(03-22-2022, 04:38 PM)Snoopy Wrote: i hope deepl dont get me wrong (the forum language is english)
to include € you have to do some work around
problem with none ascii chars is the following hashcat works with bytes and bytesized objects
the € sign is utf-8 and its byte/hex expression ist the following
€:e282ac
so the € sign is 3 bytes in length total
following simple test setup
md5(€1) - > 2677f97bb01d31b7fba26af994e087cf
maskfile (charset taken from your example)
?l == lower chars (built in)
?d == digits (built in)
Code:?l?d€$,?1
?l?d€$,?1?1
?l?d€$,?1?1?1
?l?d€$,?1?1?1?1
hashcat -a3 -m0 -O hash maskfile
this will crack the pass €1 on the fourth line, because for hashcat the pass €1 is E282AC31, 4 bytes in total so the mask has to be at least of length 4
there are some possibilitys to tune this a little bit further (--hex-charset and using all of the 4 custom charset), but i hope you get the point, to crack a "human readable" pass like -> mypassword1€ ( 12 chars for humans, your mask has to be 14 in length) or maybe a better, to crack the single sign € your mask needs to be of length 3 with the given charset, not 1
I understood correctly that I have to convert the entire set of characters, including €, into a hash file. and then write it when attacking, given this number of bytes
how in that case the request should look?
hashcat -m 13000 -a3 hash -1 maskfile ?1?1?1?1?1?1?1? but i'm probably dumb