如何在指定的时间范围内打印从日志文件中提取的行？

Posted 2021-04-25

我想从2017-12-19 19:14获取结果，直到整天从一个看起来像这样的日志文件 -

/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:00.723 Info: Saving /var/opt/MarkLogic/Forests/Meters/00001829
/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:01.134 Info: Saved 9 MB at 22 MB/sec to /var/opt/MarkLogic/Forests/Meters/00001829
/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:01.376 Info: Merging 19 MB from /var/opt/MarkLogic/Forests/Meters/0000182a and /var/opt/MarkLogic/Forests/Meters/00001829 to /var/opt/MarkLogic/Forests/Meters/0000182c, timestamp=15137318408510140
/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:02.585 Info: Merged 18 MB in 1 sec at 15 MB/sec to /var/opt/MarkLogic/Forests/Meters/0000182c
/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:05.200 Info: Deleted 15 MB at 337 MB/sec /var/opt/MarkLogic/Forests/Meters/0000182a
/var/opt/MarkLogic/Logs/ErrorLog_1.txt:2017-12-19 19:14:05.202 Info: Deleted 9 MB at 4274 MB/sec /var/opt/MarkLogic/Forests/Meters/00001829  

我是Unix的新手，熟悉grep命令。我尝试了下面的命令

date="2017-12-19 [19-23]:[14-59]"
echo "$date"
grep "$date" $root_path_values

但它会抛出无效的范围结束错误。有解决方案吗日期将来自一个变量，因此它将是不可预测的。因此，不要只是记住这个例子。 $ root_path_values是一系列错误文件，如errorLog.txt，errorLog_1.txt，errorLog_2.txt等。

答案

egrep '2017-12-19 (19|2[0-3]):(1[4-9]|[2-5][0-9]):*.*' path/to/your/file试试这个正则表达式。

如果您需要变量模式：

#!/bin/bash
date="2017-12-19 (19|2[0-3]):(1[4-9]|[2-5][0-9]):*.*"
egrep ${date} path/to/your/file