Is there way of counting the number of processes being run by a user in the unix/linux/os x terminal?
For instance, top -u taha lists my processes. I want to be able to count these.
This will show all of the users with their counts (I believe this would be close enough for you. :)
ps -u "$(echo $(w -h | cut -d ' ' -f1 | sort -u))" o user= | sort | uniq -c | sort -rn
You can use ps to output it and count the number using wc, as:
ps -u user | sed 1d | wc -l
You can also dump top output and grep it, something like:
top -u user -n1 | grep user | wc -l
I'm somewhat new to *nix, so perhaps I did not fully understand the context of your question, but here is a possible solution:
jobs | wc -l
The output of the above command is a count of all the processes reported by the jobs command. You can manipulate the parameters of the jobs command to change which processes get reported.
EDIT: Just FYI, this would only work if interested in commands originating from a particular shell. If you want more control in looking at system-wide processes you probably want to use ps as others have suggested. However, if you use wc to do your counting, make sure you take into account any extraneous white space jobs, ps or top may have generated as that will affect the output of wc.
Related
Here is to count the number of sessions by the same login user.
I could run the direct command if I know the specific user name, such as usera, as the following:
who | grep usera | wc -l
And if I don't know the current user, I need to user parameter.
But the following codes don't work:
currentuser=`whoami`
sessionnumber=`who | grep "$currentuser" | wc -l`
What's the error?
Thanks!
Grep has the -c flag so the wc -l plus the additional pipe is not needed.
who | grep -c -- "$USER"
"$LOGNAME" is also an option instead of "$USER", which one is bash specific? I don't know, all I know is that they are both on Linux and FreeBSD system. The -- is just a habit just in case the user starts with a dash grep will not interpret it as an option.
sessionnumber=`who | grep "$currentuser" | wc -l`
You are assigning the result of the who | ... command to a variable and to see its value you can use echo $sessionnumber
Looks like you are confused about parameters and variables.
What you are trying to get is likely
who | grep $(whoami) | wc -l
The $() is equivalent to the backticks you used.
When you write
sessionnumber=``
this will run whatever is within the backticks and save the output to a variable. You can then access the variable using the dollar notation:
echo "$sessionnumber"
I'm trying to tweak a bash script to pull back PID's of the individual application accounts when there are multiple applications running as a masterId. This used to run under individual user accounts, but recent changes have forced the applications to all run under a combined "masterId", but still maintain a unique application Id that I can grep against.
Normally
pgrep -u "appId"
would give me a single PID. Now I have to run:
pgrep -u "masterId"
it returns all of the PID's (each one is it's own application).
1234
2345
3456
I'm trying to come up with a command to bring me back just the PID of the appAccount(n) so I can pipe it into other useful commands. I can do a double grep (which is closer to what I want):
ps aux | grep -i "masterId" | grep -i "appAccount(n)"
and that will get me the entire single process information, but I just want the PID to do something like:
ps aux | grep -i "masterId" | grep -i "appAccount(n)" | xargs sudo -u appAccount(n) kill -9
How do I modify the initial above command to get just the PID? Is there a better way to do this?
pgrep --euid "masterId" --list-full | awk '/appAccount(n)/ {print $1}'
Output the full process command line, then select the one with the desired account and print the first field (pid).
I want to use top | grep user to know how many processes are running.
However, after I run top | grep user > temp_file, the command just keeps running.
How can I safely stop it with information being written to the temp_file?
You can use the -n 1 option to top to make it only do one iteration.
Probably the better tool to use would be ps, as in ps aux | grep user.
If count is all you're interested in:
pgrep -cu username
I am doing a bash script and i am essaying to show not logged users processes,which are typically daemon processes, for this,in the exercise, they recommend me:
To process the command line, we will use the cut command, which allows
selecting the different columns of the list through a filter.
I used:
ps -A | grep -v w
ps -A | grep -v who
ps -A | grep -v $USER
but trying all these options all the processes of all users are printed in the output file, and I only want the processes of users who are not logged.
I appreciate your help
Thank you.
grep -v w will remove lines matching the regular expression w (which is simply anything which contains the string w). To run the command w you have to say so; but as hinted in the instructions, you will also need to use cut to post-process the output.
So as not to give the answer away completely, here's rough pseudocode.
w | cut something >tempfile
ps -A | grep -Fvf tempfile
It would be nice if you could pass the post-processed results of w in a pipe, but standard input is already tied to ps -A. If you have a shell which supports process substitution, you can use that.
ps -A | grep -Fvf <(w | cut something)
Unfortunately, the output from w is not properly machine-readable -- you will properly want to cut out the header line(s), too. (On my machine, there are two header lines. Yours might differ.) You'll probably learn a bit of Awk later on in the course, but until then, maybe something like
ps -A | grep -Fvf <(w | tail -n +3 | cut something)
This still doesn't completely handle all possible situations. What if someone's account name is grep?
Suppose, one process is running and accessing OPENSSL shared library to perform some operation. Is there any way to find the pid of this process ?
Is there any way to find on which core this process is running ?
If possible, does it require any special privilege like sudo etc?
OS- Debian/Ubuntu
Depending on what exactly you want, something like this might do:
lsof | grep /usr/lib64/libcrypto.so | awk '{print $1, $2}' | sort -u
This essentially:
uses lsof to list all open files on the system
searches for the OpenSSL library path (which also catches versioned names like libcrypto.so.1.0)
selects the process name and PID
removes any duplicate entries
Note that this will also output processes using previous instances of the shared library file that were e.g. updated to a new version and then deleted. It also has the minor issue of outputting duplicates when a process has multiple threads with different names.
And yes, this may indeed require elevated privileges, depending on the permissions on your /proc directory.
If you really do need the processor core(s), you could try something like this (credit to dkaz):
lsof | grep /usr/lib64/libcrypto.so | awk '{print $2}' |
xargs -r ps -L --no-headers -o pid,psr,comm -p | sort -u
Adding the lwp variable to the ps command would also show the thread IDs:
lsof | grep /usr/lib64/libcrypto.so | awk '{print $2}' |
xargs -r ps -L --no-headers -o pid,lwp,psr,comm -p
PS: The what-core-are-the-users-of-this-library-on requirement still sounds a bit unusual. It might be more useful if you mentioned the problem that you are trying to solve in broader terms.
thkala is almost right. The problem is that the answer is half, since it doesn't give the core.
I would run that:
$ lsof | grep /usr/lib64/libcrypto.so |awk '{print $2}' | xargs ps -o pid,psr,comm -p