How to print all lines except the line before a pattern in file using awk

Question

How can we do this with awk?

Input file

Line 1
Line 2
####PATTERN####### (Line 3)
Line 4
Line 5
Line 6
####PATTERN####### (line 7)
Line 8
####PATTERN####### (Line 9)

etc..

output file

Line 1
Line 3
Line 4 
Line 5
Line 7
Line 9

Answer 1

sed '$!N;s/.*####PATTERN####### \(.*\)/\1/;P;D' text

Output

Line 1
(Line 3)
Line 4
Line 5
(line 7)
(Line 9)

The N command grabs the next line from the input into the pattern space so that we can perform a pattern matching against two lines. If we discover ####PATTERN###### in the middle, we only keep the second part so that the previous line is deleted. The P command prints the resulted pattern.

Answer 2

Code for sed :

sed -nr '/####PATTERN#######/!{x;1!p;x};$p;h' file

---

$ sed -nr '/####PATTERN#######/!{x;1!p;x};$p;h' file
Line 1
####PATTERN####### (Line 3)
Line 4
Line 5
####PATTERN####### (line 7)
####PATTERN####### (Line 9)

---

Code for GNU awk :

awk '{if ($0 ~ /PATTERN/ && NR>1) {delete l[(NR-1)]}; l[NR]=$0}END {for (a in l) print l[a]}' file

$ awk '{if ($0 ~ /PATTERN/ && NR>1) {delete l[(NR-1)]}; l[NR]=$0}END {for (a in l) print l[a]}' file
Line 1
####PATTERN####### (Line 3)
Line 4
Line 5
####PATTERN####### (line 7)
####PATTERN####### (Line 9)

Answer 3

Here is one way with awk :

awk '/PATTERN/{for(;i<NR-2;)print lines[++i];i=NR;delete lines;print $0}{lines[NR]=$0}' file

Output:

$ cat file
Line 1
Line 2
####PATTERN####### (Line 3)
Line 4
Line 5
Line 6
####PATTERN####### (line 7)
Line 8
####PATTERN####### (Line 9)
$ awk '/PATTERN/{for(;i<NR-2;)print lines[++i];i=NR;delete lines;print $0}{lines[NR]=$0}' file
Line 1
####PATTERN####### (Line 3)
Line 4
Line 5
####PATTERN####### (line 7)
####PATTERN####### (Line 9)