grep/pcregrep/sed/awk the data after the last match to the end of a file - regex

Question

I need to grab the content after the last match of ENTRY to the end of the file, and I can't seem to do it. It can be multiple lines and the data can include any character to the end of the file including (,\n, ).
I’ve tried:
tail -1 file # doesn’t work due to it not consistently being one line
grep "^(.*" # only grabs one line
pcregrep -M '\n(.*' file # I think a variation of this is the solution, but I’ve had no luck so far.
File that grows below:
TOP OF FILE
%
ENTRY
(S®s
√6ûíπ‹ôTìßÅDPˆ¬k·Ù"=ÓxF)*†‰ú˚ÃQ´¿J‘\˜©ŒG»‡∫QÆ’<πsµ-ù±ñ∞NäAOilWçk
N+P}V<ôÒ∏≠µW*`Hß”;–GØ»14∏åR"ºã
FD‘mÍõ?*ÊÎÉC)(S®s
√6ûíπ‹ôTìßÅDPˆ¬k·Ù"=ÓxF)*†‰ú˚ÃQ´¿J‘\˜©ŒG»‡∫QÆ’<πsµ-ù±ñ∞NäAOilWçk
N+P}V<ôÒ∏≠µW*`Hß”;–GØ»14∏åR"ºã
FD‘mÍõ?*ÊÎÉC)eq
{
DATA
}
ENTRY
(A® S\kÉflã1»Âbπ¯Ú∞⁄äπHZ#F◊§•Ã*‹¡‹…ÿPkJòÑíòú˛¶à˛¨¢v|u«Ùbó–Ö¶¢∂5ıÜ#¨•˘®#W´≥‡*`H∑”ı–Só¬<˙ìEçöf∞Gg±:œe™flflå)A® S\kÉflã1»Âbπ¯Ú∞⁄äπHZ#F◊§•Ã*‹¡‹…ÿPkJòÑíòú˛¶à˛¨¢v|u«Ùbó–Ö¶¢∂5ıÜ#¨•˘®#W´≥‡*`H∑”ı–Só¬<˙ìEçöf∞Gg±:œe™flflå)eq
{
DATA
}if
ENTRY
(ÌSYõ˛9°\K¬∞≈fl|”/í÷L
Ö˙h/ÜÇi"û£fi±€ÀNéÓ›bÏÿmâ[≈4J’XPü´Z
oÜlø∫…qìõ¢,ßü©cÓ{—˜e&ÚÀÓHÏÜ‚m(Œ∆⁄ˆQ˝òêpoÉÄÂ(S‘E ⁄ !ŸQ§ô6ÉH

Answer 1

$ awk '/^[(]/{s="";} {s=s"\n"$0;} END{print substr(s,2);}' file
(ÌSYõ˛9°\K¬∞≈fl|”/í÷L
Ö˙h/ÜÇi"û£fi±€ÀNéÓ›bÏÿmâ[≈4J’XPü´Z
oÜlø∫…qìõ¢,ßü©cÓ{—˜e&ÚÀÓHÏÜ‚m(Œ∆⁄ˆQ˝òêpoÉÄÂ(S‘E ⁄ !ŸQ§ô6ÉH
How it works
awk implicitly loops through files line-by-line. This script stores whatever we want to print in the variable s.
/^[(]/{s="";}
Every time that we find a line which starts with (, we set s to an empty string.
The purpose of this is to remove everything before the last occurrence of a line starting with (.
s=s"\n"$0
We add the current line onto the end of s.
END{print substr(s,2);}
After we reach the end of the file, we print s (omitting the first character which will be a surplus newline character).

Answer 2

Interesting problem. I think you can do it with just sed. When you find a match, zero the hold space and add the match line to the hold space. On the last line, print the hold space.
sed -n -e '/ENTRY/,$ { /ENTRY/ { h; n; }; H; $ { x; p; } }'
Don't print by default. From the first entry to the end of the file:
If it is an entry line; copy the new line over the hold space and move on.
Otherwise append the line to the hold space.
If it is the last line, swap the hold space and pattern space, and print the pattern space (what was in the hold space).
You might worry about what happens if the last line in the file is an ENTRY line.
Given a data file:
TOP OF FILE
not wanted
ENTRY
could be wanted
ENTRY
but it wasn't
and this isn't
because
ENTRY
this is here
EOF
The output is:
ENTRY
this is here
EOF
If you don't want ENTRY to appear, modify the script slightly:
sed -n -e '/ENTRY/,$ { /ENTRY/ { s/.*//; h; n; }; H; $ { x; s/^\n//; p; } }'

Answer 3

Using tac you could do it:
tac <file> | sed -e '/ENTRY/,$d' | tac
This will print the file with the lines reversed, then use sed to remove everything from what is now the first occurrence of ENTRY to the now end of the file, then reverse the lines again to get the original order.
As Jonathan Leffler pointed out, a faster way to do this--though probably not much because tac will still have a lot to do and it has all the overhead of rquireing 3 processes instead of just one, but the sed could be done more efficiently, but just ending when we find the ENTRY line, instead of processing the rest of the file to remove the lines:
tac <file> | sed -e '/ENTRY/q' | tac
though his answer is often going to be better still. That answer will include the ENTRY line. If you don't want that you could also do
tac <file> | sed -n '/ENTRY/q;p' | tac
to not print any ouptut by default, then quit as soon as you find the ENTRY line, but use the p command to print the lines until you get to that line.

Answer 4

This should work too (at least with gawk)
awk -vRS="ENTRY" 'END{print $0}'
set the record separator as your pattern and print the last record.

Answer 5

loadind file in memory
sed -e 'H;$!d' -e 'x;s/.*ENTRY[[:blank:]]*\n//' YourFile