AIX 中 -A/-B 替换之后/之前的 Grep 行

AIX 中 -A/-B 替换之后/之前的 Grep 行

我使用的是 AIX 6.1,它不支持 in-B-A标志:

grep: Not a recognized flag: B

假设我想运行:

cat file | grep -E -B4 'Directory entry type.*Indirect' | grep "Database name" | awk  '{print $4}'

在 AIX 中如何实现这种逻辑?

编辑

我的真实代码如下:

NAME_EXISTS=`db2 LIST DB DIRECTORY | grep -E -B5 'Directory entry type.*Remote' | grep "Database alias" | awk '{print $4}' | grep -i ${NAME} | wc -l`
if [ ${NAME_EXISTS} -gt 0 ]; then
    db2 LIST DB DIRECTORY | grep -E -A5 "Database alias.*${NAME}"
fi

这个想法是查找是否存在名为 的远程数据库$NAME,如果找到,则显示开头的 5 行Database alias.*${NAME}$NAME是独一无二的Database alias

输出db2 LIST DB DIRECTORY是这样的:

 System Database Directory

 Number of entries in the directory = 3

Database 1 entry:

 Database alias                       = OLTPA
 Database name                        = OLTPA
 Local database directory             = /db2/data
 Database release level               = 10.00
 Comment                              =
 Directory entry type                 = Indirect
 Catalog database partition number    = 0
 Alternate server hostname            =
 Alternate server port number         =

Database 2 entry:

 Database alias                       = OLTPF
 Database name                        = OLTP
 Node name                            = OLTPN
 Database release level               = 10.00
 Comment                              =
 Directory entry type                 = Remote
 Catalog database partition number    = -1
 Alternate server hostname            =
 Alternate server port number         =

Database 3 entry:

 Database alias                       = ADMIN
 Database name                        = ADMIN
 Local database directory             = /db2/data
 Database release level               = 10.00
 Comment                              =
 Directory entry type                 = Indirect
 Catalog database partition number    = 0
 Alternate server hostname            =
 Alternate server port number         =

对于NAME=OLTPF输出将是:

 Database alias                       = OLTPF
 Database name                        = OLTP
 Node name                            = OLTPN
 Database release level               = 10.00
 Comment                              =
 Directory entry type                 = Remote

因为NAME=OLTPE不会有任何输出。

答案1

ed 可能会提供一种简单的方法来完成此任务。

如果我们可以假设只有一个匹配项,那么您的管道的替代方案是使用 ed 并消除不必要的 cat 和辅助 grep:

ed -s file <<\EOED | awk '/Database name/ {print $4}'
    /Directory entry type.*Indirect/-4,//p
    q
EOED

如果有多个,不重叠匹配,可以使用 ed 的全局命令来标记它们:

ed -s file <<\EOED | awk '/Database name/ {print $4}'
    g/Directory entry type.*Indirect/-4,.p
    q
EOED

为了演示重叠匹配的情况,假设我们正在匹配字符串foo,并且第 7 行和第 9 行有匹配项,并且我们将每个匹配项的前三行作为上下文,输出将如下所示:

line 4      <--- context
line 5      <--- context
line 6      <--- context
line 7 foo  <--- matched
line 6      <--- context      <--- repeated
line 7 foo  <--- context      <--- repeated
line 8      <--- context
line 9 foo  <--- matched
line 10
line 11

答案2

我的做法略有不同。首先,运行db2 LIST DB DIRECTORY命令并将其输出保存到文本文件中。这样,您就不需要多次重新运行它。然后,对于每个目标名称,将名称传递给收集相关行的 awk 脚本:

## Run the db command
tempfile=$(mktemp)
db2 LIST DB DIRECTORY > "$tmpfile"
## I am assuming you will have a loop for the different target names
for name in OLTPF OLTPA; do
  awk -v name="$name" '{
       if(/Database alias/){n=$4; a[n]=$0; i=1}
       if (i<=6 && i>1){ a[n]=a[n]"\n"$0}
       i++;
      }END{if(name in a){print a[name]}}' $tempfile
done

答案3

$ awk -v RS= -F'\n' -v OFS='\n' '$1 ~ " OLTPF$"{for (i=1;i<=6;i++) print $i}' file
 Database alias                       = OLTPF
 Database name                        = OLTP
 Node name                            = OLTPN
 Database release level               = 10.00
 Comment                              =
 Directory entry type                 = Remote

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