how to extract version number form binary file using awk

I had tried to extract version number from a binary file.

The version number is after this string 'VeRsIoN_StRiNg'.

But how to find it using awk and print the next character I can't find out.

Someone the开发者_如何学Cr can help?

/Lasse

Do you strictly need to use awk? This seems like a better usecase for grep --binary-files=text -o 'VeRsIoN_StRiNg.' file | grep -o '.$'.

I'm not entirely sure how well a stream editor like awk will actually work with a binary file. If this is part of a larger awk script, you probably want to call the above grep formula from awk.

You can use strings command to find the printable strings in a object, or other binary, file

strings /path/to/binary | grep -o 'VeRsIoN_StRiNg.' | grep -o '.$'

why not awk ?

gawk -b/mawk/mawk2 'BEGIN { RS = "^$"; FS = "^.*VeRsIoN_StRiNg" 

    } END { print substr($2,1,1)' # mawk/mawk2 or gawk in byte mode. 
                                  # LC_ALL=C gawk -e will be here too

even in gawk unicode mode, this workaround will do

gawk -e 'BEGIN { RS = "^$"; FS = "^.*VeRsIoN_StRiNg" 

    } END { printf("%.1s\n", $2) }' # gawk in unicode mode

This is to take advantage of the fact a "precision" N specified (e.g. %.ns) for %s means

at most N items printed

But since, by definition of FS, we know the first byte of $2 is already your version number, a single character integer, then this printf will circumvent any gawk error message of complaining trying to do sub-strings on UTF8 non-compliant data.