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

sed

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…}

Best Answer

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
'

Related Solutions

Text Processing – Remove Lines Based on Pattern While Keeping First N Matches

If you want to delete all lines starting with % put preserving the first two lines of input, you could do:

sed -e 1,2b -e '/^%/d'

Though the same would be more legible with awk:

awk 'NR <= 2 || !/^%/'

Or, if you're after performance:

{ head -n 2; grep -v '^%'; } < input-file

If you want to preserve the first two lines matching the pattern while they may not be the first ones of the input, awk would certainly be a better option:

awk '!/^%/ || ++n <= 2'

With sed, you could use tricks like:

sed -e '/^%/!b' -e 'x;/xx/{h;d;}' -e 's/^/x/;x'

That is, use the hold space to count the number of occurrences of the patterns matched so far. Not terribly efficient or legible.

Keep only the first line from every sequence of consecutive lines matching a pattern

Using awk:

awk '/logical IO/ {if (!seen) {print; seen=1}; next}; {print; seen=0}' file.txt

/logical IO/ {if (!seen) {print; seen=1}; next} checks if the line contains logical IO, if found and the variable seen is false i.e. previous line does not contain logical IO, then print the line, set seen=1 and go to the next line else go to the next line as the previous line has logical IO
For any other line, {print; seen=0}, prints the line and the sets seen=0

Example:

$ cat file.txt 
select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
testing logical IO 500
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346
parsing logical IO 346
testing logical IO 12

$ awk '/logical IO/ {if (!seen) {print; seen=1}; next}; {print; seen=0}' file.txt 
select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346

Best Answer

Related Solutions

Text Processing – Remove Lines Based on Pattern While Keeping First N Matches

Keep only the first line from every sequence of consecutive lines matching a pattern

Related Question