如何在搜索模式之间获取内容

Question

我有一些这种格式的文本。

echo "Hi"
line A 
line B
echo "hello"
line C
echo "him"
line D
echo "hind legs"

当我为 "hi" grep 时，我会得到

echo "Hi"
echo "him"
echo "hind legs"

我的要求是获取搜索行之间的所有行。所需输出：

数据 1：

echo "Hi"
line A 
line B
echo "hello"
line C
echo "him"

数据2：

echo "him"
line D
echo "hind legs"

请帮忙。

Answer 1

sed -ne '/echo "Hi"/,/echo "him"/p' file > data1
sed -ne '/echo "him"/,/echo "hind legs"/p' file > data2

Answer 2

这个 awk 会将输入分成几个文件 data1.txt、data2.txt 等等：

awk 'tolower([=10=]) ~ "hi" { if(fname) { print > fname; close(fname) } ++n; fname = "data" n ".txt" } { print > fname }' filename

它的工作原理如下：

tolower([=11=]) ~ "hi" {         # If a line matches the search pattern
                             # case-insensitively,
  if(fname) {                # if there is a current file (i.e., if this is
                             # not the first delimiter line),
    print > fname            # print it to the current file
    close(fname)             # then close that file
  }
  ++n                        # increase counter
  fname = "data" n ".txt"    # build new file name
}
{
  print > fname              # print lines to that current file
}

如何在搜索模式之间获取内容

How to get content between the search patterns

shell

awk

sed

grep

sh