20 сент. 2019 г. · This is called indirect expansion in bash. The syntax to get the value of a variable whose name is determined by another variable namevar is ...

2 дек. 2015 г. · The first test ( if [ -z "$1" ]; then ... ) is true if $1 (first commandline argument) is an empty string (or was not provided).

20 авг. 2020 г. · This works for me on RHEL 6. #!/bin/bash read -rN1 -p "Enter a character: " variable; echo>&2 if [ "$variable" = 'A' ]; then echo Capital A ...

9 июл. 2021 г. · In bash, you can use variable indirection prev=prev$c current=current$c if ((${!prev} != ${!current})); then echo "prev$c is ${!prev} and current$c is ${! ...

25 июн. 2015 г. · I need to check a variable's existence in an if statement. Something to the effect of: if [ -v $somevar ] then echo "Variable somevar exists ...

24 июн. 2020 г. · If parameter is unset or null, the expansion of word is assigned to parameter. The value of parameter is then substituted. Positional parameters ...

3 февр. 2013 г. · I am trying to write a single IF statement which can check two variables at once. I have two variables, for example: $VARA $VARB Both with either contain a 0 ...

15 дек. 2022 г. · -S tests whether the given path exists and is a socket (the link documents test , but the [[ -S operator has the same meaning).

28 мар. 2023 г. · In Bash in particular, you can use [[ -v var ]] : $ foo=1; unset bar; $ varname=foo $ if [[ -v $varname ]]; then echo set; else echo not set ...

16 февр. 2020 г. · The problem is when ever I reference a variable within my if statement I get back -bash: command not found.