ps -eaf | LaunchKTRProcess | grep -v grep
this command will give me , full details of process, and i have to manually check his Running time and kill the process.
This
ps -e | sed 1d | egrep -v '^ *[^ ]+ +[^ ]+ +([^ ]|00:0.):'
gives the ps of all processes which run for more than ten minutes (I use sed 1d to remove the ps header line because not every ps has an option to suppress it); you can filter the output based on further conditions. Then
| awk '{print $1}'
extracts the PIDs (1st column).
Related
I am currenlty trying to extract all the sender domains from maillog. I am able to do some of that with the below command but the output is not quite what I desired. What would be the best approach to retrieve a unique list of sender domain from maillog?
grep from= /var/log/maillog | awk '{print $7}' | sort | uniq -c | sort -n
output
1 from=<user#test.com>,
1 from=<apache#app1.com>,
2 from=<bounceld_5BFa-bx0p-P3tQ-67Nn#example.com>,
2 from=<bounceld_19iI-HqaS-usVU-fqe5#example.com>,
12 reject:
666 from=<>,
desired output
test.com
app1.com
example.com
See useless use of grep; if you are using Awk anyway, you don't really need grep at all.
awk '$7 ~ /from=.*#/{split($7, a, /#/); ++count[a[2]] }
END { for(dom in count) print count[dom], dom }' /var/log/maillog
Collecting the counts in an associative array does away with the need to call sort and uniq, too. Obviously, if you don't care about the count, don't print count[dom] at the end.
This should give you the answer:
grep from= /var/log/maillog | awk '{print $7}' | grep -Po '(?=#).{1}\K.*(?=>)' | sort -n | uniq -c
... change last items to "| sort | uniq" to remove the counts.
References:
https://www.baeldung.com/linux/bash-remove-first-characters {1}\K use
Extract email addresses from log with grep or sed -Po grep function
I am using the system() command to run a Bash command in R, but every time I try to pipe the results of one command into the next (using '|'), I get some error.
For example:
system('grep ^SN bam_stats.txt | cut -f 2- | sed -n 8p | awk -F "\t" '{print $2}'') returns the error: Error: unexpected '{' in "system('grep ^SN bam_stats.txt | cut -f 2- | sed -n 8p | awk -F "\t" '{", and if I try to remove awk -F "\t" '{print $2}' so that I'm left with system('grep ^SN bam_stats.txt | cut -f 2- | sed -n 8p'), I get the following:
/usr/bin/grep: 2-: No such file or directory
[1] 2
I have to keep removing parts of it till I am left with only system('grep ^SN bam_stats.txt'), AKA no pipes are left, for it to work.
Here is a sample from the file 'bam_stats.txt' from which I'm extracting information:
SN filtered sequences: 0
SN sequences: 137710356
SN is sorted: 1
SN 1st fragments: 68855178
SN last fragments: 68855178
SN reads mapped: 137642653
SN reads mapped and paired: 137602018 # paired-end technology bit set + both mates mapped
SN reads unmapped: 67703
SN percentage of properly paired reads (%): 99.8
Can someone tell me why piping is not working? Apologies if this is a stupid question. Please let me know if I should provide more information.
Thank you in advance.
I don't know R but IF Rs implementation of system() just passes it's argument to a shell then, in terms of standard Unix quoting, your example
system('grep ^SN bam_stats.txt | cut -f 2- | sed -n 8p | awk -F "\t" '{print $2}'')
contains 2 strings within quotes and a string in the middle that's outside of quotes:
Inside: grep ^SN bam_stats.txt | cut -f 2- | sed -n 8p | awk -F "\t"
Outside: {print $2}
Inside: <a null string>
because the 2 quotes in the middle around '{print $2}' are ending the first quoted string then later starting a second quoted string.
You don't need sed, grep, or cut if you're using awk anyway though so try just this:
system('awk -F"\t" "/^SN/ && (++cnt==8){print \$3}" bam_stats.txt')
I have a file that has the following as the last three lines. I want to retrieve the penultimate line, i.e. 100.000;8438; 06:46:12.
.
.
.
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
I don't know the line number. How can I retrieve it using a shell script? Thanks in advance for your help.
Try this:
tail -2 yourfile | head -1
A short sed one-liner inspired by https://stackoverflow.com/a/7671772/5287901
sed -n 'x;$p'
Explanation:
-n quiet mode: dont automatically print the pattern space
x: exchange the pattern space and the hold space (hold space now store the current line, and pattern space the previous line, if any)
$: on the last line, p: print the pattern space (the previous line, which in this case is the penultimate line).
Use this
tail -2 <filename> | head -1
ed and sed can do it as well.
str='
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
'
printf '%s' "$str" | sed -n -e '${x;1!p;};h' # print last line but one
printf '%s\n' H '$-1p' q | ed -s <(printf '%s' "$str") # same
printf '%s\n' H '$-2,$-1p' q | ed -s <(printf '%s' "$str") # print last line but two
From: Useful sed one-liners by Eric Pement
# print the next-to-the-last line of a file
sed -e '$!{h;d;}' -e x # for 1-line files, print blank line
sed -e '1{$q;}' -e '$!{h;d;}' -e x # for 1-line files, print the line
sed -e '1{$d;}' -e '$!{h;d;}' -e x # for 1-line files, print nothing
You don't need all of them, just pick one.
tail +2 <filename>
This prints from second line to last line.
To clarify what has already been said:
ec2thisandthat | sort -k 5 | grep 2012- | awk '{print $2}' | tail -2 | head -1
snap-e8317883
snap-9c7227f7
snap-5402553f
snap-3e7b2c55
snap-246b3c4f
snap-546a3d3f
snap-2ad48241
snap-d00150bb
returns
snap-2ad48241
tac <file> | sed -n '2p'
I'm wondering if there is a simple way to transform a grep command such as
grep -c -f regex.txt file.txt
to return the total number of matched lines in file.txt for each line of regex.txt, instead of the sum of the matched lines found for all patterns in regex.txt as the above command does.
My current method of handling this is to use xargs (or GNU parallel interchangeably):
cat regex.txt | xargs -I{} grep -c {} file.txt
Can grep do this in one fell swoop?
grep -o -f regex.txt | sort | uniq -c
after doing the command in the terminal: forever list
i get the following output:
info: Forever processes running
data: uid command script forever pid logfile uptime
data: [0] 0ClV node enfomo-server.js 376 377 /Users/USERNAME/.forever/0ClV.log 0:0:37:26.987
i need to use grep or some alternative to give as output the following string only:
/Users/USERNAME/.forever/0ClV.log
what is the proper command?
You can do this with grep using the -o flag which only prints the matching part:
forever list | grep -o '\/Users.*log'
First you might want to isolate only the lines you want with grep, then awk would work:
grep node file | awk '{print $7}'
or cut:
grep node file | cut -d\ -f7