How to find zombie process?

  • System information as of Fri Mar  9 19:40:01 KST 2012
    
      System load:    0.59               Processes:           167
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      Swap usage:     0%                 IP address for eth1: 192.168.0.1
    
      => There is 1 zombie process.
    
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    Last login: Fri Mar  9 10:23:48 2012
    [email protected]:~$ ps auxwww | grep 'Z'
    USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND
    usera     13572  0.0  0.0   7628   992 pts/2    S+   19:40   0:00 grep --color=auto Z
    [email protected]:~$ 
    

    How to find that zombie process?

    why dont you open the system monitor and search for the zombie process?

    How to do that on a headless no-X server?

    Surprising that no answer below actually says that there's no zombie process in the system based on the above output. If there really was one, the `ps auxwww | grep 'Z'` command should have shown a process in a `Z` state. The "system information" saying `=> There is 1 zombie process.` seems to be a bug. Either that, or there's missing information in the question.

  • Duncanmoo

    Duncanmoo Correct answer

    7 years ago

    To kill a zombie (process) you have to kill its parent process (just like real zombies!), but the question was how to find it.

    Find the zombie (The question answered this part):

    [email protected]:~$ ps aux | grep 'Z'
    

    What you get is Zombies and anything else with a Z in it, so you will also get the grep:

    USER       PID     %CPU %MEM  VSZ    RSS TTY      STAT START   TIME COMMAND
    usera      13572   0.0  0.0   7628   992 pts/2    S+   19:40   0:00 grep --color=auto Z
    usera      93572   0.0  0.0   0      0   ??       Z    19:40   0:00 something
    

    Find the zombie's parent:

    [email protected]:~$ pstree -p -s 93572
    

    Will give you:

    init(1)---cnid_metad(1311)---cnid_dbd(5145)
    

    In this case you do not want to kill that parent process and you should be quite happy with one zombie, but killing the immediate parent process 5145 should get rid of it.

    Additional resources on askubuntu:

    The result you show in your answer is the the grep command itself, not the zombie process. It is the same mis-interpratation as Pablo made in his answer. The answer by Rinzwind below does actually look for the zombie process and list them. Another option could be to grep for "defunct"

    pstree -H your_desired_pid -p

    Thanks Greg for adding to the discussion, but please remember this is a help site, just pasting a command without explaining anything is not helpful to most people coming here looking for help.

    This is a great answer! It is still valid today! I was able to find my zombie process and kill its parent process without any problems. Thank you!

    if you do not have pstree installed, `ps wauxf` does the same thing

    You can also try `ps aux |grep "defunct"`

    I'd add the `-w` flag. As in `grep -w Z` to only match the word Z and avoid clutter.

  • Even though this question is old I thought everyone deserved a more reliable answer:

    ps axo pid=,stat=
    

    This will emit two whitespace-delimited columns, the first of which is a PID and the second of which is its state.

    I don't think even GNU ps provides a way to filter by state directly, but you can reliably do this with awk

    ps axo pid=,stat= | awk '$2~/^Z/ { print }'
    

    You now have a list of PIDs which are zombies. Since you know the state it's no longer necessary to display it, so that can be filtered out.

    ps axo pid=,stat= | awk '$2~/^Z/ { print $1 }'
    

    Giving a newline-delimited list of zombie PIDs.

    You can now operate on this list with a simple shell loop

    for pid in $(ps axo pid=,stat= | awk '$2~/^Z/ { print $1 }') ; do
        echo "$pid" # do something interesting here
    done
    

    ps is a powerful tool and you don't need to do anything complicated to get process information out of it.

    (Meaning of different process states here - https://unix.stackexchange.com/a/18477/121634)

    `awk` is also a powerful tool which doesn't just split text but can also match it. +1 ... the others used `grep` where it's unnecessary and imprecise.

    so now that i have list of zombie processes. how do i kill them?

    @chovy: It will depend, but generally involves killing or signalling the parent. Other answers here go in to that. From within the loop shown above you can find the parent pid like this: `ps -p "$pid" -opid=,ppid=`

    if i will the parent won't it kill all its child processes? I just want to kill the one zombie process. I know the ppid.

    @chovy: The command you want is `kill -CHLD "$ppid"` but even this may not be enough, read this entire post to understand what your options are.

    Just my 2 cents... `ps aux | grep 'Z' | grep -vE "(PID|grep)" | tr -s " " | cut -d " " -f2 | xargs -i pstree -p -s {}`

    I dunno; `grep` may just be clearer as there's only one reason `Z` would appear in the output of `ps axo pid=,stat=`. Why not `ps axo pid=,stat=|grep Z` or `ps axo pid=,stat=|sort -k 2r|head -1` or `ps axo stat=,pid=|sort -r|head -1`?

    ton: If a process name has Z in it you will get an incorrect match. If you want to run pstree on each pid you can do so by piping the output of the awk command above to xargs without resorting to two extra, needless filtering programs. @dlamblin: If you do not strip the STAT column from the output of ps it is harder to use the pid for anything other than human consumption. Why would you do either of these less-correct and more complicated things when I have already provided to you a solution which is better in every way? There are many things you *can* do but not many things you *should* do.

    I do suggest to add `ppid=` to the options list, so no need to use a separate command to obtain ppid.

    @Ding-YiChen: Yes, that would be wise. In that case I would also change to a `while read` construct, e.g.: (bash only): `while IFS=' ' read -r pid ppid ; do echo "zombie pid $pid parent $ppid" ; done < <(ps axo pid=,ppid,stat= | awk '$3~/^Z/ { print $1, $2 }')`

  • ps aux | awk '{ print $8 " " $2 }' | grep -w Z

    From: http://www.cyberciti.biz/tips/killing-zombie-process.html

    From the comments an improved one:

    for p in $(ps jauxww | grep Z | grep -v PID | awk '{print $3}'); do
        for every in $(ps auxw | grep $p | grep cron | awk '{print $2}'); do
            kill -9 $every;
        done;
    done;
    

    Careful though: this one also kills the proces.

    still returns nothing. I think my way also wasn't wrong.

    The 2nd example is hellishly unreliable and the former is needlessly verbose (try `ps axo pid=,stat= | awk '$2~/Z/ {print $1}'` instead).

  • Less is more though:

    ps afuwwx | less +u -p'^(\S+\s+){7}Z.*'
    

    That's like, give me a forest (tree) of all users' processes in a user oriented format with unlimited width on any tty and show it to me at half a screen above where it matches the case that the 8th column contains a Z, and why not highlight the whole line.

    User oriented format seems to mean: USER, PID, %CPU, %MEM, VSZ, RSS, TTY, STAT, START, TIME, COMMAND so the Zombie status will show up in the 8th column.

    You can throw in an N before the p if you want line numbers, and a J if you want an asterisk at the match. Sadly if you use G to not highlight the line that asterisk will not show, though J creates space for it.

    You end up getting something that looks like:

    …
      root      2919  0.0  0.0  61432  5852 ?      Ss Jan24 0:00 /usr/sbin/sshd -D
      root     12984  0.0  0.1 154796 15708 ?      Ss 20:20 0:00  \_ sshd: lamblin [priv]
      lamblin  13084  0.0  0.0 154796  9764 ?      S  20:20 0:00      \_ sshd: [email protected]/0
    * lamblin  13086  0.0  0.0  13080  5056 pts/0  Z  20:20 0:00          \_ -bash <defunct>
      lamblin  13085  0.0  0.0  13080  5056 pts/0  Ss 20:20 0:00          \_ -bash
      root     13159  0.0  0.0 111740  6276 pts/0  S  20:20 0:00              \_ su - nilbmal
      nilbmal  13161  0.2  0.0  13156  5004 pts/0  S  20:20 0:00                  \_ -su
      nilbmal  13271  0.0  0.0  28152  3332 pts/0  R+ 20:20 0:00                      \_ ps afuwwx
      nilbmal  13275  0.0  0.0   8404   848 pts/0  S+ 20:20 0:00                      \_ less +u -Jp^(\S+\s+){7}Z.*
    …
    

    You could follow this up with (and it'll detect if your terminal likes -U Unicode or -A Ascii):

    pstree -psS <PID LIST>
    

    OR just, you know, use the up-arrow in less to follow that tree/forest through the hierarchy; which is what I was recommending with the "Less is more" approach.

  • I suggest you this command:

    ps aux | awk '"[Zz]" ~ $8 { printf("%s, PID = %d\n", $8, $2); }'
    

    Using `aux` and munging strings out of it is needlessly unreliable when you can use `-o` and request exactly what you want. Use `ps ax -o pid=,stat= | awk '$2 ~ "[Zz]" { printf("%s, PID = %d\n", $2, $1); }'` instead.

  • To list process zombies, try this command:

    ps j | awk '$7 ~ "Z"'
    

    You may need to change $7 depending on your operating system.

    This also will return the list of their parent process ids (PPID).

    To try to kill the zombies (after testing the above command), try:

    kill -9 $(ps j | awk 'NR>1 && $7 ~ "Z" {print $2}')
    

    To identify their parents, try with pstree, like:

    $ ps j | awk 'NR>1 && $7 ~ "T" {print $2}' | xargs -L1 pstree -sg
    systemd(1)───sshd(1036)───sshd(2325)───sshd(2325)───bash(2383)───zombie(2430)
    systemd(1)───sshd(1036)───sshd(2325)───sshd(2325)───bash(2383)───zombie(2431)
    systemd(1)───sshd(1036)───sshd(2325)───sshd(2325)───bash(2383)───zombie(2432)
    

    Resorting to stripping one column out of the `j` format for this is needlessly complicated. Use `-o` to select what you want instead.

    `ps j` doesn't print all processes in the system. It only lists the current user procs (in BSD jobs style) so it may miss zombie processes.

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