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 $m2 "
break
fi
done
Related
I would like to make array which put users in a time using for loop. For example:
y[1]="user1"
y[2]="user2"
...
y[n]="usern"
I tried to do it like this
#!/bin/bash
x=$(who | cut -d " " -f1 | sort | uniq | wc -l)
for (( i=1; i<=$x; i++ )); do
y[$i]=$(who | cut -d " " -f1 | sort | uniq | sed -n '$ip')
p[$i]=$(lsof -u ${y[$i]} | wc -l)
echo "Users:"
echo ${y[$i]}
echo -e "Number of launched files:\n" ${p[$i]}
done
Most likely I'm using command "sed" wrong.
Can you help me?
Indeed your sed command seems to be a bit off. I can't really guess what you're trying to do there. Besides that, I'm wondering why you're executing who twice. You can make use of the data first obtained in the following manner.
#!/bin/bash
# define two arrays
y=()
p=()
#x=0
while read -r username; do
y+=("$username")
p+=($(lsof -u $(id -u "$username") | wc -l))
echo -e "User:\n${y[-1]}"
echo -e "Open files:\n${p[-1]}"
# The -1 index is the last index in the array, but you
# could uncomment the x=0 variable and the line below:
#((x++))
done <<< $(who | cut -d " " -f1 | sort | uniq)
echo "Amount of users: $x"
exit 0
I want to print the longest and shortest username found in /etc/passwd. If I run the code below it works fine for the shortest (head -1), but doesn't run for (sort -n |tail -1 | awk '{print $2}). Can anyone help me figure out what's wrong?
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |head -1 | awk '{print $2}'
sort -n |tail -1 | awk '{print $2}'
Here the issue is:
Piping finishes with the first sort -n |head -1 | awk '{print $2}' command. So, input to first command is provided through piping and output is obtained.
For the second command, no input is given. So, it waits for the input from STDIN which is the keyboard and you can feed the input through keyboard and press ctrl+D to obtain output.
Please run the code like below to get desired output:
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |head -1 | awk '{print $2}'
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |tail -1 | awk '{print $2}
'
All you need is:
$ awk -F: '
NR==1 { min=max=$1 }
length($1) > length(max) { max=$1 }
length($1) < length(min) { min=$1 }
END { print min ORS max }
' /etc/passwd
No explicit loops or pipelines or multiple commands required.
The problem is that you only have two pipelines, when you really need one. So you have grep | while read do ... done | sort | head | awk and sort | tail | awk: the first sort has an input (i.e., the while loop) - the second sort doesn't. So the script is hanging because your second sort doesn't have an input: or rather it does, but it's STDIN.
There's various ways to resolve:
save the output of the while loop to a temporary file and use that as an input to both sort commands
repeat your while loop
use awk to do both the head and tail
The first two involve iterating over the password file twice, which may be okay - depends what you're ultimately trying to do. But using a small awk script, this can give you both the first and last line by way of the BEGIN and END blocks.
While you already have good answers, you can also use POSIX shell to accomplish your goal without any pipe at all using the parameter expansion and string length provided by the shell itself (see: POSIX shell specifiction). For example you could do the following:
#!/bin/sh
sl=32;ll=0;sn=;ln=; ## short len, long len, short name, long name
while read -r line; do ## read each line
u=${line%%:*} ## get user
len=${#u} ## get length
[ "$len" -lt "$sl" ] && { sl="$len"; sn="$u"; } ## if shorter, save len, name
[ "$len" -gt "$ll" ] && { ll="$len"; ln="$u"; } ## if longer, save len, name
done </etc/passwd
printf "shortest (%2d): %s\nlongest (%2d): %s\n" $sl "$sn" $ll "$ln"
Example Use/Output
$ sh cketcpw.sh
shortest ( 2): at
longest (17): systemd-bus-proxy
Using either pipe/head/tail/awk or the shell itself is fine. It's good to have alternatives.
(note: if you have multiple users of the same length, this just picks the first, you can use a temp file if you want to save all names and use -le and -ge for the comparison.)
If you want both the head and the tail from the same input, you may want something like sed -e 1b -e '$!d' after you sort the data to get the top and bottom lines using sed.
So your script would be:
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n | sed -e 1b -e '$!d'
Alternatively, a shorter way:
cut -d":" -f1 /etc/passwd | awk '{ print length, $0 }' | sort -n | cut -d" " -f2- | sed -e 1b -e '$!d'
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 ${DirBogoDict} -M -k 1024 -s
done
Also look at your file names , since cat is giving an error and an option is invalid. To demonstrate this, Let say you have a file a name -3error
executing the following command
cat -3error
will gave
cat: invalid option -- '3'
cat therefore is thinking the "-" is followed by one of its command line arguments. As a result you probably get an invalid option error.
I am trying to get the max version number from a directory where i have several versions of one program
for example if output of ls is
something01_1.sh
something02_0.1.2.sh
something02_0.1.sh
something02_1.1.sh
something02_1.2.sh
something02_2.0.sh
something02_2.1.sh
something02_2.3.sh
something02_3.1.2.sh
something.sh
I am getting the max version number with the following -
ls somedir | grep some_prefix | cut -d '_' -f2 | sort -t '.' -k1 -r | head -n 1
Now if at the same time i want to check it with the version number which i already have in the system, whats the best way to do it...
in bash i got this working (if 2.5 is the current version)
(ls somedir | grep some_prefix | cut -d '_' -f2; echo 2.5) | sort -t '.' -k1 -r | head -n 1
is there any other correct way to do it?
EDIT: In the above example some_prefix is something02.
EDIT: Actual Problem here is
(ls smthing; echo more) | sort
is it the best way to merge output of two commands/program for piping into third.
I have found the solution. The best way it seems is using process substitution.
cat <(ls smthing) <(echo more) | sort
for my version example
cat <(ls somedir | grep some_prefix | cut -d '_' -f2) <(echo 2.5) | sort -t '.' -k1 -r | head -n 1
for the benefit of future readers, I recommend - please drop the lure of one-liner and use glob as chepner suggested.
Almost similar question is asked on superuser.
more info about process substitution.
Is the following code more suitable to what you're looking for:
#/bin/bash
highest_version=$(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
For example, if I execute
ps aux | awk '{print $1}' | xargs -I {} echo {}
I want to let the shell sleep for 1 second between each echo.
How can I change my shell command?
You can use the following syntax:
ps aux | awk '{print $1}' | xargs -I % sh -c '{ echo %; sleep 1; }'
Be careful with spaces and semicolons though. After every command in between brackets, semicolon is required (even after the last one).
Replace echo by some shell script named sleepecho containing
#!/bin/sh
sleep 1
echo $*
If your awk supports it:
ps aux | awk '{ system("sleep 1"); print $1 }' | xargs -I {} echo {}q
or skip awk and xargs altogether
ps aux | while read -r user rest;
echo $user
sleep 1;
done