如何用grep或awk计数,模式是否分别出现x次(一个接一个)

时间:2011-05-05 03:31:46

标签: count awk grep design-patterns

我需要很少的代码帮助。我需要的是“if语句”用grep或其他命令编码计数在文件中分别重复多少次。 文件竞赛是:

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”在每一行中重复一次(如示例文件中)并且如果达到10次,那么请完成(我会发出命令)

这部分代码在实时当前文件中监控是否有任何可能性? 我的意思是如果用crond开始一次连续监视我的文件?

修改 的 现在我看到我需要更多的脚本而不是其他的东西.. 如何实现我的脚本所有时间(实时)从文件中查看最后10行?

寻求帮助!

2 个答案:

答案 0 :(得分:1)

类似

#!/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

完成

考虑到你从其他人那里得到帮助的0%接受率,我不愿意花时间直到我知道你愿意为此付钱。请注意,接受和投票获得的帮助并不是出自您的声誉。

我希望这会有所帮助。

答案 1 :(得分:1)

当且仅当有10个或更多连续行“未找到”时,此AWK代码才会打印出该消息:

! /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"
    }