Re: create_user, what goes into %user?

[Jamie C. <jca...@we...>]

Joe Cooper wrote:
> 
> Ok... I'm almost there.  But I'm stumped now.  My perl ignorance is showing.
> 
> Is there a way for me to pass the hash to a foreign_call?
> 
> I've confirmed that everything is correct in the hashes (%user and %group):
> 
> %user =
> shell/bin/shhome/home/httpd/test7uid504realtest7gid504pass$1$85678922$2XoiZBuEZP.uSqRpjCRuP/usertest7
> 
> %group = gid504grouptest7
> 
> (Correct?)
> 
> But I don't think it is being passed to the foreign_call to create_user
> and create_group.  All is get in my passwd and group files is this:
> 
> :x:::::
> 
> and
> 
> :x::
> 
> I've tried all three of the following with no luck:
> 
> &foreign_call("useradmin", "create_user", "%user");
> &foreign_call("useradmin", "create_user", "\%user");
> &foreign_call("useradmin", "create_user", %user);
> 
> I've found references to lots of very roundabout methods of passing a
> hash to a function (by converting them to arrays or a string), all of
> which have me confused.  So will the function accept an array or a
> string and fit it back into being a correct hash, or should I, perhaps
> copy the function into my module-lib.pl?

You just have to pass it as a reference .. like so :

&foreign_call("useradmin", "create_user", \%user);

I don't think there is any way to pass an actual hash to a perl function ..

 - Jamie