dmesg将时间戳转换为人类格式

Question

我有以下dmesg示例：

 throttled log output.
57458] bar 3: test 2 on bar 8 is available
[   19.696163] bar 1403: test on bar 1405 is available
[   19.696167] foo: [   19.696168] bar 3: test 5 on bar 1405 is available
[   19.696178] foo: [   19.696179] bar 1403: test 5 on bar 1405 is available
[   20.928730] foo: [   20.928733] bar 1403: test on bar 1408 is available
[   20.928742] foo: [   20.928745] bar 3: test on bar 1408 is available
[   24.878861] foo: [   25.878861] foo: [   25.878863] bar 1403: bar 802 is present

我想将行中的all时间戳转换为人类格式（"%d/%m/%Y %H:%M:%S"）

注意：该系统没有dmesg -T，也没有安装Perl。我希望使用带sed或awk的解决方案，但也可以使用python。

我已经找到了解决这个问题的几种方法，但是没有一个能完全满足我的需要。我也不知道如何根据自己的需要进行修改。

awk -F"]" '{"cat /proc/uptime | cut -d \" \" -f 1" | getline st;a=substr( $1,2, length($1) - 1);print strftime("%d/%m/%Y %H:%M:%S",systime()-st+a)" "$0}'

或

sed -n 's/\]//;s/\[//;s/\([^.]\)\.\([^ ]*\)\(.*\)/\1\n\3/p' |  while read first; do    read second;    first=`date +"%d/%m/%Y %H:%M:%S" --date="@$(($seconds - $base + $first))"`;   printf "[%s] %s\n" "$first" "$second";  done

here中还有一个python脚本。但是输出一些错误，而我对此有零了解。

谢谢！

Answer 1

这有点麻烦，但它至少应该给您一些使用的技巧：

awk '
  {
    # tail will be the part of the line that still requires processing
    tail = $0;                               

    # Read uptime from /proc/uptime and use it to calculate the system
    # start time
    "cat /proc/uptime | cut -d \" \" -f 1" | getline st;
    starttime = systime() - st;

    # while we find matches
    while((start = match(tail, /\[[^[]*\]/)) != 0) {
      # pick the timestamp from the match
      s = substr(tail, start + 1, RLENGTH - 2);

      # shorten the tail accordingly
      tail = substr(tail, start + RLENGTH);

      # format the time to our preference
      t = strftime("%d/%m/%Y %H:%M:%S", starttime + s);

      # substitute it into the original line. [] are replaced with || so
      # the match is not re-replaced in the next iteration.
      sub(/\[[^[]*\]/, "|" t "|", $0);
    }

    # When all matches have been replaced, print the line.
    print $0
  }' foo.txt