Issue
From the bash man-page:
-c If the -c option is present, then commands are read from the first non-option argument command_string. If there are arguments after the command_string, the first argument is assigned to $0 and any remaining arguments are assigned to the positional parameters.
So I created a "script" file named foo.sh
, containing the single line
echo par 0 is $0, par 1 is $1
, set it to executable, and invoked it as
bash -c ./foo.sh x y
I expected to see par 0 is x, par 1 is y, but it was printed par 0 is ./foo.sh, par 1 is.
In what respect did I misunderstand the man-page?
UPDATE Perhaps to clarify it (since my question seems to have given rise to some confusion): My goal is to execute foo.sh
, but making it believe that its real name is not foo.sh, but x.
Solution
You would see the behaviour you wanted with this :
bash -c "$(cat foo.sh)" x y
Answered By - Philippe Answer Checked By - Marie Seifert (WPSolving Admin)