash) or newlines, and executes
the command (default is /bin/echo) one or more times with any initial-
arguments followed by items read from standard input. Blank lines on
the standard input are ignored.
I have a shell-script which lists all the file names in a directory and store them in a new file.
The problem is that when I execute this script with the nohup command, it lists the first name four times instead of listing the correct names.
Commenting the problem with other programmers they think that the problem may be the ls command.
Part of my code is the following:
for i in $( ls -1 ./Datasets/); do
awk '{print $1}' ./genes.txt | head -$num_lineas | tail -1 >> ./aux
let num_lineas=$num_lineas-1
done
Do you know an alternative to ls that works well with nohup?
Thanks.
Don't use ls to feed the loop, use:
for i in ./Datasets/*; do
or if subdirectories are of interest
for i in ./Datasets/*/*; do
Lastly, and more correctly, use find if you need the entire tree below Datasets:
find ./Datasets -type f | while IFS= read -r file; do
(do stuff with $file)
done
Others frown, but there is nothing wrong with also using find as:
for file in $(find ./Datasets -type f); do
(do stuff with $file)
done
Just choose the syntax that most closely meets your needs.
First of all, don't parse ls! A simple glob will suffice. Secondly, your awk | head | tail chain can be simplified by only printing the first column of the line that you're interested in using awk. Thirdly, you can redirect the output of your loop to a file, rather than using >>.
Incorporating all of those changes into your script:
for i in Datasets/*; do
awk -v n="$(( num_lineas-- ))" 'NR==n{print $1}' genes.txt
done > aux
Every time the loop goes round, the value of $num_lineas will decrease by 1.
In terms of your problem with nohup, I would recommend looking into using something like screen, which is known to be a better solution for maintaining a session between logins.
Linux command: I am using following command which returns the latest file name in the directory.
ls -Art | tail -n 1
When i run this command it returns me latest file changed which is actually soft link, i wants to ignore soft link in my result, and wants to get file names other then soft link how can i do that any quick help appreciated.
May be can i specify regex matched latest file file name is
rum-12.53.2.war
-- Latest file in directory without softlink
ls -ArtL | tail -n 1
-- Latest file without extension
ls -ArtL | sed 's/\(.*\)\..*/\1/' | tail -n 1
The -L option for ls does dereference the link, i.e. you'll see the information of the reference instead of the link. Is this what you want? Or would you like to completely ignore links?
If you want to ignore links completely you can use this solution, although I am sure there exists an easier one:
a=$( ls -Artl | grep -v "^l" | tail -1 )
aa=()
for i in $(echo $a | tr " " "\n")
do
aa+=($i)
done
aa_length=${#aa[#]}
echo ${aa[aa_length-1]}
First you store the output of your ls in a variable called a. By grepping for "^l" you chose only symbolic links and with the -v option you invert this selection. So you basically have what you want, only downside is that you need to use the -l option for ls, as otherwise there's no grepping for "^l". So in the second part you split the variable a by " " and fill an array called aa (sorry for the bad naming). Then you need only the last item in aa, which should be the filename.