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How do I kill all processes returned from the following?
lsof -i
I have tried the following to no avail:
lsof -i | awk '{print $2}' | kill
kill takes PID's as its arguments whereas when you pipe it, it goes to the stdin of kill.
Pass them as arguments:
kill $(lsof -i | awk '{print $2}')
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How to get from a file exactly what I want in Linux?
I have: 123456789012,refid2141,test1,test2,test3 and I want this: 123456789012 or 123456789012 test3.
$ echo "123456789012,refid2141,test1,test2,test3" | awk -F "," '{print $1}'
123456789012
$ echo "123456789012,refid2141,test1,test2,test3" | awk -F "," '{printf("%s, %s", $1,$5)}'
123456789012, test3
foo.csv:
123456789012,refid2141,test1,test2,test3
import csv
with open("foo.csv", "rt") as fd:
data = list(csv.reader(fd))
print(data[0][0])
For a bash solution:
cat foo.csv | cut -d',' -f1
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Closed 6 years ago.
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I want find highest cpu utilisation process.I am using
ps -aux|awk -F " " '{print $2" ,"$3}'|sort -r | head -5
Please help me if this is right or wrong command.I am getting 'Warning: bad syntax, perhaps a bogus '-'? See /usr/share/doc/procps-3.2.7/FAQ'
ps aux --sort %cpu | tail -n 1
user 5627 7.6 16.0 1928396 1331680 ? Sl Mar12 120:58 /opt/firefox/firefox
-n 1 gives the highest, adjust number to give highest x processes. Tail because default (+) for --sort is lowest to highest.
To get just the top cpu itself though that's not particularly useful:
ps aux --sort %cpu | tail -n 1 |awk '{print $3}'
7.6
To get it with the headers use highest to lowest (-) sort:
ps aux --sort -%cpu | head -n 2
USER PID %CPU %MEM VSZ RSS TTY STAT START TIME COMMAND
user 5627 7.6 16.0 1928396 1331680 ? Sl Mar12 120:58 /opt/firefox/firefox
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Closed 7 years ago.
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The ls -ali shell command shows the following output:
933442 -rwxrw-r-- 10 root root 2048 Jan 13 07:11 afile.exe
What are all the fields in the preceding display?
+--------------+------------------+-----------------+-------+-------+------+-------+-----+-------+-----------+
| index number | file permissions | number of links | owner | group | size | month | day | time | filename |
+--------------+------------------+-----------------+-------+-------+------+-------+-----+-------+-----------+
| 933442 | -rwxrw-r-- | 10 | root | root | 2048 | Jan | 13 | 07:11 | afile.exe |
+--------------+------------------+-----------------+-------+-------+------+-------+-----+-------+-----------+
Note: month, day and time is the date of last modification.
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What command can I use to find a process that's using a lot of CPU? Can I do this without installing something new?
Or using a few other utils you could do:
ps aux | sort -rk 3,3 | head -n 5
Change the value of head to get the number of processes you want to see.
Try doing this :
top -b -n1 -c
And if you want the process that takes the most %CPU times :
top -b -n1 -c | awk '/PID *USER/{print;getline;print}'
or
top -b -n1 -c | grep -A 2 '^$'
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How can extract only "name" and "available space" from df linux command.
You can use...
df | tail -n +2 | awk '{ print $1, $4 }'
...assuming you don't want the headers too. If you do, remove the tail.
We are piping df's output into tail, where we cut the first line off (the headers), and then pipe that into awk, using it to print the first and fourth columns.
Assuming name and available space are 1st and 4th columns:
df | awk '{print $1, $4}'
the traditional approach would be
df | awk '{printf "%-15s%-8s\n",$1,$5}'