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在 awk/sed/grep 中删除匹配的第 n 行直到空行

发布于 2024-10-08 06:49:16 字数 59 浏览 9 评论 0原文

我需要删除文件中从匹配到下一个空白行的第 n 个匹配行（即从第 n 个匹配开始的一大块空白行分隔文本）。

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喜爱纠缠 2024-10-15 06:49:16

这将删除以第四个空行开始和结束的空白行的文本块。它还删除那些分隔线。

sed -n '/^$/!{p;b};H;x;/^\(\n[^\n]*\)\{4\}/{:a;n;/^$/!ba;d};x;p' inputfile

更改第一个 /^$/ 以更改开始匹配。更改第二个即可更改最终匹配。

给定以下输入：

aaa
---
bbb
---
ccc
---
ddd delete me
eee delete me
===
fff
---
ggg

此版本的命令：

sed -n '/^---$/!{p;b};H;x;/^\(\n[^\n]*\)\{3\}/{:a;n;/^===$/!ba;d};x;p' inputfile

将给出以下结果：

aaa
---
bbb
---
ccc
fff
---
ggg

编辑：

我从 sed 命令中删除了无关的 b 指令多于。

这是一个带注释的版本：

sed -n '      # don't print by default
  /^---$/!{   # if the input line doesn't match the begin block marker
    p;        # print it
    b};       # branch to end of script and start processing next input line
  H;          # line matches begin mark, append to hold space
  x;          # swap pattern space and hold space
  /^\(\n[^\n]*\)\{3\}/{    # if what was in hold consists of 3 lines
                           # in other words, 3 copies of the begin marker
    :a;       # label a
    n;        # read the next line
    /^===$/!ba;    # if it's not the end of block marker, branch to :a
    d};       # otherwise, delete it, d branches to the end automatically
  x;          # swap pattern space and hold space
  p;          # print the line (it's outside the block we're looking for)
' inputfile   # end of script, name of input file

任何明确的模式都应该适用于开始和结束标记。它们可以相同或不同。

This will delete a chunk of text that starts and ends with a blank line starting with the fourth blank line. It also deletes those delimiting lines.

sed -n '/^$/!{p;b};H;x;/^\(\n[^\n]*\)\{4\}/{:a;n;/^$/!ba;d};x;p' inputfile

Change the first /^$/ to change the start match. Change the second one to change the end match.

Given this input:

aaa
---
bbb
---
ccc
---
ddd delete me
eee delete me
===
fff
---
ggg

This version of the command:

sed -n '/^---$/!{p;b};H;x;/^\(\n[^\n]*\)\{3\}/{:a;n;/^===$/!ba;d};x;p' inputfile

would give this as the result:

aaa
---
bbb
---
ccc
fff
---
ggg

Edit:

I removed an extraneous b instruction from the sed commands above.

Here's a commented version:

sed -n '      # don't print by default
  /^---$/!{   # if the input line doesn't match the begin block marker
    p;        # print it
    b};       # branch to end of script and start processing next input line
  H;          # line matches begin mark, append to hold space
  x;          # swap pattern space and hold space
  /^\(\n[^\n]*\)\{3\}/{    # if what was in hold consists of 3 lines
                           # in other words, 3 copies of the begin marker
    :a;       # label a
    n;        # read the next line
    /^===$/!ba;    # if it's not the end of block marker, branch to :a
    d};       # otherwise, delete it, d branches to the end automatically
  x;          # swap pattern space and hold space
  p;          # print the line (it's outside the block we're looking for)
' inputfile   # end of script, name of input file

Any unambiguous pattern should work for the begin and end markers. They can be the same or different.

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纵山崖 2024-10-15 06:49:16

perl -00 -pe 'if (/pattern/) {++$count == $n and $_ = "\n";}' file

-00 是以“段落”模式读取文件（记录分隔符是一个或多个空行）

$` 是 Perl 的“预匹配”特殊变量（文本在匹配模式的前面）

perl -00 -pe 'if (/pattern/) {++$count == $n and $_ = "\n";}' file

-00 is to read the file in "paragraph" mode (record separator is one or more blank lines)

$` is Perl's special variable for the "prematch" (text in front of the matching pattern)

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囍笑 2024-10-15 06:49:16

在 AWK

/m1/  {i++};

(i==3)  {while (getline temp > 0 && temp != "" ){}; if (temp == "") {i++;next}};

{print}

中将其转换

m1 1
first

m1 2
second

m1 3
third delete me!

m1 4
fourth

m1 5
last

为

m1 1
first

m1 2
second

m1 4
fourth

m1 5
last

：删除“m1”的第三个块...

在此处在 ideone 上运行

哈！

In AWK

/m1/  {i++};

(i==3)  {while (getline temp > 0 && temp != "" ){}; if (temp == "") {i++;next}};

{print}

Transforms this:

m1 1
first

m1 2
second

m1 3
third delete me!

m1 4
fourth

m1 5
last

into this:

m1 1
first

m1 2
second

m1 4
fourth

m1 5
last

deleting the third block of "m1" ...

Running on ideone here

HTH!

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夏九 2024-10-15 06:49:16

强制性 awk 脚本。只需将 n=2 更改为您的第 n 个匹配项即可。

n=2; awk -v n=$n '/^HEADER$/{++i==n && ++flag} !flag; /^$/&&flag{flag=0}' ./file

输入

$ cat ./file
HEADER
line1a
line2a
line3a

HEADER
line1b
line2b
line3b

HEADER
line1c
line2c
line3c

HEADER
line1d
line2d
line3d

输出

$ n=2; awk -v n=$n '/^HEADER$/{++i==n&&++flag} !flag; /^$/&&flag{flag=0}' ./file
HEADER
line1a
line2a
line3a

HEADER
line1c
line2c
line3c

HEADER
line1d
line2d
line3d

Obligatory awk script. Just change n=2 to whatever your nth match should be.

n=2; awk -v n=$n '/^HEADER$/{++i==n && ++flag} !flag; /^$/&&flag{flag=0}' ./file

Input

$ cat ./file
HEADER
line1a
line2a
line3a

HEADER
line1b
line2b
line3b

HEADER
line1c
line2c
line3c

HEADER
line1d
line2d
line3d

Output

$ n=2; awk -v n=$n '/^HEADER$/{++i==n&&++flag} !flag; /^$/&&flag{flag=0}' ./file
HEADER
line1a
line2a
line3a

HEADER
line1c
line2c
line3c

HEADER
line1d
line2d
line3d

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