bash quotes inside quotes site:unix.stackexchange.com - Axtarish в Google
8 июн. 2015 г. · You don't need to escape the nested quotes inside. They get parsed properly, surprisingly! DATA="$(cat file.hex | xxd -r | tr -d "$(cat trim.txt)")"
14 июн. 2016 г. · A double-quoted string that starts inside a command substitution will never extend outside of it or vice versa.
17 сент. 2020 г. · Double quotes do not nest. That's why you escaped the "inner" quotes in your question. You could use one pair of single quotes and one pair of ...
4 мар. 2021 г. · The data you want to insert into the argument sent to curl is a JSON-encoded string encoding a JSON document. This string can be created separately.
10 мая 2024 г. · Generally, to properly quote a variable within single quotes in Bash, use '"$var"' (single quote followed by double quote).
23 нояб. 2014 г. · Enclosing characters in double quotes ('"') preserves the literal value of all characters within the quotes, with the exception of '$', '`', '\' ...
16 июн. 2015 г. · You can't expand variables in single quotes. You can end single quotes and start double quotes, though: echo 'visit:"'"$site"'"'.
17 сент. 2020 г. · Once inside a command substitution, the shell begins an entirely new quoting context. That is, double quotes inside the substitution do not match up with ...
6 февр. 2017 г. · Bash will add single quotes (in eg the set -x tracing output) when you use \"$ARGUMENT\" because you're making the double quotes part of the argument's value.
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