UNIX for Dummies Questions & Answers

Hello,

Can somebody tell me the differnce between a call of a ksh with the dot :

>. script.ksh

and the call without the dot:

>script.ksh

In my script I have writen a test for the number of parameters:
if [ $# -ne 0 ]
then echo 'Usage : '$0
exit 1
fi

as my script doesn't need any parameter

if I call like this:
>. script.ksh
$# =
$0 = -ksh
which is wrong
I can't do a correct test of my parameters, but if I remove the parameter test I can give value to variables, and these variables are known and have a value in my parent shell and can be used by other scripts

If I call like this:
>script.ksh
$# = 0
$0 = script.ksh
which is OK, but I want to export variables that must be used in other script and my variables remains unknown in my parent shell
but it seems that I can execute other script that can see the variables....

I don't understand anything....

see you
Cécile

When you run a script as ". ./scriptname", the script executes in the context of the current shell. All your shell variables (such as $0, $#, etc) will also be those of the current shell. So showing $0 to be -ksh is what should happen.

Since the script that you are using is basically to setup env. variables, don't check for things like $0, just export your variables and be done with it.

Thank you for your answer.

I understand a little bit more.

If a call with the dot I stay in the parent shell ?
In fact I don't really execute the script but I read it??
Do I execute another shell when I call a script with the dot ?

You 're right I don't need to check parameters here.

Ok, it works when I call like that :

. Script1.ksh

. Script2.ksh

Absolutely correct. As far as I know, sourced scripts don't even need to be set executable. Nope, it runs in the existing one. They call it 'sourcing' but a better name might be 'include' or 'import'.