How i can see memory usage by user in linux centos 6
For example:
USER USAGE
root 40370
admin 247372
user2 30570
user3 967373
This one-liner worked for me on at least four different Linux systems with different distros and versions. It also worked on FreeBSD 10.
ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" "int(a[i]/1024+0.5);}' | sort -rnk2
About the implementation, there are no shell loop constructs here; this uses an associative array in awk to do the grouping & summation.
Here's sample output from one of my servers that is running a decent sized MySQL, Tomcat, and Apache. Figures are in MB.
mysql 1566
joshua 1186
tomcat 353
root 28
wwwrun 12
vbox 1
messagebus 1
avahi 1
statd 0
nagios 0
Caveat: like most similar solutions, this is only considering the resident set (RSS), so it doesn't count any shared memory segments.
EDIT: A more human-readable version.
echo "USER RSS PROCS" ; echo "-------------------- -------- -----" ; ps hax -o rss,user | awk '{rss[$2]+=$1;procs[$2]+=1;}END{for(user in rss) printf "%-20s %8.0f %5.0f\n", user, rss[user]/1024, procs[user];}' | sort -rnk2
And the output:
USER RSS PROCS
-------------------- -------- -----
mysql 1521 1
joshua 1120 28
tomcat 379 1
root 19 107
wwwrun 10 10
vbox 1 3
statd 1 1
nagios 1 1
messagebus 1 1
avahi 1 1
Per-user memory usage in percent using standard tools:
for _user in $(ps haux | awk '{print $1}' | sort -u)
do
ps haux | awk -v user=${_user} '$1 ~ user { sum += $4} END { print user, sum; }'
done
or for more precision:
TOTAL=$(free | awk '/Mem:/ { print $2 }')
for _user in $(ps haux | awk '{print $1}' | sort -u)
do
ps hux -U ${_user} | awk -v user=${_user} -v total=$TOTAL '{ sum += $6 } END { printf "%s %.2f\n", user, sum / total * 100; }'
done
The first version just sums up the memory percentage for each process as reported by ps. The second version sums up the memory in bytes instead and calculates the total percentage afterwards, thus leading to a higher precision.
If your system supports, try to install and use smem:
smem -u
User Count Swap USS PSS RSS
gdm 1 0 308 323 820
nobody 1 0 912 932 2240
root 76 0 969016 1010829 1347768
or
smem -u -t -k
User Count Swap USS PSS RSS
gdm 1 0 308.0K 323.0K 820.0K
nobody 1 0 892.0K 912.0K 2.2M
root 76 0 937.6M 978.5M 1.3G
ameskaas 46 0 1.2G 1.2G 1.5G
124 0 2.1G 2.2G 2.8G
In Ubuntu, smem can be installed by typing
sudo apt install smem
This will return the total ram usage by users in GBs, reverse sorted
sudo ps --no-headers -eo user,rss | awk '{arr[$1]+=$2}; END {for (i in arr) {print i,arr[i]/1024/1024}}' | sort -nk2 -r
You can use the following Python script to find per-user memory usage using only sys and os module.
import sys
import os
# Get list of all users present in the system
allUsers = os.popen('cut -d: -f1 /etc/passwd').read().split('\n')[:-1]
for users in allUsers:
# Check if the home directory exists for the user
if os.path.exists('/home/' + str(users)):
# Print the current usage of the user
print(os.system('du -sh /home/' + str(users)))
Related
I need to merge two outputs in Linux.
This:
lsblk -n -b --output KNAME,NAME,SIZE,MOUNTPOINT | grep -v "fd0" | grep -v "loop" | grep -v "sr0" | grep -v "hdc" | grep -v "cdrom"
In a result I have:
sda sda 53687091200
sda1 └─sda1 53684994048
dm-3 └─dockerVG-rootLV 53682896896 /
sdb sdb 2147483648000
sdb1 └─sdb1 2147482599424
dm-1 ├─hddVG-dockerLV 536866717696 /var/lib/docker
dm-2 └─hddVG-hddLV 1610612736000 /dockerhdd
sdc sdc 536870912000
sdc1 └─sdc1 536869863424
dm-0 └─ssdVG-ssdLV 536866717696 /dockerssd
And this:
df --exclude={tmpfs,devtmpfs,squashfs,overlay} | sed -e /^Filesystem/d | awk '{print $6 " " $1 " " $3 " " $4 " " $5}'
In a result I have:
/ /dev/mapper/dockerVG-rootLV 8110496 40591632 17%
/dockerssd /dev/mapper/ssdVG-ssdLV 214133656 274642488 44%
/dockerhdd /dev/mapper/hddVG-hddLV 83278236 1385191240 6%
/var/lib/docker /dev/mapper/hddVG-dockerLV 76046204 412729940 16%
So, I want to Join via these points /, /var/lib/docker, /dockerhdd, /dockerssd.
Important! I want to check this in another place, where we will have another mount points. Also I have to save structure of first output without sorting.
In a result I have to receive something like this:
sda sda 53687091200
sda1 └─sda1 53684994048
dm-3 └─dockerVG-rootLV 53682896896 / /dev/mapper/dockerVG-rootLV 8110496 40591632 17%
sdb sdb 2147483648000
sdb1 └─sdb1 2147482599424
dm-1 ├─hddVG-dockerLV 536866717696 /var/lib/docker /dev/mapper/hddVG-dockerLV 76046204 412729940 16%
dm-2 └─hddVG-hddLV 1610612736000 /dockerhdd /dev/mapper/hddVG-hddLV 83278236 1385191240 6%
sdc sdc 536870912000
sdc1 └─sdc1 536869863424
dm-0 └─ssdVG-ssdLV 536866717696 /dockerssd /dev/mapper/ssdVG-ssdLV 214133656 274642488 44%
Of course better to have one-liner, but if it is not possible, we can send output to separate files and join them. Could You please help me in this ?
Using awk:
awk '!/^\/&^fd0&^loop&^sr0&^hdc&^cdrom/ { print $0" "arr[$4] } /^Filesystem/ { mrk=1;next } mrk==1 && /^\// { arr[$1]=$0 }' <<< $(df --exclude={tmpfs,devtmpfs,squashfs,overlay};lsblk -n -b --output KNAME,NAME,SIZE,MOUNTPOINT)
Redirect the two commands back into awk, stripping out any grep and sed processing. We process the df command first and where we find a line beginning with "Filesystem" we set a marker (mrk) to 1 and move to the next line. We then create an array (arr) indexed with the mountpoint and containing the line returned from the df command. We move onto the lsblk command and search for the lines starting with the KNAMEs required. We print the line from the lsblk command and append the value in the arr array indexed by the mount point ($4)
I am running a cron-ed bash script to extract cache hits and bytes served per IP address. The script (ProxyUsage.bash) has two parts:
(uniqueIP.awk) find unique IPs and create a bash script do add up the hits and bytes
run the hits and bytes per IP
ProxyUsage.bash
#!/usr/bin/env bash
sudo gawk -f /home/maxg/scripts/uniqueIP.awk /var/log/squid3/access.log.1 > /home/maxg/scripts/pxyUsage.bash
source /home/maxg/scripts/pxyUsage.bash
uniqueIP.awk
{
arrIPs[$3]++;
}
END {
for (n in arrIPs) {
m++; # count arrIPs elements
#print "Array elements: " m;
arrAddr[i++] = n; # fill arrAddr with IPs
#print i " " n;
}
asort(arrAddr); # sort the array values
for (i = 1; i <= m; i++) { # write one command line per IP address
#printf("#!/usr/bin/env bash\n");
printf("sudo gawk -f /home/maxg/scripts/proxyUsage.awk -v v_Var=%s /var/log/squid3/access.log.1 >> /home/maxg/scripts/pxyUsage.txt\n", arrAddr[i])
}
}
pxyUsage.bash
sudo gawk -f /home/maxg/scripts/proxyUsage.awk -v v_Var=192.168.1.13 /var/log/squid3/access.log.1 >> /home/maxg/scripts/pxyUsage.txt
sudo gawk -f /home/maxg/scripts/proxyUsage.awk -v v_Var=192.168.1.14 /var/log/squid3/access.log.1 >> /home/maxg/scripts/pxyUsage.txt
sudo gawk -f /home/maxg/scripts/proxyUsage.awk -v v_Var=192.168.1.22 /var/log/squid3/access.log.1 >> /home/maxg/scripts/pxyUsage.txt
TheProxyUsage.bash script runs as scheduled and creates the pxyUsage.bash script.
However the pxyUsage.text file is not amended with the latest values when the script runs.
So far I run pxyUsage.bash every day myself, as I cannot figure out, why the result is not written to file.
Both bash scripts are set to execute. Actually the file permissions are below:
-rwxr-xr-x 1 maxg maxg 169 Mar 14 08:40 ProxySummary.bash
-rw-r--r-- 1 maxg maxg 910 Mar 15 17:15 proxyUsage.awk
-rwxrwxrwx 1 maxg maxg 399 Mar 17 06:10 pxyUsage.bash
-rw-rw-rw- 1 maxg maxg 2922 Mar 17 07:32 pxyUsage.txt
-rw-r--r-- 1 maxg maxg 781 Mar 16 07:35 uniqueIP.awk
Any hints appreciated. Thanks.
The sudo(8) command requires a pseudo-tty and you do not have one allocated under cron(8); you do have one allocated when logged in the usual way.
Instead of mucking about with sudo(8), just run the script as the correct user.
If you cannot do that, then in the root crontab, do something like this:
su - username /path/to/mycommand arg1 arg2...
This will work because root can use su(1) without neding a password.
I have a RHEL (Redhat Enterprise Linux) v6.5 (Santiago) server. On this server if i do a df -help there are list of options available. I am interested in the option --total
However there is an older version of RHEL (v5.5). In which there is no --total option.
My question is, I have a command like this:
df -h --total | grep total | awk 'NR==1{print$2}+NR==1{print$3}+NR==1{print$4}+NR==1{print$5}'
which gives the output as
62G
39G
21G
66%
Where
62G is Total size of the Disk
39G is Used
21G is remaining
61% Total usage %
The above command is working fine in RHEL v6.5. But fails in RHEL v5.5 since it does not have a --total option for df command.
When i run the same command on RHEL v5.5 i get the below error:
df: unrecognized option `--total'
Try `df --help' for more information.
So is there a command that can give me the output in the following way:
Total Disk Space
Used Space
Remaining Disk space
Usage %
Ex:
62G
39G
21G
66%
You'll have to do the calculation work yourself.
Something like this awk script should work.
$ cat dftotal.awk
BEGIN {
map[0] = "K"
map[1] = "M"
map[2] = "G"
map[3] = "T"
}
function fmt(val, c) {
c=0
while (val > 1024) {
c++
val = val / 1024
}
return val map[c]
}
{
for (i=2;i<5;i++) {
sum[i]+=$i
}
}
END {
print fmt(sum[2]) ORS fmt(sum[3]) ORS fmt(sum[4])
print ((sum[3] / sum[2]) * 100) "%"
}
$ df -P | awk -f dftotal.awk
I need to loop trough n lines of a file and for any i between 1 and n - 1 to get the difference line(n - 1) - line(n).
And here is the source file:
root#syncro:/var/www# cat cron.log | grep "/dev/vda"
/dev/vda 20418M 14799M 4595M 77% /
/dev/vda 20418M 14822M 4572M 77% /
/dev/vda 20418M 14846M 4548M 77% /
/dev/vda 20418M 14867M 4527M 77% /
/dev/vda 20418M 14888M 4506M 77% /
/dev/vda 20418M 14910M 4484M 77% /
/dev/vda 20418M 14935M 4459M 78% /
/dev/vda 20418M 14953M 4441M 78% /
/dev/vda 20418M 14974M 4420M 78% /
/dev/vda 20418M 15017M 4377M 78% /
/dev/vda 20418M 15038M 4356M 78% /
root#syncro:/var/www# cat cron.log | grep "/dev/vda" | cut -b 36-42 | tr -d " M"
4595
4572
4548
4527
4506
4484
4459
4441
4420
4377
4356
those /dev/vda... lines are logged hourly with df -BM in cron.log file and the difference between lines will reveal the hourly disk consumption.
So, the expected output will be:
23 (4595 - 4572)
24 (4572 - 4548)
...
43 (4420 - 4377)
21 (4377 - 4356)
I don't need the text between ( and ), I put it here for explanation only.
I'm not sure if I got you correctly, but the following awk script should work:
awk '{if(NR>1){print _n-$4};_n=$4}' your.file
Output:
23
24
21
21
22
25
18
21
43
21
You don't need the other programs in the pipe. Just:
awk '/\/dev\/vda/ {if(c++>0){print _n-$4};_n=$4}' src/checkout-plugin/a.txt
will be enough. The regex on start of the awk scripts tells awk to apply the following block only to lines which match the pattern. A side effect is that NR can't be used anymore to detect the "second line" in which the calculation starts. I introduced a custome counter c for that purpose.
Also note that awk will remove the M on it's own, because the column has been used in a numeric calculation.
I'm writing script is ksh. Need to find all users who has over N process and echo them in shell.
N reads from ksh.
I know what I should use ps -elf but how parse it, find users with >N process and create array with them. Little troubles with array in ksh. Please help. Maybe simple solutions can help me instead of array creating.
s162103#helios:/home/s162103$ ps -elf
0 S s153308 4804 1 0 40 20 ? 17666 ? 11:03:08 ? 0:00 /usr/lib/gnome-settings-daemon --oa
0 S root 6546 1327 0 40 20 ? 3584 ? 11:14:06 ? 0:00 /usr/dt/bin/dtlogin -daemon -udpPor
0 S webservd 15646 485 0 40 20 ? 2823 ? п╪п╟я─я ? 0:23 /opt/csw/sbin/nginx
0 S s153246 6746 6741 0 40 20 ? 18103 ? 11:14:21 ? 0:00 iiim-panel --disable-crash-dialog
0 S s153246 23512 1 0 40 20 ? 17903 ? 09:34:08 ? 0:00 /usr/bin/metacity --sm-client-id=de
0 S root 933 861 0 40 20 ? 5234 ? 10:26:59 ? 0:00 dtgreet -display :14
...
when i type
ps -elf | awk '{a[$3]++;}END{for(i in a)if (a[i]>N)print i, a[i];}' N=1
s162103#helios:/home/s162103$ ps -elf | awk '{a[$3]++;}END{for(i in a)if (a[i]>N)print i, a[i];}' N=1
root 118
/usr/sadm/lib/smc/bin/smcboot 3
/usr/lib/autofs/automountd 2
/opt/SUNWut/lib/utsessiond 2
nasty 31
dima 22
/opt/oracle/product/Oracle_WT1/ohs/ 7
/usr/lib/ssh/sshd 5
/usr/bin/bash 11
that is not user /usr/sadm/lib/smc/bin/smcboot
there is last field in ps -elf ,not user
Something like this(assuming 3rd field of your ps command gives the user id):
ps -elf |
awk '{a[$3]++;}
END {
for(i in a)
if (a[i]>N)
print i, a[i];
}' N=3
The minimal ps command you want to use here is ps -eo user=. This will just print the username for each process and nothing more. The rest can be done with awk:
ps -eo user= |
awk -v max=3 '{ n[$1]++ }
END {
for (user in n)
if (n[user]>max)
print n[user], user
}'
I recommend to put the count in the first column for readability.
read number
ps -elfo user= | sort | uniq -c | while read count user
do
if (( $count > $number ))
then
echo $user
fi
done
That is best solution and it works!