如何解析固定行数，如果关键字匹配则显示前2行

Question

我是bash的新手，但真的需要帮助，我正在解析一个具有相同结构的日志文件，我需要每7行解析一次，如果关键字“失败”匹配，则显示前2行，然后继续下7行。

如果我尝试手动操作，结果会是这样的，但我只能使用关键字"cat multipath_output | grep -B3‘故障’| grep -A1 DGC“

日志文件是

mpathes (3600601601a003200eca1345cf13ce511) dm-84 DGC,VRAID
size=800G features='0' hwhandler='1 alua' wp=rw
|-+- policy='round-robin 0' prio=0 status=enabled
| |- 0:0:1:147 sdtu  65:704   failed faulty running
| `- 1:0:0:147 sdtx  65:752   failed faulty running
`-+- policy='round-robin 0' prio=0 status=enabled
  |- 1:0:1:147 sdadh 65:816   failed faulty running
  `- 0:0:0:147 sdadi 65:832   failed faulty running
mpathdn (3600601601a00320024631529381ee511) dm-50 DGC,VRAID
size=200G features='0' hwhandler='1 alua' wp=rw
|-+- policy='round-robin 0' prio=0 status=enabled
| |- 0:0:0:114 sdhi  133:128  failed faulty running
| `- 1:0:1:114 sdib  134:176  failed faulty running
`-+- policy='round-robin 0' prio=0 status=enabled
  |- 0:0:1:114 sdqn  132:368  failed faulty running
  `- 1:0:0:114 sdrs  134:352  failed faulty running
mpathf (3600601601a00320078b040c86b38e211) dm-19 DGC,RAID 5
size=60G features='0' hwhandler='1 alua' wp=rw
|-+- policy='round-robin 0' prio=130 status=active
| |- 0:0:1:0   sdim  135:96   active ready  running
| `- 1:0:0:0   sdke  66:288   active ready  running
`-+- policy='round-robin 0' prio=10 status=enabled
  |- 0:0:0:0   sda   8:0      active ready  running
  `- 1:0:1:0   sdj   8:144    active ready  running
matchg (3600601601a003200e2efb15f8442e511) dm-257 DGC,VRAID
size=200G features='0' hwhandler='1 alua' wp=rw
|-+- policy='round-robin 0' prio=0 status=enabled
| |- 0:0:1:205 sdye  128:736  failed faulty running
| `- 1:0:0:205 sdyk  129:576  failed faulty running
`-+- policy='round-robin 0' prio=0 status=enabled
  |- 1:0:1:205 sdaht 128:880  failed faulty running
  `- 0:0:0:205 sdahu 128:896  failed faulty running

输出应为：

mpathes (3600601601a003200eca1345cf13ce511) dm-84 DGC,VRAID size=800G features='0' hwhandler='1 alua' wp=rw

mpathdn (3600601601a00320024631529381ee511) dm-50 DGC,VRAID
size=200G features='0' hwhandler='1 alua' wp=rw

matchg (3600601601a003200e2efb15f8442e511) dm-257 DGC,VRAID
size=200G features='0' hwhandler='1 alua' wp=rw

"cat multipath_output | grep -B3   'faulty'  | grep -A1  DGC"

Script :-

FILES="multipath.txt"
for f in $FILES
do
        echo "$f"
        cat $f
        # Logic needs to go here

done

Answer 1

如果在awk (从bash shell，LOL)中可以做到这一点，那么下面的方法应该可以。我会稍微重写一下规则：

1. “块”的开始是当第一个字符是“m”
2. 时我们要捕获块的第1行和第2行，当块中的任何行包含单词“

”时

1. 如果前一个块有故障，我们将显示第1行和第2行。

作为一行程序：

awk '/^m/ {if (fault) {print line1 " " line2} ; line1=$0 ; fault=0} ; /^size/ {line2=$0} ; /faulty/ {fault=1} ; END {if (fault=1) {print line1 " " line2}}' t

或者是逐行

打印错误/^m/ {if () {print line1“”line2}；line1=$0；fault=0}

1. /^size/ {line2=$0}
2. /faulty/ {fault=1}
3. END {if (Line2) {print line1“”line2}}

如果行以m开头，则显示1=Grab line1。如果上一块出现故障，则显示line1和line2。当我们得到第一个块时，错误就不是真的了。

当行以大小开始时，使用2=Grab line2。

3=If块中有故障，请将标志设置为true

4=Since我们在一个块的末尾显示故障，在最后一个块之后检查是否有故障。

mpathes (3600601601a003200eca1345cf13ce511) dm-84 DGC,VRAID size=800G features='0' hwhandler='1 alua' wp=rw
mpathdn (3600601601a00320024631529381ee511) dm-50 DGC,VRAID size=200G features='0' hwhandler='1 alua' wp=rw
matchg (3600601601a003200e2efb15f8442e511) dm-257 DGC,VRAID size=200G features='0' hwhandler='1 alua' wp=rw