how to get substring from - bash

Question

how to get substring from
42 45 47 49 4e 21 40 23 47 68 6a 6b 2c 47 68 6a BEGIN!##Ghjk,Ghj 6b 45 4e 44 23 40 21 kEND##!
to be
BEGIN!##Ghjk,GhjkEND##!
Note: there is whitespaces at end of lines, I tried removing whitespaces at end of lines but I cant.
I tried
#!/bin/bash
s=$(awk '/BEGIN!##/,/END##!/' switch.log )
while IFS= read -r line
do
h=$(echo "$line" | awk '{$1=$1;print}')
for i in {0..100}
do
zzz=$(echo "$h" | awk '{print $(NF-$i)}')
if [ ! -z "$zzz" -a "$zzz" != " " ]; then
hh=$(echo "$h" | awk '{print $(NF-$i)}')
echo "$zzz"
echo -e "$zzz" >> ggg.txt
break
fi
done
done <<< "$s"
I got
BEGIN!##Ghjk,Ghj

Answer 1

Another option is using sed with the normal substitute method storing the text you want to keep as the first two backreferences. For example:
sed -E 's/^.*(BEGIN[^[:space:]]+).*(kEND[^[:space:]]+)/\1\2/' <<< 'your string`
Example Use/Output
(note: updated to handle whitespace at the end)
$ sed -E 's/^.*(BEGIN[^[:space:]]+).*(kEND[^[:space:]]+)/\1\2/' <<< '42 45 47 49 4e 21 40 23 47 68 6a 6b 2c 47 68 6a BEGIN!##Ghjk,Ghj 6b 45 4e 44 23 40 21 kEND##!'
BEGIN!##Ghjk,GhjkEND##!
(note: single-quoting the string is required due to '!')

Answer 2

Using sed
$ sed -E 's/[0-9]+[a-z]? +| +//g' input_file
BEGIN!##Ghjk,GhjkEND##!

Answer 3

UPDATED, to fix an error:
You have not defined precisely in your question, how the string to be extracted looks like in general, but based on your example, this would do:
if [[ $line =~ (BEGIN[^ ]+)\ .*([^ ]+END[^ ]+) ]]
then
substring=${BASH_REMATCH[1]}${BASH_REMATCH[2]}
else
echo Pattern not found in line 1>&2
fi

Answer 4

I would harness GNU AWK for this task following way, let file.txt content be
42 45 47 49 4e 21 40 23 47 68 6a 6b 2c 47 68 6a BEGIN!##Ghjk,Ghj 6b 45 4e 44 23 40 21 kEND##!
then
awk 'BEGIN{FPAT="[^[:space:]]*(BEGIN|END)[^[:space:]]*";OFS=""}{$1=$1;print}' file.txt
gives output
BEGIN!##Ghjk,GhjkEND##!
Explanation: I inform GNU AWK using field pattern (FPAT) that field is BEGIN or (|) END, prefixed and suffixed by zero-or-more (*) non (^)-whitespace ([:space:]) characters and output field separator (OFS) is empty string, then for each line I do $1=$1 to trigger line rebuilt and print it. If you are sure only space characters are used in line you might elect to replace [^[:space:]] using [^ ]
(tested in gawk 4.2.1)

Answer 5

s=$(awk '/BEGIN!##/,/END##!/' switch.log)
echo "$s" > ggg.txt
ss=$(sed -E 's/[0-9]+[a-z]? +| +//g' ggg.txt )
echo "$ss" > ddd.txt
sss=$(awk '{print $1}' ddd.txt)
echo "$sss" > hhhh.txt
ssss=$(awk '/BEGIN!##/,/END##!/' hhhh.txt)
echo "$ssss" > hhh.txt
aaa=$(<hhh.txt)
aaa=$(cat hhh.txt | tr -d '\n' )

Answer 6

By setting the awk record separator RS to " ", awk processes a white-spaced-separated portion of your file at a time (with each record containing only one field). So the two parts that are needed can be extracted with simple awk condition patterns /BEGIN/ and /END/. There can be no white space in any record since this was the delimiter.
If printed, the pattern-filtered records would normally be separated by a new line (the default output record-separator ORS) but this can be changed to an empty string ORS="" to make the two print statements run into one another with no space.
Thus this simple awk command will return the required fields as a concatentated string with no white space:
awk ' BEGIN{RS=" ";ORS=""} /BEGIN/{print} /END/{print}' file.txt
output:
BEGIN!##Ghjk,GhjkEND##!

Answer 7

$ grep -oE '(BEGIN|END)\S*' file | paste -sd'\0'
BEGIN!##Ghjk,GhjEND##!

Answer 8

echo ' 42 45 47 49 4e 21 40 23 47 68 6a 6b 2c 47 68 ' \
'6a BEGIN!##Ghjk,Ghj 6b 45 4e 44 23 40 21 kEND##!' |
{m,g}awk NF=NF FS='[ \t]*([^ \t][^ \t][ \t]+)+[ \t]*' OFS=
BEGIN!##Ghjk,GkEND##!