Question

我在一个文件夹中有多个文件。这是一个文件的样子 File1.txt

ghfgh gfghh
  dffd  kjkjoliukjkj
  sdf ffghf
  sf 898575
  sfkj utiith

## 
my data to be extracted

我想从所有文件中提取“ ##”模式正下方的行，并将其写入输出文件。我也希望文件名也附加在输出文件中。所需的输出

>File1
My data to be extracted
>File2
My data to be extracted
>File3
My data to be extracted 

This is what i tried 
awk '/##/{getline; print FILENAME; print ">"; print}' *.txt > output.txt

Answer 1

假设每个文件有一个摘录（否则将重复文件名头）

UPDATE wpp_posts join wp_term_relationships on (wp_posts.ID = wp_term_relationships.object_id)
SET post_content = '' 
WHERE wp_term_relationships.term_taxonomy_id = 410

Answer 2

抢救Perl！

Camera

perl -ne 'print ">$ARGV\n", scalar <> if /^##/' -- *.txt > output.txt逐行读取输入
-n包含当前输入文件名
$ARGV从输入中读取一行

Answer 3

使用grep的快速方法：

grep -A1 '##' *.txt|grep -v '##' > output.txt

Answer 4

POSIX或GNU sed：

$ sed -n '/^##/{n;p;}' file
my data to be extracted

grep和sed：

$ grep -A 1 '##' file | sed '1d'
my data to be extracted

awk / sed解决方案，仅在匹配模式后才打印下一行

4 个答案: