I have a simple script test.sh
#!/bin/bash
echo $0
When I run the following from csh terminal:
bash -c 'test.sh'
Then the output is test.sh
But when I run:
bash -c 'source test.sh'
The output is bash
Does anybody know how to print the script name in this case?
#!/bin/bash
declare -r SCRIPT_NAME=$(readlink -f ${BASH_SOURCE[0]})
Use $BASH_SOURCE. From the man-page:
BASH_SOURCE
An array variable whose members are the source filenames where
the corresponding shell function names in the FUNCNAME array
variable are defined. The shell function ${FUNCNAME[$i]} is
defined in the file ${BASH_SOURCE[$i]} and called from
${BASH_SOURCE[$i+1]}.
You can simply refer to $BASH_SOURCE instead of ${BASH_SOURCE[0]} since dereferencing an array variable in bash without an index will give you the first element.
This is a common pattern in my scripts to allow different behavior for sourcing versus executing a script:
foo() {
some cool function
}
if [[ "$BASH_SOURCE" == "$0" ]]; then
# actually run it
foo "$@"
fi
When a script is run using source it runs within the existing shell, any variables created or modified by the script will remain available after the script completes. In contrast if the script is run just as filename, then a separate subshell (with a completely separate set of variables) would be spawned to run the script.
Also you want to modify the shell scripts
This can not work for the same reason that you can not use a child shell to modify the enviroment of the parent shell. The environment of the child process is private and cannot affect the environment of its parent.
The only way to accomplish what you are attempting might be to have make compose a standard output stream that contains a list of shell variable assignments. The standard output could then be used as the input to the parent 'source' or '.' command. Using a 'noisy' program such as make to do this will be significantly challenging.
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