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 $vmax
}
lastversion "something02"
It will print: 30102
Related
I would like to make array which put users in a time using for loop. For example:
y[1]="user1"
y[2]="user2"
...
y[n]="usern"
I tried to do it like this
#!/bin/bash
x=$(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
This is a code that shows my all user names.
-q user | grep -A 0 -B 2 -e uid:\ 5'[0-9][0-9]' | grep ^name | cut -d " " -f2-
For example, the output is like...
usernameone
hello
whoami
Then, I hope that I want to check a length of all user names.
Like this output...
11 //usernameone
5 //hello
6 //whoami
How can I get a length of pipeline code?
Given some command cmd that produces the list of users, you can do this pretty easily with xargs:
$ cat x
usernameone
hello
whoami
$ cat x | xargs -L 1 sh -c 'printf "%s //%s\n" "$(echo -n "$1" | wc -c)" "$1"' '{}'
11 //usernameone
5 //hello
6 //whoami
To get a piped command might not be possible, so here's a one liner that uses a split and a while loop to accomplish this:
-q user | grep -A 0 -B 2 -e uid:\ 5'[0-9][0-9]' | grep ^name | cut -d " " -f2-|tr " " "\n"|while read user; do echo $(echo $user|wc -c) '//'$user;done|tr "\n" " ";echo
This should give you an output in the desired format. I used user as a file hence the cat
i=0;for token in $(cat user); do echo -n "${#token} //$token";echo;i=$((i+1));done;echo;
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 ${DirBogoDict} -M -k 1024 -s
done
Also look at your file names , since cat is giving an error and an option is invalid. To demonstrate this, Let say you have a file a name -3error
executing the following command
cat -3error
will gave
cat: invalid option -- '3'
cat therefore is thinking the "-" is followed by one of its command line arguments. As a result you probably get an invalid option error.
I get the following error:
> echo "${$(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
If I have
days="1 2 3 4 5 6"
func() {
echo "lSecure1"
echo "lSecure"
echo "lSecure4"
echo "lSecure6"
echo "something else"
}
and do
func | egrep "lSecure[1-6]"
then I get
lSecure1
lSecure4
lSecure6
but what I would like is
lSecure2
lSecure3
lSecure5
which is all the days that doesn't have a lSecure string.
Question
My current idea is to use awk to split the $days and then loop over all combinations.
Is there a better way?
Note that grep -v inverts the sense of a plain grep and does not solve the problem as it does not generate the required strings.
I usually use the -f flag of grep for similar purposes. The <( ... ) code generates a file with all possibilities, grep only selects those not present in the func.
func | grep 'lSecure[1-6]' | grep -v -f- <( for i in $days ; do echo lSecure$i ; done )
Or, you may prefer it the other way round:
for i in $days ; do echo lSecure$i ; done | grep -vf <( func | grep 'lSecure[1-6]' )
F=$(func)
for f in $days; do
if ! echo $F | grep -q lSecure$f; then
echo lSecure$f
fi
done
An awk solution:
$ func | awk -v i="${days}" 'BEGIN{split(i,a," ")}{gsub(/lSecure/,"");
for(var in a)if(a[var] == $0){delete a[var];break}}
END{for(var in a) print "lSecure" a[var]}' | sort
We store it in an awk array a then while reading a line, get the last number, if it is present in array, then remove that from the array. So at the end, in the array, only those element which have not been found remains. Sort is just to present in a sorted manner :)
I am not sure exactly what you are trying to achieve, but you might consider using uniq -u which deletes repeated sequences. For example you can do this with it:
( echo "$days" | tr -s ' ' '\n'; func | grep -oP '(?<=lSecure)[1-6]' ) | sort | uniq -u
Output:
2
3
5