删除匹配的第n行,直到空行的awk / SED / grep的 [英] Delete matching nth line until blank line in awk/sed/grep
问题描述
我需要从比赛中删除一个文件中第n个匹配的行到下一个空行(即开始第n匹配空行分隔符的文本中一个模块)。
I need to delete the nth matching line in a file from the match up to the next blank line (i.e. one chunk of blank line delimited text starting with the nth match).
推荐答案
这将删除文本块启动,并以一个空行开始的第四个空行结束。它还会删除那些划定线路。
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
更改第一个 / ^ $ /
来改变开始的比赛。更改第二个改变比赛结束
Change the first /^$/
to change the start match. Change the second one to change the end match.
鉴于这种输入:
aaa
---
bbb
---
ccc
---
ddd delete me
eee delete me
===
fff
---
ggg
该版本的命令:
sed -n '/^---$/!{p;b};H;x;/^\(\n[^\n]*\)\{3\}/{:a;n;/^===$/!ba;d};x;p' inputfile
会给这个作为结果:
would give this as the result:
aaa
---
bbb
---
ccc
fff
---
ggg
编辑:
我删除了上述 SED
命令外来 B
指令。
I removed an extraneous b
instruction from the sed
commands above.
下面是一个注释版本:
sed -n ' # don't print by default
/^---$/!{ # 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
任何明确的模式应该工作的开始和结束标记。它们可以是相同或不同的
Any unambiguous pattern should work for the begin and end markers. They can be the same or different.
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