How to display processes which have been running for more than a X hours?

UNIX for Beginners Q & A
1

Hi,

Is it possible to display processes which have been running for more than a 5hrs using a variation of the ps -ef command?

Regards,

Manny

2

The ps command output varies per platform, so could you specify what OS you are using?

On Linux you could try:

ps -eo pid,etimes | awk '$2/3600>=5{print $1}'
3 
ps -eo pid,etime

and the "no header" variant

ps -e -o pid= -o etime=

are quite portable (Posix standard).
The usual output format for times is D-HH:MM:SS
where the D- (=days) prefix is only printed if D > 0

1 Like
4

Sorry, i should have said, this command needs to run on a unix (solaris) operating system.

Thanks.

5

So, since you're using a Solaris/SunOS system, instead of using:

ps -eo pid,etime

and the "no header" variant

ps -e -o pid= -o etime=

use:

/usr/xpg4/bin/ps -eo pid,etime

and the "no header" variant

/usr/xpg4/bin/ps -e -o pid= -o etime=
6

Have to add that even the HH: is usually omitted if HH is zero.
A suitable postprocessing with awk:

ps -e -o pid= -o etime= | /usr/xpg4/bin/awk '{ n=split($NF, A, /[-:]/); seconds=(A[n]+60*(A[n-1]+60*(A[n-2]+24*A[n-3]))) } (seconds > 5*3600) { print $1 }'

The { print $1 } prints only the pids.
Another example with two more columns and printing everything:

ps -e -o pid= -o user= -o args= -o etime= | /usr/xpg4/bin/awk '{ n=split($NF, A,  /[-:]/); seconds=(A[n]+60*(A[n-1]+60*(A[n-2]+24*A[n-3]))) } (seconds > 5*3600)'

BTW Sun made the /usr/bin/ps Posix compliant; there is no /usr/xpg4/bin/ps. But df, find, awk, and >60 more Posix compliant commands live in /usr/xpg4/bin/