How to find PID's user name in linux

Question

Can you help me to find the PID's user name, Some time my server got high load. When i top -c, I cannot even find the PID's user name who / which is causing load on the server.

Solution 1

I'm surprised nobody has put this up yet:

Try the -p option to the ps command.

For instance, if you have PID 1234, run:

ps -u -p 1234

(The -u was added to include the username in the output)

You can the use grep or awk, etc. to extract the info you want.

Solution 2

/proc/processID/status will have the information about user's ID which you can use to find the username.

This does the same:

uid=$(awk '/^Uid:/{print $2}' /proc/YOUR_PROCESS_ID/status)
getent passwd "$uid" | awk -F: '{print $1}'

Replace YOUR_PROCESS_ID with your process ID number.

Solution 3

Get only username from a PID:

PID=136323
USERNAME="$( ps -o uname= -p "${PID}" )"

You can also combine it with a pgrep. In this example we show all usernames executing some .php file:

pgrep -f '\.php' | xargs -r ps -o uname= -p | sort -u

Find only one username running a certain unique process:

USERNAME="$( pgrep -nf 'script\.php' | xargs -r ps -o uname= -p )

Solution 4

I think the shortest way is:

id -nu </proc/<pid>/loginuid

The /proc/<pid>/loginuid file has the uid number of the user running the process; id -nu reads uid from stdin and returns a user name.

Solution 5

What do you want exactly? On my system, if I run 'top -c' I get:

  PID USER      PR  NI  VIRT  RES  SHR S   %CPU %MEM    TIME+  COMMAND                                                                                                                        
  2873 matt      20   0 3022m 1.6g 1.6g S     22 21.6   2245:42 /usr/lib/virtualbox/VirtualBox --comment ESX5-1 --startvm 4fd78ee9-739a-4d53-a0ce-4f9819ab9411 --no-startvm-errormsgbox        
  29764matt      20   0 2779m 1.4g 1.3g S      5 18.4 210:33.51 /usr/lib/virtualbox/VirtualBox --comment win2008-2 --startvm 202ec2b7-ae12-40e9-af76-2be429e553d7 --no-startvm-errormsgbox     
  17281root      20   0     0    0    0 S      2  0.0   0:05.90 [kworker/u:2]                                                                                                              

So the PID (processus/task identifier) is the first column, and the user account the processus runs under is the second column

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Comments

[Admin]

We were experiencing server load issue due to bulk php process, so that i had this question, we can then find them using 'lsof -p xxxx'.

[ott--]

You were a tick faster than me. You're waking up earlier? Depending on the Linux distrbution, ps u 1234 (Debian) or just ps 1234 (Android with Busybox) also works.

[Cory Robinson]

Nice. Is there a file of uid to user names that you can map the loginuid to a string user name?

[Jdamian]

That does not work

[dess]

Good answer, one of the best, but needs correction a bit. This will work (bash example): id -nu $(< /proc/<pid>/loginuid) (to work with any POSIX shell sholud be escaped with backticks instead of "$( )" bash construction). The answer with ps is more flexible, you can get also real and saved user name. But this answer command is several times faster than ps --no-headers -o euser -p <pid>.

[dess]

The best answer, since it's the fastest, especially if changed to : getent passwd $(< /proc/"$uid"/loginuid) | sed 's/\:.*$//'

[Community]

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