Portable way to remove the first line from the pattern space (when multiple lines are present)

up vote
3
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What's the portable way to delete the first line¹ from the pattern space ?

With gnu sed I can do

s/[^n]*n//

but as far as I know this (using n in a bracket expression) is not portable.

Practical example: here, sed prints the last section of the file including the delimiter via portable code. I'd like to remove the first line from the pattern space so as to exclude the delimiter and do that in a portable manner. With gnu sed it's simple:

sed 'H;/===/h;$!d;//d;x;s/[^n]*n//' infile

^{1: Obviously this should be done without restarting the cycle of commands...}

asked 2 days ago

don_crissti

47k15124154

Indeed. POSIX sed only supports BREs (Basic Regular Expressions), and in BREs Ã¢Â€Âthe special characters ., *, [, and `\` [Ã¢Â€Â¦] shall lose their special meaning within a bracket expression.Ã¢Â€Âœ
â€“Â myrdd
2 days ago

add a commentÂ |Â

up vote
3
down vote

favorite

What's the portable way to delete the first line¹ from the pattern space ?

With gnu sed I can do

s/[^n]*n//

but as far as I know this (using n in a bracket expression) is not portable.

sed 'H;/===/h;$!d;//d;x;s/[^n]*n//' infile

^{1: Obviously this should be done without restarting the cycle of commands...}

asked 2 days ago

don_crissti

47k15124154

Indeed. POSIX sed only supports BREs (Basic Regular Expressions), and in BREs Ã¢Â€Âthe special characters ., *, [, and `\` [Ã¢Â€Â¦] shall lose their special meaning within a bracket expression.Ã¢Â€Âœ
â€“Â myrdd
2 days ago

add a commentÂ |Â

up vote
3
down vote

favorite

What's the portable way to delete the first line¹ from the pattern space ?

With gnu sed I can do

s/[^n]*n//

but as far as I know this (using n in a bracket expression) is not portable.

sed 'H;/===/h;$!d;//d;x;s/[^n]*n//' infile

^{1: Obviously this should be done without restarting the cycle of commands...}

asked 2 days ago

don_crissti

47k15124154

What's the portable way to delete the first line¹ from the pattern space ?

With gnu sed I can do

s/[^n]*n//

but as far as I know this (using n in a bracket expression) is not portable.

sed 'H;/===/h;$!d;//d;x;s/[^n]*n//' infile

^{1: Obviously this should be done without restarting the cycle of commands...}

sed

asked 2 days ago

don_crissti

47k15124154

asked 2 days ago

don_crissti

47k15124154

asked 2 days ago

don_crissti

47k15124154

asked 2 days ago

don_crissti

47k15124154

asked 2 days ago

don_crissti

47k15124154

Indeed. POSIX sed only supports BREs (Basic Regular Expressions), and in BREs Ã¢Â€Âthe special characters ., *, [, and `\` [Ã¢Â€Â¦] shall lose their special meaning within a bracket expression.Ã¢Â€Âœ
â€“Â myrdd
2 days ago

add a commentÂ |Â

Indeed. POSIX sed only supports BREs (Basic Regular Expressions), and in BREs Ã¢Â€Âthe special characters ., *, [, and `\` [Ã¢Â€Â¦] shall lose their special meaning within a bracket expression.Ã¢Â€Âœ
â€“Â myrdd
2 days ago

Indeed. POSIX sed only supports BREs (Basic Regular Expressions), and in BREs Ã¢Â€Âthe special characters ., *, [, and `\` [Ã¢Â€Â¦] shall lose their special meaning within a bracket expression.Ã¢Â€Âœ
â€“Â myrdd
2 days ago

add a commentÂ |Â

3 Answers
3

active

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votes

up vote
2
down vote

accepted

One way to portably do this sort of a thing is as follows:

sed -e '
 # ... assuming prev sed cmds made pattern space carry newline(s)
 y/n_/_n/ ;# exchange newlines with an underscore
 s/^[^_]*_// ;# remove up till the first underscore, ummm newline
 y/n_/_n/ ;# revert the transformation
'

answered yesterday

Rakesh Sharma

55513

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up vote
2
down vote

I don't have a suitable sed to try, but you can always do interior loops testing for and removing characters until you get to the newline:

sed 'H
/===/h
$!d
//d
x

 :rpt
 s/^n//
 t done
 s/.//
 t rpt

:done
'

The second test-branch, t rpt, needs to be a t rather than a b in order to reset the internal flag that says an s has suceeded since the last read. You don't need the , they are just to show the loop better.

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

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up vote
1
down vote

Silly, but working:

sed 'h;G;s/n/&&/;s/^(.*)n(.*)n12$/2/'

What's that? You double the whole content, then replace the first newline with two newlines. So you have the same content twice, with one additional newline after the first line. With backreferences you can now identify the different parts.

If you don't want to use the hold buffer:

sed 's/.*/&&/;s/n/&&/;s/^(.*)n(.*)12$/2/'

And no, I don't like that. If there is a way to avoid it, avoid it.

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

One way to portably do this sort of a thing is as follows:

sed -e '
 # ... assuming prev sed cmds made pattern space carry newline(s)
 y/n_/_n/ ;# exchange newlines with an underscore
 s/^[^_]*_// ;# remove up till the first underscore, ummm newline
 y/n_/_n/ ;# revert the transformation
'

answered yesterday

Rakesh Sharma

55513

add a commentÂ |Â

up vote
2
down vote

accepted

One way to portably do this sort of a thing is as follows:

sed -e '
 # ... assuming prev sed cmds made pattern space carry newline(s)
 y/n_/_n/ ;# exchange newlines with an underscore
 s/^[^_]*_// ;# remove up till the first underscore, ummm newline
 y/n_/_n/ ;# revert the transformation
'

answered yesterday

Rakesh Sharma

55513

add a commentÂ |Â

up vote
2
down vote

accepted

One way to portably do this sort of a thing is as follows:

sed -e '
 # ... assuming prev sed cmds made pattern space carry newline(s)
 y/n_/_n/ ;# exchange newlines with an underscore
 s/^[^_]*_// ;# remove up till the first underscore, ummm newline
 y/n_/_n/ ;# revert the transformation
'

answered yesterday

Rakesh Sharma

55513

One way to portably do this sort of a thing is as follows:

sed -e '
 # ... assuming prev sed cmds made pattern space carry newline(s)
 y/n_/_n/ ;# exchange newlines with an underscore
 s/^[^_]*_// ;# remove up till the first underscore, ummm newline
 y/n_/_n/ ;# revert the transformation
'

answered yesterday

Rakesh Sharma

55513

answered yesterday

Rakesh Sharma

55513

answered yesterday

Rakesh Sharma

55513

answered yesterday

Rakesh Sharma

55513

add a commentÂ |Â

up vote
2
down vote

I don't have a suitable sed to try, but you can always do interior loops testing for and removing characters until you get to the newline:

sed 'H
/===/h
$!d
//d
x

 :rpt
 s/^n//
 t done
 s/.//
 t rpt

:done
'

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

add a commentÂ |Â

up vote
2
down vote

I don't have a suitable sed to try, but you can always do interior loops testing for and removing characters until you get to the newline:

sed 'H
/===/h
$!d
//d
x

 :rpt
 s/^n//
 t done
 s/.//
 t rpt

:done
'

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

add a commentÂ |Â

up vote
2
down vote

I don't have a suitable sed to try, but you can always do interior loops testing for and removing characters until you get to the newline:

sed 'H
/===/h
$!d
//d
x

 :rpt
 s/^n//
 t done
 s/.//
 t rpt

:done
'

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

I don't have a suitable sed to try, but you can always do interior loops testing for and removing characters until you get to the newline:

sed 'H
/===/h
$!d
//d
x

 :rpt
 s/^n//
 t done
 s/.//
 t rpt

:done
'

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

edited 2 days ago

answered 2 days ago

meuh

29.8k11751

answered 2 days ago

meuh

29.8k11751

answered 2 days ago

meuh

29.8k11751

add a commentÂ |Â

up vote
1
down vote

Silly, but working:

sed 'h;G;s/n/&&/;s/^(.*)n(.*)n12$/2/'

If you don't want to use the hold buffer:

sed 's/.*/&&/;s/n/&&/;s/^(.*)n(.*)12$/2/'

And no, I don't like that. If there is a way to avoid it, avoid it.

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

5,93211546

add a commentÂ |Â

up vote
1
down vote

Silly, but working:

sed 'h;G;s/n/&&/;s/^(.*)n(.*)n12$/2/'

If you don't want to use the hold buffer:

sed 's/.*/&&/;s/n/&&/;s/^(.*)n(.*)12$/2/'

And no, I don't like that. If there is a way to avoid it, avoid it.

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

5,93211546

add a commentÂ |Â

up vote
1
down vote

Silly, but working:

sed 'h;G;s/n/&&/;s/^(.*)n(.*)n12$/2/'

If you don't want to use the hold buffer:

sed 's/.*/&&/;s/n/&&/;s/^(.*)n(.*)12$/2/'

And no, I don't like that. If there is a way to avoid it, avoid it.

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

5,93211546

Silly, but working:

sed 'h;G;s/n/&&/;s/^(.*)n(.*)n12$/2/'

If you don't want to use the hold buffer:

sed 's/.*/&&/;s/n/&&/;s/^(.*)n(.*)12$/2/'

And no, I don't like that. If there is a way to avoid it, avoid it.

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

5,93211546

edited 2 days ago

don_crissti

47k15124154

edited 2 days ago

don_crissti

47k15124154

edited 2 days ago

don_crissti

47k15124154

answered 2 days ago

Philippos

5,93211546

answered 2 days ago

Philippos

5,93211546

answered 2 days ago

Philippos

5,93211546

add a commentÂ |Â

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