Hello,
I am looking for a method to use in my bash script which allows me to use long strings with all special characters.
I have found that printf method could be helpful for me but unfortunately, when I trying
root@machine:~# tevar=`printf "%s%c" "'``fF/0,g_V/Y)>^.(a0(,%<A.`8N:q*Rd|lCy0t9N'FDPs|S,$2(^$/*>5Ush*w87L#m@t~:[Ple+=Od633Z3m3xGV.zX+a-N%x1K=J7gOO=?1c$HZU78iVtJG1N[s@-eH#DwH]V5l}??XWI.YJ;4%:LmAEML9%4jm'Y6GXTc{DT7Iia$!+/[x.),b]5`iO3E,-+,44+9Fke45Z=m^ba.EP^)+GqZ.uQNF*sU9'Kxq6i19+ie*6rZjjCC.2`rSYVqlmHlHqeFBCapXXSxXROXUrljMWGsicEdCdPJuvbXnPXlENaBzpMBnRgtGsFtTYbHsyilugLrTSTMvGDGqSIhXhJISqnIuBwxfqr`'" ; echo $tevar
-bash: bd skadni przy nieoczekiwanym znaczniku `)'
I am getting error
-bash: bd skadni przy nieoczekiwanym znaczniku `)'
What is important - I need to pass different length strings, I don't know the length, and the positions of special characters.
This is a password from my password list and I need find working pass to archive.
What can I do to pass very long stream with any special character as a string and put it as variable in bash script?
Or - maybe I can make it outside the bash ? I am using 7 zip command line tool where I need to put the password as parameter like
7z t -pP@55w0rd!
where P@55w0rd! is my password.
What was my fault before (long time ago)- I have had used almost 5000 characters in passwords. I have lot of possible passes in my dictionary, and I am sure, that one of them will be correct.