I have the below output, i want to get rid of /data & / as well in the output.
cat /etc/fstab | egrep -v '^#' | awk '{print $2}'| grep -i "^/" | egrep -v '/etc/fstab|proc|sys|shm|pts|/apps|/boot|home|/opt|/var|/var|/crash|/tmp|"' > /tmp/mounts.txt
Output:
/
/data
/data/logs/mount1
/data/logs/mount2
I just need /data/logs/mount1 & /data/logs/mount2 to be displayed. Any suggestions ?
Thanks,
KG
awk can do all that your pipeline does:
awk '
# skip comments and empty lines
$1 ~ /^#/ || NF == 0 {next}
# skip mountpoints not beginning with a slash
$2 ~/^[^/]/ {next}
# skip / and /data
$2 == "/" || $2 == "/data" {next}
{print $2}
' /etc/fstab
or, if you're a fan of linenoise:
awk 'NF&&$1!~/^#/&&$2!~/^[^/]/&&$2!="/"&&$2!="/data" {print $2}' /etc/fstab
Add " | grep /data/" that will not match / and /data.
For your example that is:
cat /etc/fstab | egrep -v '^#' | awk '{print $2}'| grep -i "^/" | egrep -v '/etc/fstab|proc|sys|shm|pts|/apps|/boot|home|/opt|/var|/var|/crash|/tmp|"' | grep /data/ > /tmp/mounts.txt
Related
Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 is my string and the result I want is vm-1.0.3
What is the best way to do this
Below is what I tried
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | awk -F - {'print $1,$2'}
vm 1.0.3
I also tried
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | awk -F - {'print $1"-",$2'}
vm- 1.0.3
Here I do not need space in between
I tried using cut and I got the expected result
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | cut -c 1-8
vm-1.0.3
What is the best way to do the same?
Making assumptions from the 1 example you provided about what the general form of your input will be so it can handle that robustly, using any sed:
$ echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' |
sed 's/^[^-]*-[^-]*-[^_]*_\(.*\)-[^-]*$/\1/'
vm-1.0.3
or any awk:
$ echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' |
awk 'sub(/^[^-]+-[^-]+-[^_]+_/,"") && sub(/-[^-]+$/,"")'
vm-1.0.3
You don't need 2 calls to awk, but your syntax with the single quotes outside the curly's, including printing the hyphen:
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 |
awk -F_ '{print $2}' | awk -F- '{print $1 "-" $2}'
If your string has the same format, let the field separator be either - or _
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F"[-_]" '{print $4 "-" $5}'
Or split the second field on - and print the first 2 parts
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F_ '{
split($2,a,"-")
print a[1] "-" a[2]
}'
Or with gnu-awk a bit more specific match with a capture group:
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 |
awk 'match($0, /^Apps-[^_]*_(vm-[0-9]+\.[0-9]+\.[0-9]+)/, a) {print a[1]}'
Output
vm-1.0.3
This is the easiest I can think of:
echo "Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2" | cut -c 25-32
Obviously you need to be sure about the location of your characters. In top of that, you seem to be have two separators: '_' and '-', while both characters also are part of the name of your entry.
echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' | sed -E 's/^.*_vm-([0-9]+).([0-9]+).([0-9]+)-.*/vm-\1.\2.\3/'
I have a linux script for selecting the node.
For example:
4
40*r13n15:40*r10n61:40*r11n18:40*r09n15
The correct result should be:
r13n15
r10n61
r11n18
r09n15
My linux script content is like:
hostNum=`bjobs -X -o "nexec_host" $1 | grep -v NEXEC`
hostSer=`bjobs -X -o "exec_host" $1 | grep -v EXEC`
echo $hostNum
echo $hostSer
for i in `seq 1 $hostNum`
do
echo $hostSer | awk -F ':' '{print '$i'}' | awk -F '*' '{print $2}'
done
But unlucky, I got nothing about node information.
I have tried:
echo $hostSer | awk -F ':' '{print "'$i'"}' | awk -F '*' '{print $2}'
and
echo $hostSer | awk -F ':' '{print '"$i"'}' | awk -F '*' '{print $2}'
But there are wrong. Who can give me a help?
One more awk:
$ echo "$variable" | awk 'NR%2==0' RS='[*:\n]'
r13n15
r10n61
r11n18
r09n15
By setting the record separtor(RS) to *:\n , the string is broken into individual tokens, after which you can just print every 2nd line(NR%2==0).
You can use multiple separators in awk. Please try below:
h='40*r13n15:40*r10n61:40*r11n18:40*r09n15'
echo "$h"| awk -F '[:*]' '{ for (i=2;i<=NF;i+=2) print $i }'
**edited to make it generic based on the comment from RavinderSingh13.
I have a task which asks to write a script which displays all partitions formatted with a specific file system, given as parameter.
I have written the script but when i run it it displays '0'. What am i doing wrong?
This is my code:
#!/bin/bash
n=sudo parted -l | tail -n +8 | awk '{print $5}' | wc | awk '{print $2}'
m=sudo parted -l | tail -n +8 | awk '{print $5}'
q=sudo parted -l | tail -n +8
for i in $n; do
if [ "[ $m | sed -n ip ]" = "$1" ]; then
echo "$q | sed -n ip"
fi
done
Different approach from yours, but does it do what you need?
lsblk -f | awk '$0 ~ fs {print $NF}' fs=ext2
i have a growing test.log like this:
abc ID1
aaa ID2
abb ID3
ccc ID4
and i want to save corresponding ID of ".*a.b." to log file like this
$ tail -f test.log | grep --line-buffered '.*a.*b.*' | awk '{print $2}' > a_ID.log
i tried
$ tail -f test.log | grep '.*a.*b.*'
$ tail -f test.log | grep --line-buffered '.*a.*b.*' > a.log
both work fine, but what should be done with awk?
# No output
$ tail -f test.log | awk '{print $2}'
# Obviously nothing in ID.log
$ tail -f test.log | awk '{print $2}' > ID.log
does awk have a '--line-buffered' like grep? how about sed?
Can you try
$ tail -f test.log | awk '{ print $2; fflush(); }'
from the man page it said it will flush the stdout.
This seems to work with mawk.
$ tail -f test.log | awk -W interactive '{print $2}' > ID.log
The -W interactive makes mawk write unbuffered to stdout.
I am running below script:-
#!/bin/bash
threshold="20"
i=2
result=`df -kh |grep -v “Filesystem” | awk ‘{ print $5 }’ | sed ‘s/%//g’`
for percent in $result; do
if ((percent > threshold))
then
partition=`df -kh | head -$i | tail -1| awk ‘{print $1}’`
echo “$partition at $(hostname -f) is ${percent}% full”
fi
let i=$i+1
done
But I get the following error:
awk: ‘{
awk: ^ invalid char '▒' in expression
sed: -e expression #1, char 1: unknown command: `▒'
Please help me to resolve this.
What awk does not work? (your script does work fine on my Ubuntu)
This line:
result=`df -kh |grep -v "Filesystem" | awk '{ print $5 }' | sed 's/%//g'`
could be changed to:
result=$(df -kh | awk '!/Filesystem/ {print $5+0}')
Avoid using old and outdated backtics if parentheses works like this: var=$(code...)
This:
partition=`df -kh | head -$i | tail -1| awk '{print $1}'`
could be changed to:
partition=$(df -kh | awk -v line="$i" 'NR==line {print $1}')
This
let i=$i+1
could be change to:
((i++))
This would then give some like this:
#!/bin/bash
threshold="20"
i=2
result=$(df -kh | awk '!/Filesystem/ {print $5+0}')
for percent in $result; do
if ((percent > threshold))
then
partition=$(df -kh | awk -v line="$i" 'NR==line {print $1}')
echo "$partition at $(hostname -f) is ${percent}% full"
fi
((i++))
done
You're using ‘ for a single quote not '. Try re-encoding your file with an editor.
You got the answer to your syntax error, now re-write the whole script as just:
#!/bin/bash
df -kh |
awk -v t=20 -v h="$(hostname -f)" '(NR>1)&&($5+0>t){printf "%s at %s is %s full\n",$1,h,$5}'