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edotan
Asked: 2012-03-31 06:37:53 +0800 CST

Linux Centos with dmesg timestamp

  • 772

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

linux centos dmesg

6 Answers

  1. Best Answer

    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
    • 11

  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
    • 9

  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. :)

    • 2

  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.
    • 0

  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
    • 0

  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.
    • 0