辨識日誌文件的持續時間

如何使用 awk 或 perl 腳本在以下日誌文件中計算（刷新時間 - 聚合時間）的持續時間

09/03/2020 00:05:03.364 Aggregated 0 NMEs at a rate of 0 NMEs/sec
09/03/2020 00:05:03.366 Scheme S20_SessionClassAggregation tree contained 0 nmes, 0 flushed, 0 remain.
09/03/2020 00:05:03.582 Flushed 0 NMEs at a rate of 0 NMEs/sec
09/03/2020 00:20:03.598 Aggregated 0 NMEs at a rate of 0 NMEs/sec
09/03/2020 00:20:03.602 Scheme S20_SessionClassAggregation tree contained 0 nmes, 0 flushed, 0 remain.
09/03/2020 00:20:03.860 Flushed 0 NMEs at a rate of 0 NMEs/sec

例子：

我需要與第 3 行 (009/03/2020 00:05:03.582) - 第 1(09/03/2020 00:05:03.364) 和第 6 行(09/03/2020 00:20:03.860) 的區別 -第 4 行 (09/03/2020 00:20:03.598)

預期成績：

0 min 0 sec 218 ms
0 min 0 sec 262 ms
.
.
.

假設您的時間戳總是成對出現，您可以使用GNU sedand來做到這一點GNU coreutils：

# Extract the relevant timestamps and convert them to secs + ns
<infile \
sed -nE 's/(.*) (Aggregated|Flushed).*/\1/; T; s/^/date -d "/; s/$/" +%s.%N/ep' |

# Find the time difference
sed '1~2 s/^/-/' |
paste -d+ - -    |
bc               |

# Print the difference in the desired format
while read dt; do
 date -u -d "1970/01/01 + $(printf "%.3f" $dt) sec" +'%_M min %S sec %3N ms'
done

輸出：

0 min 00 sec 218 ms
0 min 00 sec 262 ms

引用自：https://unix.stackexchange.com/questions/608394