What does it mean in shell when we put a command inside dollar sign and parentheses: $(command)
BashShellTerminalShKshBash Problem Overview
I just want to understand following line of code in shell. It is used to get the current working directory. I am aware that $(variable) name return the value inside the variable name, but what is $(command) supposed to return? Does it return the value after executing the command? In that case, we can use ` to execute the command.
CWD="$(cd "$(dirname $0)"; pwd)"
Same output can be taken from the following line of code also in different version of shell
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
I am unable to understand the meaning of $(cd.. and $(dirname.
Could anybody help me to figure out how this command get executed?
Bash Solutions
Solution 1 - Bash
Usage of the $ like ${HOME} gives the value of HOME. Usage of the $ like $(echo foo) means run whatever is inside the parentheses in a subshell and return that as the value. In my example, you would get foo since echo will write foo to standard out
Solution 2 - Bash
> DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )" > > Could anybody help me to figure out how this command get executed?
Let's look at different parts of the command. BASH_SOURCE is a bash array variable containing source filenames. So "${BASH_SOURCE[0]}" would return you the name of the script file.
dirname is a utility provided by GNU coreutils that remove the last component from the filename. Thus if you execute your script by saying bash foo, "$( dirname "${BASH_SOURCE[0]}" )" would return .. If you said bash ../foo, it'd return ..; for bash /some/path/foo it'd return /some/path.
Finally, the entire command "$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )" gets the absolute directory containing the script being invoked.
$(...) allows command substitution, i.e. allows the output of a command to replace the command itself and can be nested.