I would like to know how I could get the number of processes for each user that is currently logged in.
You could try some variation of this:
ps haux Ou | cut '-d ' -f1 | uniq -c
It gives you the number of processes for each users (being logged in or not). Now you could filter those results using the output of the w command or another way of determining who is logged in.
Give this a try:
ps -u "$(echo $(w -h | cut -d ' ' -f1 | sort -u))" o user= | sort | uniq -c | sort -rn
In order to properly handle usernames that may be longer than eight characters, use users instead of w. The latter truncates usernames.
ps -u "$(echo $(printf '%s\n' $(users) | sort -u))" o user= | sort | uniq -c | sort -rn
ps -u aboelnour | awk 'END {print NR}'
will show number of process which user aboelnour running it
If you are ever concerned about nearing the user process limit shown by ulimit -a, the you want to get ALL the processes (including LWPs). In such a case you should use:
ps h -Led -o user | sort | uniq -c | sort -n
On one system doing this:
ps haux Ou | cut '-d ' -f1 | uniq -c
yields:
# ps haux Ou | cut '-d ' -f1 | uniq -c
30 user1
1 dbus
3 user2
1 ntp
1 nut
1 polkitd
2 postfix
124 root
2 serv-bu+
where doing the former yields the true process count:
# ps h -Led -o user | sort | uniq -c | sort -n
1 ntp
1 nut
2 dbus
2 postfix
2 serv-builder
3 user2
6 polkitd
141 root
444 user1
Just try:
lslogins -o USER,PROC
If you just want a count of processes you can use procfs directly like this:
(requires linux 2.2 or greater)
you can use wc:
number_of_processes=`echo /proc/[0-9]* | wc -w`
or do it in pure bash (no external commands) like this
procs=( /proc/[0-9]* )
number_of_proccesses=${#procs[*]}
If you only want the current userid
procs=( /proc/[0-9]*/fd/. )
number_of_proccesses=${#procs[*]}
userlist=$(w|awk 'BEGIN{ORS=","}NR>2{print $1}'|sed 's/,$//' )
ps -u "$userlist"
Following links contain useful ps commands options including your requirements:
Displaying all processes owned by a specific user
Show All Running Processes in Linux
Here is my solution, for Linux:
$ find /proc –user $USER -maxdepth 1 -name '[0-9]*' | wc –l
This solution will not fail when the number of processes is larger than the command line limit.
Related
I want to list all shells with number of how many users have this shell set as default
to get example output like this :
13 /bin/bash
6 /sbin/nologin
1 /usr/sbin/nologin
the only command that I was managed to create is like this:
cut -d: -f1,7 /etc/passwd | grep -c bash
which returns me only the number of users with set bash as default
Can anyone tell me how should I modify this to get output as I mentioned before?
First, you only want column 7 from the passwd file. The usernames will just get in the way. Then sort and use the uniq command to count them:
$ cut -d: -f7 /etc/passwd | sort | uniq -c
2 /bin/bash
24 /bin/false
1 /bin/sync
1 /usr/local/bin/fish
16 /usr/sbin/nologin
Working on a script that disables accounts that have been inactive for 90 days. Couldn't really find an answer after researching my problem for a few days, but I did find this command on a forum:
lastlog -t 10000 > temp1; lastlog -t 90 > temp2; diff temp1 temp2; rm temp1; rm temp2
This command outputs the users that have been inactive for 90 days. I think the solution to my problem would be to:
Filter the output of this command so only the usernames are displayed (in a list, with 1 username per line).
Take this output and write it to a text file.
Run a for-loop that for each line in the file, the contents of the line (which should be just a single username) are stored in a variable called "inactiveUser". Then the command usermod -L $inactiveUser will be executed.
Would my proposed solution work? If so, how could it be achieved? Is there a much easier method to lock inactive accounts that I am not aware of?
you can simplify this with:
lastlog -b 90
which directly lists users who have not logged in in the past 90 days.
however, it also has a header row, and lists lots of system users.
use tail to skip the header row:
lastlog -b 90 | tail -n+2
then you could use grep to filter out system users:
lastlog -b 90 | tail -n+2 | grep -v 'Never log'
although perhaps there is a safer way to find real, non-system users, e.g.:
cd /home; find * -maxdepth 0 -type d
that issue aside, you can get just the usernames out with awk:
lastlog -b 90 | tail -n+2 | grep -v 'Never log' | awk '{print $1}'
then either output the list to a file, or else directly run usermod via while read loop or xargs:
lastlog -b 90 | tail -n+2 | grep -v 'Never log' | awk '{print $1}' |
xargs -I{} usermod -L {}
perhaps you should also log what you've done:
lastlog -b 90 | tail -n+2 | grep -v 'Never log' | awk '{print $1}' |
tee -a ~/usermod-L.log | xargs -I{} usermod -L {}
While the other answer works, it can be made much cleaner by using awk instead of tail | grep | awk
lastlog -b 90 | awk '!/Never log/ {if (NR > 1) print $1}' | xargs -I{} usermod -L {}
The awk command checkes for lines that don't have the expression 'Never log' in it (!/Never log/).
NR > 1 emulates tail -n +2.
print $1 prints the first column.
Is there any way to specify a field delimiter for more spaces with the cut command? (like " "+) ?
For example: In the following string, I like to reach value '3744', what field delimiter I should say?
$ps axu | grep jboss
jboss 2574 0.0 0.0 3744 1092 ? S Aug17 0:00 /bin/sh /usr/java/jboss/bin/run.sh -c example.com -b 0.0.0.0
cut -d' ' is not what I want, for it's only for one single space.
awk is not what I am looking for either, but how to do with 'cut'?
thanks.
Actually awk is exactly the tool you should be looking into:
ps axu | grep '[j]boss' | awk '{print $5}'
or you can ditch the grep altogether since awk knows about regular expressions:
ps axu | awk '/[j]boss/ {print $5}'
But if, for some bizarre reason, you really can't use awk, there are other simpler things you can do, like collapse all whitespace to a single space first:
ps axu | grep '[j]boss' | sed 's/\s\s*/ /g' | cut -d' ' -f5
That grep trick, by the way, is a neat way to only get the jboss processes and not the grep jboss one (ditto for the awk variant as well).
The grep process will have a literal grep [j]boss in its process command so will not be caught by the grep itself, which is looking for the character class [j] followed by boss.
This is a nifty way to avoid the | grep xyz | grep -v grep paradigm that some people use.
awk version is probably the best way to go, but you can also use cut if you firstly squeeze the repeats with tr:
ps axu | grep jbos[s] | tr -s ' ' | cut -d' ' -f5
# ^^^^^^^^^^^^ ^^^^^^^^^ ^^^^^^^^^^^^^
# | | |
# | | get 5th field
# | |
# | squeeze spaces
# |
# avoid grep itself to appear in the list
I like to use the tr -s command for this
ps aux | tr -s [:blank:] | cut -d' ' -f3
This squeezes all white spaces down to 1 space. This way telling cut to use a space as a delimiter is honored as expected.
I am going to nominate tr -s [:blank:] as the best answer.
Why do we want to use cut? It has the magic command that says "we want the third field and every field after it, omitting the first two fields"
cat log | tr -s [:blank:] |cut -d' ' -f 3-
I do not believe there is an equivalent command for awk or perl split where we do not know how many fields there will be, ie out put the 3rd field through field X.
Shorter/simpler solution: use cuts (cut on steroids I wrote)
ps axu | grep '[j]boss' | cuts 4
Note that cuts field indexes are zero-based so 5th field is specified as 4
http://arielf.github.io/cuts/
And even shorter (not using cut at all) is:
pgrep jboss
One way around this is to go:
$ps axu | grep jboss | sed 's/\s\+/ /g' | cut -d' ' -f3
to replace multiple consecutive spaces with a single one.
Personally, I tend to use awk for jobs like this. For example:
ps axu| grep jboss | grep -v grep | awk '{print $5}'
As an alternative, there is always perl:
ps aux | perl -lane 'print $F[3]'
Or, if you want to get all fields starting at field #3 (as stated in one of the answers above):
ps aux | perl -lane 'print #F[3 .. scalar #F]'
If you want to pick columns from a ps output, any reason to not use -o?
e.g.
ps ax -o pid,vsz
ps ax -o pid,cmd
Minimum column width allocated, no padding, only single space field separator.
ps ax --no-headers -o pid:1,vsz:1,cmd
3443 24600 -bash
8419 0 [xfsalloc]
8420 0 [xfs_mru_cache]
8602 489316 /usr/sbin/apache2 -k start
12821 497240 /usr/sbin/apache2 -k start
12824 497132 /usr/sbin/apache2 -k start
Pid and vsz given 10 char width, 1 space field separator.
ps ax --no-headers -o pid:10,vsz:10,cmd
3443 24600 -bash
8419 0 [xfsalloc]
8420 0 [xfs_mru_cache]
8602 489316 /usr/sbin/apache2 -k start
12821 497240 /usr/sbin/apache2 -k start
12824 497132 /usr/sbin/apache2 -k start
Used in a script:-
oldpid=12824
echo "PID: ${oldpid}"
echo "Command: $(ps -ho cmd ${oldpid})"
Another way if you must use cut command
ps axu | grep [j]boss |awk '$1=$1'|cut -d' ' -f5
In Solaris, replace awk with nawk or /usr/xpg4/bin/awk
I still like the way Perl handles fields with white space.
First field is $F[0].
$ ps axu | grep dbus | perl -lane 'print $F[4]'
My approach is to store the PID to a file in /tmp, and to find the right process using the -S option for ssh. That might be a misuse but works for me.
#!/bin/bash
TARGET_REDIS=${1:-redis.someserver.com}
PROXY="proxy.somewhere.com"
LOCAL_PORT=${2:-6379}
if [ "$1" == "stop" ] ; then
kill `cat /tmp/sshTunel${LOCAL_PORT}-pid`
exit
fi
set -x
ssh -f -i ~/.ssh/aws.pem centos#$PROXY -L $LOCAL_PORT:$TARGET_REDIS:6379 -N -S /tmp/sshTunel$LOCAL_PORT ## AWS DocService dev, DNS alias
# SSH_PID=$! ## Only works with &
SSH_PID=`ps aux | grep sshTunel${LOCAL_PORT} | grep -v grep | awk '{print $2}'`
echo $SSH_PID > /tmp/sshTunel${LOCAL_PORT}-pid
Better approach might be to query for the SSH_PID right before killing it, since the file might be stale and it would kill a wrong process.
how to list all of the users whom has at least one running process.
The user name should not be duplicated.
The user name should be sorted.
$ ps xau | cut -f1 -d " "| sort | uniq | tail -n +2
You may want to weed out names starting with _ as well like so :
ps xau | cut -f1 -d " "| sort | uniq | grep -v ^_ | tail -n +2
users does what is requested. From the man page:
users lists the login names of the users currently on the system, in
sorted order, space separated, on a single line.
Try this:
w -h | cut -d' ' -f1 | sort | uniq
The w -h displays all users in system, without header and some output. The cut part removes all other information without username. uniq ignores duplicate lines.
an Unix shell script with only purpose - count the number of running processes of qmail (could be anything else). Easy thing, but there must be some bug in code:
#!/bin/bash
rows=`ps aux | grep qmail | wc -l`
echo $rows
Because
echo $rows
always shows greater number of rows (11) than if I just count rows in
ps aux | grep qmail
There are just 8 rows. Does it work this way on your system too?
Nowadays with linux, there is pgrep. If you have it on your system, you can skip grep -v grep
$ var=$(pgrep bash) # or `pgrep bash | wc -l`
$ echo $var
2110 2127 2144 2161 2178 2195 2212 2229
$ set -- $var; echo ${#}
8
also, if your ps command has -C option, another way
$ ps -C bash -o pid= | wc -l
if not, you can set a character class in your grep pattern
$ ps aux|grep [q]mail | wc -l
It appears that you're counting the grep process itself and the header line that ps normally prints before its output.
I'd suggest something more like:
qprocs=$(ps auxwww | grep -c "[q]mail")
... note that GNU grep has a "-c" switch to have it print a "count" of matches rather than the lines themselves. The trick with the regular expression here is to match qmail without matching the literal string that's on the grep command line. So we take any single character in the string and wrap it in square brackets such that it is a single character "class." The regexp: [q]mail matches the string qmail without matching the string [q]mail.
Note that even with this regex you may still find some false positive matches. If you really want to be more precise then you should supply a custom output format string to your ps command (see the man pages) or you should feed it through a pipemill or you should parse the output of the ps command based on fields (using awk or cut or a while read loop). (The -o option to ps is by far the easiest among these).
No, since I'm not running qmail. However, you will want to, at a bare minimum, exclude the process running your grep:
ps aux | grep qmail | grep -v grep
For debugging, you may want to do:
rows=`ps aux | grep qmail`
echo $rows >debug.input
od -xcb debug.input
(to see your input to the script in great detail) and then rewrite your script temporarily as:
#!/bin/bash
rows=`cat debug.input | wc -l`
echo $rows
That way, you can see the input and figure out what effect it's having on your code, even as you debug it.
A good debugger will eventually learn to only change one variable at a time. If your changing your code to get it working, that's the variable - don't let the input to your code change as well.
Use
$ /sbin/pidof qmail
A few ways...
ps -e | grep ' [q]mail' | wc -l
ps -C qmail -opid= | wc -l
pidof qmail | tr ' ' '\n' | wc -l
pgrep is on many Linux distributions, and I imagine available for other Unices.
[dan#khorium ~]$ whatis pgrep
pgrep (1) - look up or signal processes based on name and other attributes
[dan#khorium ~]$ pgrep mingetty
1920
1921
1922
1923
1924
In your case, pgrep qmail | wc -l should do the trick.