如何用 grep 或 awk 进行计数，模式是否分别出现 x 次（一个接一个）

Question

我需要一些代码的帮助。我需要的是“if 语句”代码，用 grep 或其他命令分别计算文件中两个单词重复的次数。 比赛的文件是：

2011/05/04 21:47:37 45992299 c blabla (xxx/xxx/xx:xxx): invalid (0 ms) (of 0 avail 0) - CHANNEL 1 (0,8W) (invalid SID)
2011/05/04 21:47:46 45992299 c blabla (xxx/xxx/xx:xxx): invalid (0 ms) (of 0 avail 0) - 0000:0000 unknown (invalid SID)
2011/05/04 21:48:11 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2010 ms) by blablabla (of 1 avail 1) - CHANNEL
2011/05/04 21:48:21 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:48:42 45992299 c blabla (xxx/xxx/xx:xxx): invalid (0 ms) (of 0 avail 0) - 0000:1000 unknown (invalid SID)
2011/05/04 21:49:19 45992299 c blabla (xxx/xxx/xx:xxx): not found (2007 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:49:30 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL CHANNEL (0,8
2011/05/04 21:49:32 45992299 c blabla (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:50:01 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2007 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:50:18 45992299 c blabla (xxx/xxx/xx:xxx): invalid (0 ms) (of 0 avail 0) - 0000:0333 unknown (invalid SID)
2011/05/04 21:50:21 45992299 c blabla (xxx/xxx/xx:xxx): invalid (0 ms) (of 0 avail 0) - 0000:0333 unknown (invalid SID)
2011/05/04 21:51:06 45992299 c blabla (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:51:08 45992299 c blabla (xxx/xxx/xx:xxx): not found (2010 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:51:18 45992299 c blabla (xxx/xxx/xx:xxx): not found (2007 ms) by blablabla (of 1 avail 1) - CHANNEL
2011/05/04 21:52:19 45992299 c blabla (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:52:45 45992299 c blabla (xxx/xxx/xx:xxx): not found (2008 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:53:56 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2007 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:53:58 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2008 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:01 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2008 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:04 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2010 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:07 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2010 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:10 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2010 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:13 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2009 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:15 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2007 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:18 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2011 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)
2011/05/04 21:54:20 45992299 c blabla (xxx/xxx/xx:xxx): not found (2763 ms) by blablabla (of 1 avail 1) - CHANNEL (0.8W)
2011/05/04 21:54:29 05002299 c blabla2 (xxx/xxx/xx:xxx): not found (2008 ms) by blablabla (of 1 avail 1) - CHANNEL (0,8W)

如果单词（模式）“未找到”在每一行中一一重复（如示例文件中），并且如果达到 10 次，则执行某些操作（我将输入命令）

这部分代码有没有可能实时监控当前文件？ 我的意思是如果用 crond 启动一次来持续监控我的文件？

编辑 现在我发现我需要更多的脚本而不是其他的东西.. 如何实现我的脚本始终（实时）监视文件中的最后 10 行？

感谢帮助！

Answer 1

当且仅当有 10 个或更多连续行显示“未找到”时，此 AWK 代码才会打印消息：

! /not found/ {
    if (count >= 10) {
        print "More than 10 lines found"
    }
    count = 0
}
/not found/ {
    ++count
}

END {
    if (count >= 10) {
        print "More than 10 lines found"
    }

Answer 2

类似的东西

#!/bin/bash

while ${cleanLogFile:-true} ; do
   if (( $(grep -c "not found" logFile) > 9 )) ; then
       echo "Too many 'not found's in logFile"
       # do your stuff here
       cleanLogFile=false
    fi
done