sed - Remove previous line and current line based on pattern - sed

Question

I want to delete the previous line and current line based on pattern match on next line
This is my sample.txt
This is test line 11
This is test line 999
This is test line 12
This is test line 13
This is test line 16
This is test line 999
This is test line 17
This is test line 18
I want to match for pattern 999 and delete both itself and previous line
I am trying this command but i get no output
sed -Ene ':a;N;/999/{d;}; ba; P' sample.txt

Answer 1

This might work for you (GNU sed):
sed 'N;/\n.*999/d;P;D' file
Open a running window of two lines throughout the length of the file.
If the second line of the window contains 999 delete both lines.
Otherwise, print the first line of the window, delete the first line and repeat.
An alternative solution for line 1 or 2 or more contiguous lines containing 999:
sed -n ':a;$!N;/\n.*999/{:b;n;/999/bb;ba};/999/!P;D' file

Answer 2

tried on gnu sed
sed -Ez 's/[^\n]*\n[^\n]*999\n//g' sample.txt

Answer 3

Could you please try following(if ok with awk).
awk 'prev && $NF!=999{print prev ORS FNR,$0;prev="";next} $NF==999{prev=""} $NF!=999{prev=FNR FS $0}' Input_file
Or if you have even number of lines and you want to take care of printing last odd even.
awk 'prev && $NF!=999{print prev ORS FNR,$0;prev="";next} $NF==999{prev=""} $NF!=999{prev=FNR FS $0} END{if(prev){print prev}}' Input_file

Answer 4

With a more comprehensive sample input of:
$ cat file
This is test line 999
This is test line 11
This is test line 999
This is test line 12
This is test line 13
This is test line 999
This is test line 999
This is test line 999
This is test line 14
This is test line 15
This is test line 16
This is test line 999
This is test line 17
This is test line 18
Try this:
$ cat tst.awk
$NF == 999 {
prev = ""
next
}
{
printf "%s", prev
prev = $0 ORS
}
END {
printf "%s", prev
}
$ awk -f tst.awk file
This is test line 12
This is test line 14
This is test line 15
This is test line 17
This is test line 18
or if you favor brevity over clarity:
$ awk '$NF==999{p="";next} {printf "%s",p; p=$0 ORS} END{printf "%s",p}' file
This is test line 999
This is test line 11
This is test line 999
This is test line 12
This is test line 13
This is test line 999
This is test line 999
This is test line 999
This is test line 14
This is test line 15
This is test line 16
This is test line 999
This is test line 17
This is test line 18
Notice that the above will work even if some other part of your line than the last field contained 999 or if the last field as 9999 instead of your target 999, it doesn't require 999 to be written/tested multiple times in the script, if you wanted to test, say, the 3rd field in the line instead of the last field you could just change $NF to $3 =, if you WANTED to test the whole line for a regexp you'd just change $NF==999 to /999/, it'll work even if your target string contains regexp metacharacters, and it will work in any awk in any shell on any UNIX box.

Answer 5

For a sed1 solution, that handles all edge cases (999 in first 2 rows or consecutive rows of 999):
sed '
1{
/999/d # Special case needed for line 1. Delete if it contains 999.
}
$!N # Append next line. $!N stops exit w/o printing at EOF.
/999/d # If pattern space contains 999, d & begin next cycle.
P # If we get to here, there is no 999. Print to first newline.
D # Delete to first newline.
' FILE
Output:
This is test line 12
This is test line 13
This is test line 17
This is test line 18
1 Tested on both BSD (Mac OS X) & GNU sed.