If torrent has a problem like deleted data on hard drive in id column it has number like "ID*".
I want to filter IDs of torrents in torrent list which have a symbol "*" at the end of id(LIKE ID* or 1*,2*,25*) and delete them from torrent list.
Full command is:
/usr/bin/transmission-remote 127.0.0.1:9091 --auth ts:ts -l | grep "*" | awk '{print $1}' \
| xargs -n 1 -I % /usr/bin/transmission-remote 127.0.0.1:9091 --auth ts:ts -t% -r
I expected result:
/usr/bin/transmission-remote 127.0.0.1:9091 --auth ts:ts -t ID* -r
But something went wrong.
Transmission said that:
127.0.0.1:9091/transmission/rpc/ responded: "success"
But torrent didn't delete from list.
How I can see the final result to compare with expected?
To get IDs :
transmission-remote -l | grep '*' | awk '{print $1}' | grep -o '[0-9]*'
The full command :
transmission-remote -l | grep '*' | awk '{print $1}' | grep -o '[0-9]*' | tr "\\n" "," | xargs -n 1 -I \% transmission-remote -t \% -r
Done and done (:
With the added improvement of using "tr" to join all torrent IDs and avoid running everything in a loop ( Transmission-RPC is extremely resource intensive to call repeatedly )
Related
When I am trying to run the below Script it says invalid option 3 for cat..Whats the problem?
I am tried to use index file which specifies which file is ham and which is spam...to read the files and train spamfilter
#!bin/bash
DirBogoDict=$1
BogoFilter=/home/gunna/Downloads/bogofilter-1.2.4/src/bogofilter
x=0
for i in 'cat index | fgrep spam | head -300 | awk -F "/" '{print$2"/"$3}''
do
x=$((x+1)) ; echo $x
cat /home/gunna/Downloads/db-6.1.19.NC/build_unix/ceas08-1/$i| $BogoFilter -d $DirBogoDict -M -k 1024 -s
done
for i in 'cat index | fgrep ham | head -300 | awk -F "/" '{print$2"/"$3}''
do
x=$((x+1)) ; echo $x
cat /home/gunna/Downloads/db-6.1.19.NC/build_unix/ceas08-1/$i | $BogoFilter -d $DirBogoDict -M -k 1024 -n
done
This part
'cat index | fgrep spam | head -300 | awk -F "/" '{print$2"/"$3}''
needs to be in back-ticks, not single quotes
`cat index | fgrep spam | head -300 | awk -F "/" '{print$2"/"$3}'`
And you could probably simplify it a little with
for i in `fgrep spam index | head -300 | awk "/" '{print$2"/"$3}'`
Kdopen has explained the error you got , here is the improved code for similar for-loop function.
DirBogoDict=$1
BogoFilter=/home/gunna/Downloads/bogofilter-1.2.4/src/bogofilter
awk '/spam/&&++myctr<=300{print $2 FS $3}' FS="/" index |while read i
do
cat /home/gunna/Downloads/db-6.1.19.NC/build_unix/ceas08-1/"$i"| $BogoFilter -d ${DirBogoDict} -M -k 1024 -s
done
awk '/ham/&&++myctr<=300{print $2 FS $3}' FS="/" index |while read i
do
cat /home/gunna/Downloads/db-6.1.19.NC/build_unix/ceas08-1/"$i"| $BogoFilter -d ${DirBogoDict} -M -k 1024 -s
done
Also look at your file names , since cat is giving an error and an option is invalid. To demonstrate this, Let say you have a file a name -3error
executing the following command
cat -3error
will gave
cat: invalid option -- '3'
cat therefore is thinking the "-" is followed by one of its command line arguments. As a result you probably get an invalid option error.
How do I get the five last created users in order?
I can sort them by userId with this command but is there a file where you can see the date when the user is created?
tail -5 /etc/passwd | sort -r | grep '/home' | cut -d: -f1
You can look at the files in the system and sort the users based on the least recently modified files they own:
find /home | xargs -n 1 -I {} stat -c '%Y %U' "{}" | \
awk '$1 < d[$2] || !d[$2] { d[$2]=$1 } \
END{ for (u in d) { print d[u], u } }' | \
sort -n -k1
I get the following error:
> echo "${$(qstat -a | grep kig):0:7}"
-bash: ${$(qstat -a | grep kig):0:7}: bad substitution
I'm trying to take the number before. of
> qstat -a | grep kig
1192530.perceus- kigumen lr_regul pbs.sh 27198 2 16 -- 24:00:00 R 00:32:23
and use it as an argument to qdel in openPBS so that I can delete all process that I started with my login kigumen
so ideally, this should work:
qdel ${$(qstat -a | grep kig):0:7}
so far, only this works:
str=$(qstat -a | grep kig); qdel "${str:0:7}"
but I want a clean one-liner without a temporary variable.
The shell substring construct you're using (:0:7) only works on variables, not command substitution. If you want to do this in a single operation, you'll need to trim the string as part of the pipeline, something like one of these:
echo "$(qstat -a | grep kig | sed 's/[.].*//')"
echo "$(qstat -a | awk -F. '/kig/ {print $1}')"
echo "$(qstat -a | awk '/kig/ {print substr($0, 1, 7)}')"
(Note that the first two print everything before the first ".", while the last prints the first 7 characters.) I don't know that any of them are particularly cleaner, but they do it without a temp variable...
qstat -u palle | cut -f 1 -d "." | xargs qdel
Kills all my jobs... normally I grep out the jobname(s) before cut'ing...
So I use a small script "idlist":
qstat -u palle | grep -E "*.in" | grep -E "$1" | cut -f 1 -d "." | xargs
To see all my "map_..." jobs:
idlist "map_*"
For killing all my "map_...." jobs:
idlist "map_*" | xargs qdel
yet another ways :
foreach m1 in $(qstat -a );do
if [[ $m1 =~ kig ]];then
m2=${m1%.kig}
echo "kig found $m2 "
break
fi
done
I am trying to get the max version number from a directory where i have several versions of one program
for example if output of ls is
something01_1.sh
something02_0.1.2.sh
something02_0.1.sh
something02_1.1.sh
something02_1.2.sh
something02_2.0.sh
something02_2.1.sh
something02_2.3.sh
something02_3.1.2.sh
something.sh
I am getting the max version number with the following -
ls somedir | grep some_prefix | cut -d '_' -f2 | sort -t '.' -k1 -r | head -n 1
Now if at the same time i want to check it with the version number which i already have in the system, whats the best way to do it...
in bash i got this working (if 2.5 is the current version)
(ls somedir | grep some_prefix | cut -d '_' -f2; echo 2.5) | sort -t '.' -k1 -r | head -n 1
is there any other correct way to do it?
EDIT: In the above example some_prefix is something02.
EDIT: Actual Problem here is
(ls smthing; echo more) | sort
is it the best way to merge output of two commands/program for piping into third.
I have found the solution. The best way it seems is using process substitution.
cat <(ls smthing) <(echo more) | sort
for my version example
cat <(ls somedir | grep some_prefix | cut -d '_' -f2) <(echo 2.5) | sort -t '.' -k1 -r | head -n 1
for the benefit of future readers, I recommend - please drop the lure of one-liner and use glob as chepner suggested.
Almost similar question is asked on superuser.
more info about process substitution.
Is the following code more suitable to what you're looking for:
#/bin/bash
highest_version=$(ls something* | sort -V | tail -1 | sed "s/something02_\|\.sh//g")
current_version=$(echo $0 | sed "s/something02_\|\.sh//g")
if [ $current_version > $highest_version ]; then
echo "Uh oh! Looks like we need to update!";
fi
You can try something like this :
#! /bin/bash
lastversion() { # prefix
local prefix="$1" a=0 b=0 c=0 r f vmax=0
for f in "$prefix"* ; do
test -f "$f" || continue
read a b c r <<< $(echo "${f#$prefix} 0 0 0" | tr -C '[0-9]' ' ')
v=$(((a*100+b)*100+c))
if ((v>vmax)); then vmax=$v; fi
done
echo $vmax
}
lastversion "something02"
It will print: 30102
I have a jar file, i need to execute the files in it in Linux.
So I need to get the result of the unzip -l command line by line.
I have managed to extract the files names with this command :
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 ;
But i can't figure out how to obtain the file names one after another to execute them.
How can i do it please ?
Thanks a lot.
If all you need the first row in a column, add a pipe and get the first line using head -1
So your one liner will look like :
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 |head -1;
That will give you first line
now, club head and tail to get second line.
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 |head -2 | tail -1;
to get second line.
But from scripting piont of view this is not a good approach. What you need is a loop as below:
for class in `unzip -l el-api.jar | awk '{print $NF}' | grep javax/el/[A-Za-Z] | cut -d "/" -f3`; do echo $class; done;
you can replace echo $class with whatever command you wish - and use $class to get the current class name.
HTH
Here is my attempt, which also take into account Daddou's request to remove the .class extension:
unzip -l package.jar | \
awk -F'/' '/com\/tests\/[A-Za-z]/ {sub(/\.class/, "", $NF); print $NF}' | \
while read baseName
do
echo " $baseName"
done
Notes:
The awk command also handles the tasks of grep and cut
The awk command also handles the removal of the .class extension
The result of the awk command is piped into the while read... command
baseName represents the name of the class file, with the .class extension removed
Now, you can do something with that $baseName