Confusion while extracting required part of a file using awk - linux

Question

I have a script making use of awk,sed,grep and other shell features.
I have stuck at a place so need your help ...
This is the input file for the my problem
udit#udit-Dabba ~/ah $ cat decrypt.txt
60 00 00 00 00 17 3a 20 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 02 *00 00 e0 f9 6a 61 61 6e
65 6b 61 68 61 6e 67 61 79 65 77 6f 64 69 6e* 00
00 00 03 29
My purpose is to extract 00 00 e0 f9 6a 61 61 6e
65 6b 61 68 61 6e 67 61 79 65 77 6f 64 69 6e from the above mentioned file
,also marked between *'s above
Although obvious but these *'s are shown to clear the situation here , they are not actually present in the file.
The last five units of the file as shown above are ..
00 00 00 03 29
These 00 are simple pad bytes and 03 specify their pad length
and now here is the part of script to extract the required part :
size=`wc -w decrypt.txt`
padlen=3 // calculated by some other mechanism
awk -v size=$size -v padlen=$padlen 'BEGIN {RS=" ";ORS=" ";} {if (NR > 40
&& NR <=size-padlen-2) print $0}' decrypt.txt | sed '1,1s/ //'
output :
00 00 e0 f9 6a 61 61 6e
65 6b 61 68 61 6e 67 61 79 65 77 6f 64 69
My problem :
last unit 6e missing
Also tried through terminal ...
size=68,padlen=3 so loop should go from NR=40 to NR<=63
udit#udit-Dabba ~/ah $ awk 'BEGIN {RS=" ";ORS=" ";} {if (NR > 40 && NR <= 65)
print $0}' decrypt.txt | sed '1,1s/ //'
00 00 e0 f9 6a 61 61 6e
65 6b 61 68 61 6e 67 61 79 65 77 6f 64 69 6e 00
00
Working fine if loop goes upto 65.So should also work upto 63
udit#udit-Dabba ~/ah $ awk 'BEGIN {RS=" ";ORS=" ";} {if (NR > 40 && NR <= 64)
print $0}' decrypt.txt | sed '1,1s/ //'
00 00 e0 f9 6a 61 61 6e
65 6b 61 68 61 6e 67 61 79 65 77 6f 64 69 6e
But what is this ???? when I decrease 65 to 64 , there is loss of two 00 units.Why this is happening ???
Also tried this one but could not find a reason why this weird output.
udit#udit-Dabba ~/ah $ awk 'BEGIN {RS="[ \n]";ORS=" ";} {if (NR > 40
&& NR <=65)print $0}' decrypt.txt | sed '1,1s/ //'
0002 00 00 e0 f9 6a 61 61 6e 65 6b 61 68 61 6e 67 61 79 65 77 6f 64
Plase help me out ...
May be I have explained the problem more than the required but really need it .
I am new to all these shell and awk things and so there may be a silly mistake which I could not find out .
Please help me on this ..
Thnx in advance ..
EDIT :
60 00 00 00 00 17 3a 20 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 02
These are fixed 40 units of ipv6 header,will always remain same.
The portion between *'s is of variable length that is why I need to work in that way otherwise it would have been a simple task .

Answer 1

_padlen=3 _length=23
awk '{
for (i = NF - l - p - 2; i < NF - p - 2; i++)
printf "%s", ($i (i < NF - p - 2 - 1 ? OFS : ORS))
}' l="$_length" p="$_padlen" RS= ORS='\n' decrypt.txt

Answer 2

I made some small changes in the code and able to get till 6e*
size=68; padlen=3 ;awk -v size=$size -v padlen=$padlen 'BEGIN {RS=" ";ORS=" ";} {if (NR > 40 && NR <=size-padlen-1) print $0}' decrypt.txt | sed '1,1s/ //'
I made size as 68 becos wc wis printing the size and file name and you have to remove it when u are passing the same to the awk script.
Note: I havent understood your requirement fully

Answer 3

If I understand the problem as being: discard the first 40 values and the last n values (where n is the padding + 2 i.e. in this case 3 + 2 = 5), this might work:
header=40;padding=5;
tr -d '\n' <decrypt.txt |
sed -r 's/\s+/ /g;s/^(\S+\s+){'"$header"'}//;s/(\S+\s*){'"$padding"'}$//'
The trick is to unroll the data and then pick the bits you want.