sed：删除以标记结尾的行后面的空白行
如何使用 sed 执行此操作？
Input: Expected output:
1 | 1
2 MARKER | 2 MARKER
| 3 MARKER
| 4
|
| 5 MARKER
| 6
|
3 MARKER |
| 7 MARKER
4 | 8
| 9
5 MARKER |
6 |
|
|
7 MARKER |
|
8 |
9 |
首先我尝试了这个：它不起作用，因为“# t 或 b 都不起作用，因为 d 导致脚本中断。”
/MARKER$/ {
# i -- Line with MARKER
:my_branch
# i -- in branch
# write current patt space to output and read next line to pattern space
n
# if non blank goto EO script
/^$/! b
# if blank line delete it
/^$/ d
# loop back for more blank lines.
t my_branch
# niether t or b work because d causes the script to break.
b my_branch
}
然后我尝试了这个，这几乎有效:((( -- 它不会删除 3 和 4 之间的线，我认为这是因为 3 在处理 2 的过程中被消耗，因此它的标记被遗漏了。
/MARKER$/ {
# i -- Line with MARKER
:my_branch
# i -- in branch
N
s/MARKER\n/MARKER/
t my_branch
# I added the following but it doesn't help..
/MARKER$/ b my_branch
}
有 4-5我尝试过其他版本，但没有一个有效。
我这样做是因为我试图回答这个问题：在 shell 命令行中将两个换行符替换为一个换行符，所以我从 http://sed.sourceforge.net/sed1line.txt 和 http ://www.grymoire.com/Unix/Sed.html
IOW 我知道有使用 awk、perl 等的解决方案，但我只是想学习使用 sed
谢谢。
How do I do this using sed?
Input: Expected output:
1 | 1
2 MARKER | 2 MARKER
| 3 MARKER
| 4
|
| 5 MARKER
| 6
|
3 MARKER |
| 7 MARKER
4 | 8
| 9
5 MARKER |
6 |
|
|
7 MARKER |
|
8 |
9 |
First I tried this: It doesn't work coz "# niether t or b work because d causes the script to break."
/MARKER$/ {
# i -- Line with MARKER
:my_branch
# i -- in branch
# write current patt space to output and read next line to pattern space
n
# if non blank goto EO script
/^$/! b
# if blank line delete it
/^$/ d
# loop back for more blank lines.
t my_branch
# niether t or b work because d causes the script to break.
b my_branch
}
then I tried this, which almost works :((( -- it doesn't remove the line between 3 and 4, which I think is because 3 is consumed during the processing of 2 and thus it's marker is missed.
/MARKER$/ {
# i -- Line with MARKER
:my_branch
# i -- in branch
N
s/MARKER\n/MARKER/
t my_branch
# I added the following but it doesn't help..
/MARKER$/ b my_branch
}
there 4-5 other versions I tried, but none worked.
I did all this because I was trying to answer this question: replace two newlines to one in shell command line, so I started learning sed from http://sed.sourceforge.net/sed1line.txt and http://www.grymoire.com/Unix/Sed.html
IOW I know there are solutions using awk, perl etc but I just wanted to learn using sed.
Thanks.
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见下文，我想这就是你想要的：（它在我的 Linux 机器上使用 gnu sed 运行）
see below, I think it is what you want: (it ran on my linux box with gnu sed)