Two ways:
1. use the printf method to embed punctuation and other odd characters. Unless you are using windows do not put \r in anything. expect gets a return from bash reading the script -- and it is a newline, \n
2. you can escape characters ( \ escape in front of each character, so \\ is read by the shell as a backslash, not escape). The shell takes them as literal characters, not some shell command -- example \*\# is an asterisk and an octothorpe (pound sign). Not a wild card and not a start of a comment.
Thank You Jim for your help - and sorry for my late reply. The problem is still actual for me.
So, I suppose, that only proper way for me will be the 1st method. Due to diversity of the passkeys it would be easier than putting backslashes in 2000-marks passkey's for example. The lenght of the keys can differ, and the special chars qty is also random.
About the method you mentioned:
1. I would like to use only in linux, and on linux I also can prepare passlist (unix-type line endings and non viible characters on passlist).
2. If that method (printf) could help - maybe 'expect' will not be essential?
I have tried using expect due to problems with using 7z x -p$klucz (or $passkey) archive.7z
The $ mark is only for calling a variable. That is possible, that on the begginning will be Dollar sign - $ .
Maybe your solution would allow me to use passkeys without bash-processing it in native 7z commands? So, if I can use it as a full text in script - 7z x -p$passkeyfromprintf
Unfortunately - I am starting with bash programming - and I really do not have idea how to use your snippet. Would you like to show me any example how to start using this ?
So, if I would use it my simpler script: (could you please help me with that one?)
"7z t" - 7z TEST, which wll exit when the passkey is bad.
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