Linux Centos with dmesg timestamp

Question

I would like to read my Centos 5.x dmesg with timestamp, how do I do this?

Answer 1

dmesg reads the Kernel log ring buffer. It doesn't do timestamps. What you should do is configure syslog to grab the kernel logs from that buffer and send them to a file (if it isn't already set to do so). Note, default CentOS 5.x syslog config sends kernel logs to /var/log/messages, as I recall.

If you'd like to send all kernel (dmesg) logs to /var/log/kern.log, using the default syslog daemon, you'd add a line like the following to /etc/syslog.conf

kern.*                         /var/log/kern.log

Answer 2

There is solution "Enabling Timestamps for dmesg/Kernel Ring Buffer"

You could add:

printk.time=1

to kernel cmdline.

As for me, I have added to rc.local on all machines with puppet. It's easier for me) :

if test -f /sys/module/printk/parameters/time; then
   echo 1 > /sys/module/printk/parameters/time
fi

Answer 3

I've written this simple script. Yes, it's slow. If you want something faster you either actually write a script on perl, python or something else. I'm sure this simple script can give you the hang of how it can be calculated.

Please note I ignored the seconds fraction registered in each line (after the . in the timestamp).

#!/bin/bash
localtime() {
 perl -e "print(localtime($1).\"\n\");";
}

upnow="$(cut -f1 -d"." /proc/uptime)"
upmmt="$(( $(date +%s) - ${upnow} ))"

dmesg | while read line; do
 timestamp="$(echo "${line}" | sed "s/^\[ *\([0-9]\+\).*/\1/g")"
 timestamp=$(( ${timestamp} + ${upmmt} ))
 echo "${line}" | sed "s/^[^]]\+]\(.*\)/$(localtime "${timestamp}") -\1/g"
done

I hope it helps. :)

Answer 4

A little perl script as below. It's a general way, and I'm not the author.

dmesg|perl -ne 'BEGIN{$a= time()- qx!cat /proc/uptime!};s/(\d+)\.\d+/localtime($1 + $a)/e; print $_;'

• perl -n is a way to read standard input and read into variable $_.
• The body (not the BEGIN section) is then run once for each line.
• BEGIN runs the code in {} once.
• $a is the start of dmesg since the epoch
• The s/... command takes the value in $_ and substitutes the #####.###### part of the timestamp with the "localtime" version of the dmesg offset ( $1) added to the system start time ($a)
• print $a prints the dmesg with the locale friendly time stamp substituted for the "seconds since boot" time stamp.

Answer 5

Script modification in case line do not begin with a "["

#!/bin/bash
localtime() {
 perl -e "print(localtime($1).\"\n\");";
}

upnow=$(cut -f1 -d"." /proc/uptime)
upmmt=$(( $(date +%s) - ${upnow} ))

dmesg \
| while read LINE; do
    if [ "$(echo ${LINE} | egrep -v "^\[")" == "" ] ; then
        timestamp=$(echo "${LINE}" | sed "s/^\[ *\([0-9]\+\).*/\1/g")
        timestamp=$(( ${timestamp} + ${upmmt} ))
        echo "${LINE}" | sed "s/^[^]]\+]\(.*\)/$(localtime "${timestamp}") -\1/g"
    else
        echo "${LINE}"
    fi
done

Answer 6

This is an update on Plutoid's suggestion, removing the leading spaces from the timestamp.

dmesg|perl -ne 'BEGIN{$a= time()- qx!cat /proc/uptime!};s/( *)(\d+)\.\d+/localtime($2 + $a)/e; print $_;' 

• perl -n is a way to read standard input and read into variable $_.
• The body (not the BEGIN section) is then run once for each line.
• BEGIN runs the code in {} once.
• $a is the start of dmesg since the epoch (seconds)
• The s/... command takes the value in $_ and substitutes the \s*#####.###### part of the timestamp with the "localtime" version of the dmesg offset ( $1) added to the system start time ($a)
• print $a prints the dmesg with the locale friendly time stamp substituted for the "seconds since boot" time stamp.