removes bytes within file when binary string is found

Question

so I'd like to remove parts within (NOT at beginning or the end, or I would have used dd) a binary file, when I encounter a certain binary string value. ie :

if FF FB FF FB A4 is found, remove 2048 bytes before the second FF FB here (meaning the first FF FB is deleted as well as the 2046 bytes preceeding it)

repeat till the end of the file.(no need to test/prevent "eating itself", range between that string occurences always much larger than 2048)

how can I do that in bash ? thanks in advance

PasteBT · Accepted Answer · 2013-12-03 03:49:54Z

1

You can use grep to locate the "FF FB FF FB A4", then use dd to cut file:

pos=$(grep --only-matching --byte-offset --binary --text --perl-regexp "\xFF\xFB\xFF\xFB\xA4" filename|head -1|cut -d ':' -f1)

It will tell you where the string.

dd if=filename of=output bs=1 ibs=1 count=$pos

This get the leading part. I think you know how to deal with left staff

answered Dec 3, 2013 at 3:49

PasteBT

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