Passing variables to a bash script when sourcing it

Question

Suppose I have in main.sh:

$NAME="a string"
if [ -f $HOME/install.sh ]
    . $HOME/install.sh $NAME
fi

and in install.sh:

echo $1

This is supposed to echo "a string", but it echoes nothing. Why?

Answer 1

Michael Mrozek covers most of the issues and his fixes will work since you are using Bash.

You may be interested in the fact that the ability to source a script with arguments is a bashism. In sh or dash your main.sh will not echo anything because the arguments to the sourced script are ignored and $1 will refer to the argument to main.sh.

When you source the script in sh, it is as if you just copy and pasted the text of the sourced script into the file from which it was sourced. Consider the following (note, I've made the correction Michael recommended):

$ bash ./test.sh
A String
$ sh ./test.sh

$ sh ./test.sh "HELLO WORLD"
HELLO WORLD

Answer 2

I see three errors:

1. Your assignment line is wrong:

   $NAME="a string"
   

   When you assign to a variable you don't include the $; it should be:

   NAME="a string"
   

2. You're missing then; the conditional line should be:

   if [ -f $HOME/install.sh ]; then
   

3. You're not quoting $NAME, even though it has spaces. The source line should be:

   . $HOME/install.sh "$NAME"
   

Answer 3

simply set your parametrs before sourcing the script !

main.sh

#!/bin/bash
NAME=${*:-"a string"}
if [[ -f install.sh ]];
then
    set -- $NAME ;
    . install.sh ;
fi
exit;

install.sh

#!/bin/bash
echo  " i am sourced by [ ${0##*/} ]";
echo  " with [ $@ ] as parametr(s) ";
exit;

test

u@h$ ./main.sh some args
 i am sourced by [ main.sh ]
 with [ some args ] as parametr(s) 
u@h$