Without the aid of the escape mechanism, how to put a raw string into a variable? Post: 303045752

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Top Forums UNIX for Advanced & Expert Users Without the aid of the escape mechanism, how to put a raw string into a variable? Post 303045752 by RudiC on Sunday 12th of April 2020 05:52:15 AM

04-12-2020

Registered User

Quote:

Originally Posted by wisecracker

.
.
.

STRING=' !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~'

Your STRING assignment ends with the second single quote after the & :

Code:

STRING=' !"#$%&'

and thus the following ( is not recognized by the shell:

Quote:

./rawstring.sh: 33: ./rawstring.sh: Syntax error: "(" unexpected

Quote:

# The line below does not even get executed!
echo "${STRING}"
.
.
.

Of course not - it is part of the single quoted literal string started with the very last character two lines above.

I'm afraid you won't escape escaping in some way. Try e.g. using this mechanism man bash:

Quote:

Quoting:
Words of the form $'string' are treated specially. The word expands to string, with backslash-escaped characters replaced as specified by the ANSI C standard. Backslash escape sequences, if present, are decoded as follows: ...

Code:

STRING=$' !"#$%&\47()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~'

Last edited by RudiC; 04-12-2020 at 07:01 AM..

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