how to get substring from

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#@!