How to dump part of binary file

Question

I have binary and want to extract part of it, starting from know byte string (i.e. FF D8 FF D0) and ending with known byte string (AF FF D9)

In the past I've used dd to cut part of binary file from beginning/ending but this command doesn't seem to support what I ask.

What tool on terminal can do this?

Answer 1

This should work with standard tools (xxd, tr, grep, awk, dd). This correctly handles the "pattern split across line" issue, also look for the pattern only aligned at byte offset (not nibble).

file=<yourfile>
outfile=<youroutputfile>
startpattern="ff d8 ff d0"
endpattern="af ff d9"
xxd -g0 -c1 -ps ${file} | tr '\n' ' ' > ${file}.hex 
start=$((($(grep -bo "${startpattern}" ${file}.hex\
    | head -1 | awk -F: '{print $1}')-1)/3))
len=$((($(grep -bo "${endpattern}" ${file}.hex\
    | head -1 | awk -F: '{print $1}')-1)/3-${start}))
dd ibs=1 count=${len} skip=${start} if=${file} of=${outfile}

Note: The script above use a temporary file to prevent having the binary>hex conversion twice. A space/time trade-off is to pipe the result of xxd directly into the two grep. A one-liner is also possible, at the expense of clarity.

One could also use tee and named pipe to prevent having to store a temporary file and converting output twice, but I'm not sure it would be faster (xxd is fast) and is certainly more complex to write.

Answer 2

See this link for a way to do binary grep. Once you have the start and end offset, you should be able with dd to get what you need.

Answer 3

In a single pipe:

xxd -c1 -p file |
  awk -v b="ffd8ffd0" -v e="aaffd9" '
    found == 1 {
      print $0
      str = str $0
      if (str == e) {found = 0; exit}
      if (length(str) == length(e)) str = substr(str, 3)}
    found == 0 {
      str = str $0
      if (str == b) {found = 1; print str; str = ""}
      if (length(str) == length(b)) str = substr(str, 3)}
    END{ exit found }' |
  xxd -r -p > new_file
test ${PIPESTATUS[1]} -eq 0 || rm new_file

The idea is to use awk between two xxd to select the part of the file that is needed. Once the 1st pattern is found, awk prints the bytes until the 2nd pattern is found and exit.

The case where the 1st pattern is found but the 2nd is not must be taken into account. It is done in the END part of the awk script, which return a non-zero exit status. This is catch by bash's ${PIPESTATUS[1]} where I decided to delete the new file.

Note that en empty file also mean that nothing has been found.

Answer 4

Locate the start/end position, then extract the range.

$ xxd -g0 input.bin | grep -im1 FFD8FFD0  | awk -F: '{print $1}'
0000cb0
$ ^FFD8FFD0^AFFFD9^
0009590
$ dd ibs=1 count=$((0x9590-0xcb0+1)) skip=$((0xcb0)) if=input.bin of=output.bin

Answer 5

Another solution in sed, but using less memory:

xxd -c1 -p file |
  sed -n -e '1{N;N;N}' -e '/ff\nd8\nff\nd0/{:begin;p;s/.*//;n;bbegin}' -e 'N;D' | 
  sed -n -e '1{N;N}' -e '/aa\nff\nd9/{p;Q1}' -e 'P;N;D' |
  xxd -r -p > new_file
test ${PIPESTATUS[2]} -eq 1 || rm new_file

The 1st sed prints from ff d8 ff d0 till the end of file. Note that you need as much N in -e '1{N;N;N}' as there is bytes in your 1st pattern less one.

The 2nd sed prints from the beginning of the file to aa ff d9. Note again that you need as much N in -e '1{N;N}' as there is bytes in your 2nd pattern less one.

Again, a test is needed to check if the 2nd pattern is found, and delete the file if it is not.

Note that the Q command is a GNU extension to sed. If you do not have it, you need to trash the rest of the file once the pattern is found (in a loop like the 1st sed, but not printing the file), and check after hex to binary conversion that the new_file end with the wright pattern.

Answer 6

A variation on the awk solution that assumes that your binary file, once converted in hex with spaces, fits in memory:

xxd -c1 -p file |
  tr "\n" " " |
  sed -n -e 's/.*\(ff d8 ff d0.*aa ff d9\).*/\1/p' |
  xxd -r -p > new_file