Bash string replace on command result - bash

I have a simple bash script which is getting the load average using uptime and awk, for example
LOAD_5M=$(uptime | awk -F'load averages:' '{ print $2}' | awk '{print $2}')
However this includes a ',' at the end of the load average
e.g.
0.51,
So I have then replaced the comma with a string replace like so:
LOAD_5M=${LOAD_5M/,/}
I'm not an awk or bash wizzkid so while this gives me the result I want, I am wondering if there is a succinct way of writing this, either by:
Using awk to get the load average without the comma, or
Stripping the comma in a single line

You can do that in same awk command:
uptime | awk -F 'load averages?: *' '{split($2, a, ",? "); print a[2]}'
1.32

The 5 min load is available in /proc/loadavg. You can simply use cut:
cut -d' ' -f2 /proc/loadavg
With awk you can issue:
awk '{print $2}' /proc/loadavg
If you are not working on Linux the file /proc/loadavg will not being present. In this case I would suggest to use sed, like this:
uptime | sed 's/.*, \(.*\),.*,.*/\1/'

uptime | awk -F'load average:' '{ print $2}' | awk -F, '{print $2}'
0.38
(My uptime output has 'load average:' singular)

The load average numbers are always the last 3 fields in the 'uptime' output so:
IFS=' ,' read -a uptime_fields <<<"$(uptime)"
LOAD_5M=${uptime_fields[#]: -2:1}

Related

How to count duplicates in Bash Shell

Hello guys I want to count how many duplicates there are in a column of a file and put the number next to them. I use awk and sort like this
awk -F '|' '{print $2}' FILE | sort | uniq -c
but the count (from the uniq -c) appears at the left side of the duplicates.
Is there any way to put the count on the right side instead of the left, using my code?
Thanks for your time!
Though I believe you shouls show us your Input_file so that we could create a single command or so for this requirement, since you have't shown Input_file so trying to solve it with your command itself.
awk -F '|' '{print $2}' FILE | sort | uniq -c | awk '{for(i=2;i<=NF;i++){printf("%s ",$i)};printf("%s%s",$1,RS)}'
You can just use awk to reverse the output like below:
awk -F '|' '{print $2}' FILE | sort | uniq -c | awk {'print $2" "$1'}
awk -F '|' '{print $2}' FILE | sort | uniq -c| awk '{a=$1; $1=""; gsub(/^ /,"",$0);print $0,a}'
You can use awk to calculate the amount of duplicates, so your command can be simplified as followed,
awk -F '|' '{a[$2]++}END{for(i in a) print i,a[i]}' FILE | sort
Check this command:
awk -F '|' '{c[$2]++} END{for (i in c) print i, c[i]}' FILE | sort
Use awk to do the counting is enough. If you do not want to sort by browser, remove the pipe and sort.

Need to use awk to get a specific word or value after another specific word?

I need to use awk to get a specific word or value after another specific word, I tried some awk commands already but after many other filters like grep and sed. The file that I need to get the word from is having the same line more than one time like the below line:
Configuration: number=6 model=MSA SNT=4 IC=8 SIZE=16384MB NRF=24 meas=2.00
If need 24 I used
grep IC file | awk 'NF>1{print $NF}'
If need 16384MB I used
grep IC file | awk -F'SIZE=' '{ print $2 }'|awk '{ print $1 }'
We need to get any word from that line using awk? what I used can get what is needed but we still need a minimized awk command.
I am sure we can use one single awk to get the needed info from one line minimized command?
sed -r 's/.*SIZE=([^ ]+).*/\1/' input
16384MB
sed -r 's/.*NRF=([^ ]+).*/\1/' input
24
grep way :
grep -oP 'SIZE=\K[^ ]+' imput
16384MB
awk way :
awk '{for(i=1;i<=NF;i++) if($i ~ /SIZE=/) split($i,a,"=");print a[2]}' input
You could use an Awk with multi-character de-limiter as below to get this done. Loop through the fields, match the pattern you need and print the next field which contains the field value.
awk -F'[:= ]' -v option="${match}" '{for(i=1;i<=NF;i++) if ($i ~ option) {print $(i+1)}}' file
Examples,
match="number"
awk -F'[:= ]' -v option="${match}" '{for(i=1;i<=NF;i++) if ($i ~ option) {print $(i+1)}}' file
6
match="model"
awk -F'[:= ]' -v option="${match}" '{for(i=1;i<=NF;i++) if ($i ~ option) {print $(i+1)}}' file
MSA
match="meas"
awk -F'[:= ]' -v option="${match}" '{for(i=1;i<=NF;i++) if ($i ~ option) {print $(i+1)}}' file
2.00
here is a more general approach
... | awk -v k=NRF '{for(i=2;i<=NF;i++) {split($i,a,"="); m[a[1]]=a[2]} print m[k]}'
code will stay the same just change the key k.
If you have GNU awk you could use the third parameter of match:
$ awk 'match($0,/( IC=)([^ ]*)/,a)&& $0=a[2]' file
8
Or get the meas:
$ awk 'match($0,/( meas=)([^ ]*)/,a)&& $0=a[2]' file
2.00
Should you use some other awk, you could use this combination of split, substr and match:
$ awk 'split(substr($0,match($0,/ IC=[^ ]*/),RLENGTH),a,"=") && $0=a[2]' file
8

AWK Print Second Column of Last Line

I'm trying to use AWK to post the second row of last line of this command (the total disk space):
df --total
The command I'm using is:
df --total | awk 'FNR == NF {print $2}'
But it does not get it right.
Is there another way to do it?
You're using the awk variable NF which is Number of Fields. You might have meant NR, Number of Rows, but it's easier to just use END:
df --total | awk 'END {print $2}'
You can use tail first then use awk:
df --total | tail -1 | awk '{print $2}'
One way to do it is with a tail/awk combination, the former to get just the last line, the latter print the second column:
df --total | tail -1l | awk '{print $2}'
A pure-awk solution is to simply store the second column of every line and print it out at the end:
df --total | awk '{store = $2} END {print store}'
Or, since the final columns are maintained in the END block from the last line, simply:
df --total | awk 'END {print $2}'
awk has no concept of "this is the last line". sed does though:
df --total | sed -n '$s/[^[:space:]]\+[[:space:]]\+\([[:digit:]]\+\).*/\1/p'

SSH call inside ruby, using %x

I am trying to make a single line ssh call from a ruby script. My script takes a hostname, and then sets out to return the hostname's machine info.
return_value = %x{ ssh #{hostname} "#{number_of_users}; #{number_of_processes};
#{number_of_processes_running}; #{number_of_processes_sleeping}; "}
Where the variables are formatted like this.
number_of_users = %Q(users | wc -w | cat | awk '{print "Number of Users: "\$1}')
number_of_processes = %Q(ps -el | awk '{print $2}' | wc -l | awk '{print "Number of Processes: "$1}')
I have tried both %q, %Q, and just plain "" and I cannot get the awk to print anything before the output. I either get this error (if I include the colon)
awk: line 1: syntax error at or near :
or if I don't include the slash in front of $1 I just get empty output for that line. Is there any solution for this? I thought it might be because I was using %q, but it even happens with just double quotes.
Use backticks to capture the output of the command and return the output as a string:
number_of_users = `users | wc -w | cat | awk '{print "Number of Users:", $1}'`
puts number_of_users
Results on my system:
48
But you can improve your pipeline:
users | awk '{ print "Number of Users:", NF }'
ps -e | awk 'END { print "Number of Processes:", NR }'
So the solution to this problem is:
%q(users | wc -w | awk '{print \"Number of Users: \"\$1}')
Where you have to use %q, not %, not %Q, and not ""
You must backslash double quotes and the dollar sign in front of any awk variables
If somebody could improve upon this answer by explaining why, that would be most appreciated
Though as Steve pointed out I could have improved my code using users | awk '{ print \"Number of Users:\", NF }'
In which case there is no need to backslash the NF.

bash awk first 1st column and 3rd column with everything after

I am working on the following bash script:
# contents of dbfake file
1 100% file 1
2 99% file name 2
3 100% file name 3
#!/bin/bash
# cat out data
cat dbfake |
# select lines containing 100%
grep 100% |
# print the first and third columns
awk '{print $1, $3}' |
# echo out id and file name and log
xargs -rI % sh -c '{ echo %; echo "%" >> "fake.log"; }'
exit 0
This script works ok, but how do I print everything in column $3 and then all columns after?
You can use cut instead of awk in this case:
cut -f1,3- -d ' '
awk '{ $2 = ""; print }' # remove col 2
If you don't mind a little whitespace:
awk '{ $2="" }1'
But UUOC and grep:
< dbfake awk '/100%/ { $2="" }1' | ...
If you'd like to trim that whitespace:
< dbfake awk '/100%/ { $2=""; sub(FS "+", FS) }1' | ...
For fun, here's another way using GNU sed:
< dbfake sed -r '/100%/s/^(\S+)\s+\S+(.*)/\1\2/' | ...
All you need is:
awk 'sub(/.*100% /,"")' dbfake | tee "fake.log"
Others responded in various ways, but I want to point that using xargs to multiplex output is rather bad idea.
Instead, why don't you:
awk '$2=="100%" { sub("100%[[:space:]]*",""); print; print >>"fake.log"}' dbfake
That's all. You don't need grep, you don't need multiple pipes, and definitely you don't need to fork shell for every line you're outputting.
You could do awk ...; print}' | tee fake.log, but there is not much point in forking tee, if awk can handle it as well.

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