linux - 在同一行打印 awk 的两个匹配变量

Question

这是我写的脚本。

Date=$(date +'%Y-%m-%d')

for i in `awk -F"[.: ]" '/start/{start=($4 * 3600) + ($5 * 60) + $6;date=$4$5} /end/{print date;print (($4 * 3600) + ($5 * 60) + $6)-start;start=""}' logs.txt`;  do
    echo "$i"
done

logs.txt 包括:

11.04.2018 09:21:35 aaaaa: start_time
11.04.2018 09:22:35 aaaaa: end_time
11.04.2018 10:45:00 aaaaa: start_time
11.04.2018 11:00:00 aaaaa: end_time

在此预期输出是:

2018-04-11 09:21:35,60
2018-04-11 10:45:00,900

但我得到的输出是:

2018-04-13,0921
2018-04-13,60
2018-04-13,1045
2018-04-13,900

谁能纠正错误是什么？

Answer 1

在 GNU awk 中:

awk '
{
    split($1 " " $2,a,"[.: ]")                                  # split for mktime
    s=mktime(a[3] " " a[2] " " a[1] " " a[4] " " a[5] " " a[6]) # to secs
    if($4=="end_time")                                          # at end_time
        print strftime("%Y-%m-%d %H:%M:%S,",p) s-p              # print start time and difference
    p=s                                                         # set secs to prev
}' file
2018-04-11 09:21:35,60
2018-04-11 10:45:00,900