I am trying to get back the time in the format hh:mm without the seconds,
running this:
ps -p $pid -o time=
gets back (for example) 00:03:19 but I want just (for example) 00:03
I have tried looking at the manual and searched on the net, couldn't find anything that i can understand. Can someone help me please. Thanks in advance.
ps itself doesn't offer to configure the time format. However, you can pipe to cut:
ps -p $pid -o time= | cut -d: -f1,2
Related
I want to obtain all the process that are running in the system, but only from yesterday.
I am using this, ps -eo etime,pid
but i need only list the process from yesterday, any idea?
INFO: active from yesterday, process actually running from yesterday
Thanks in advance
If you want all the pids that have been running for more than 1 day but less than 2:
ps -e -o pid= -o etime= | sed 's/^ *//' | awk -F '[ -]+' 'NF>2 && $2==1 {print $1}'
if you want just more than 1 day, change it to $2>=1
It is not clear what exactly you need
But if you needed it from a specific time, just add it to crontab.
For example:
0 2 * * * ps -eo etime,pid>/tmp/processes.txt
This way you will have a snapshot from the processes running on 2:00
If you didnt mean that please be more specific of what exactly do you need
I'm writing a bash script that will print on the screen all the latest logs from a service that has already died (or still lives, both situations must work). I know its name and don't have to guess.
I'm having difficulty getting the latest PID for a process that has already died from journalctl. I'm not talking about this:
journalctl | grep "<processname>"
This will give me all the logs that include processname in their text.
I've also tried:
journalctl | pgrep -f "<processname>"
This command gave me a list of numbers which supposedly should include the pid of my process. It was not there.
These ideas came from searching for previous questions. I haven't found a question that answers specifically what I asked.
How can I extract the latest PID from journalctl for a specific process?
I figured this out.
First, you must be printing your PID in your logs. It doesn't appear there automatically. Then, you can use grep -E and awk to grab exactly the expression you want from your log:
Var=$(journalctl --since "24 hours ago" | grep -E "\[([0-9]+)\]" | tail -n 1 | awk '{print $5}' | awk -F"[][{}]" '{print $2}'
This one-liner script takes the logs from the last 24 hours, grep with -E to use an expression, tail -n 1 to grab the last most updated line from those results and then, using awk to delimit the line and grab the exact expression you need from it.
I have a bash script with contents-
#!/bin/bash
while true;do
netstat -antp | grep LISTEN | tr -s ' ' | cut -d ' ' -f 4 > /tmp/log
sleep 100
done
Say I create a service which executes the script on boot.But when I use ps -eo command I'm able to see the commands being executed.For eg -
netstat -antp
grep LISTEN
tr -s ' '
cut -d ' ' -f 4
But I wish to suppress this output and hide the execution of these commands.Is there a way to do it?
Any other suggestions are welcome too.Thanks in advance!
You can't hide running processes from the system, at least not without some kernel hooks. Doing so is something typically only found in malware, so you'll not likely get much help.
There's really no reason to hide those processes from the system. If something in your script gets hung up, you'll want to see those processes to give you an idea of what's happening.
If there's a specific problem the presence of those processes is causing, you need to detail what that is.
I am working on unix.
I want to knwo the current cpu usage of a process.
I understood that ps give the average of cpu used till the process is up - it is not the current usage.
Is there a way to print only the cpu from the top command without 10 more parameters and
headers? I know how to do it with awk - this is not the way i want to do it.
top -p 20705 -bc -n 1 | tail -n 2 | awk '{ print $9}' | head -n 1
If there is another simple way to do it, not reading /proc/stat...
If there is a simple way doing it from c++, it is also ok.
Most likely, you will need to read /proc/stat, However, here is an interesting article with C code that may help you out. To understand and use the output from the program you should do man 5 proc. And here is the source code.
The bottom line is that you will need to read from /proc/stat to do what you want.
to see cpu usage of a proccess whose pid is 24556
ps -p 24556 -o \%cpu=
to see mem usage of a process named syslogd
ps -C syslogd -o \%mem=
I have the following bash script, that lists the current number of httpd processes, and if it is over 60, it should email me. This works 80% of the time but for some reason sometimes it emails me anyways when it is not over 60. Any ideas?
#!/bin/bash
lines=`ps -ef|grep httpd| wc -l`
if [ "$lines" -gt "60" ]
then
mailx -s "Over 60 httpd processes" me#me.com < /dev/null
fi
There is a delay between checking and emailing. In that time, some httpd processes might finish, or start, or both. So, the number of processes can be different.
You are including the grep process in the processes (most of the time, it could happen that the ps finishes before grep starts). An easy way to avoid that is to change your command to ps -ef | grep [h]ttpd. This will make sure that grep doesn't match grep [h]ttpd.
On linux, you have pgrep, which might be better suited for your purposes.
grep ... | wc -l can usually be replaced with grep -c ....
If you want to limit the number of httpd requests, I am sure you can set it in apache configuration files.
You've probably thought of this, but ...
At time t0, there are 61.
At time t1, when you read the email, there are 58.
Try including the value of $lines in the email and you'll see.
Or try using /proc/*/cmdline, it might be more reliable.
grep httpd finds all processes that include httpd in their name, including possibly grep httpd itself, and perhaps other ones.
"ps -ef|grep httpd" doesn't find just httpd processes, does it? It finds processes whose full (-f) listing in ps includes the string "httpd".
This probably doesn't solve your issue but you could simplify things by using pgrep instead.
you can do it this way too, reducing the use of grep and wc to just one awk.
ps -eo args|awk '!/awk/&&/httpd/{++c}
END{
if (c>60){
cmd="mailx -s \047Over 60\047 root"
cmd | getline
}
}'