Copying a value from an element in the ARGV array to a variable...

joelw has asked for the wisdom of the Perl Monks concerning the following question:

Hello... Please take a look at the following snippet of code...

if( $#ARGV == 0 )
{
   my ($date) = $ARGV[0];
}
else
{
   # obtain the date from the previous day in the format YYYYMMDD
   my($yr,$mo,$day) = (localtime( time - 60*60*(12+(localtime)[2])))[5
+,4,3];
   #my $date = sprintf "%04d%02d%02d",1900+$yr,1+$mo,$day;
 
   # for test purposes!!!
   my $date = "20010912";
}
 
# Output the header to the info file.
# -----------------------------------
print OUTREP "\nThe following trades from " .  $date . " finished with
+ a ticket status of 11\n\n";
[download]

Perl gives me the following error:

Global symbol "date" requires explicit package name at ebOrdHist.pl li
+ne 60.
Execution of ebOrdHist.pl aborted due to compilation errors.
[download]

Comment on Copying a value from an element in the ARGV array to a variable... Select or Download Code

Replies are listed 'Best First'.
Re: Copying a value from an element in the ARGV array to a variable... by arturo (Vicar) on Sep 17, 2001 at 19:54 UTC
Just a note: you can test whether `@ARGV` has any members with the idiomatic `my $date; if (@ARGV) { $date = $ARGV[0]; }` [download] Two things: `@ARGV` gets evaluated in a scalar context, which means that what is checked is the number of elements in the array; if that number is not 0 (i.e. if there are any elements in the array), the test evaluates to true, and the block will be executed. Second note : the declaration of `$date` occurs outside the braces, so when you set `$date` inside the braces, that change will be visible to other code outside the braces. When you use `my` inside a set of braces, you're creating a new variable that is only visible to code inside the braces. You can see how this works with `my $date = "bar"; print "Outside 1 : $date\n"; { my $date = "foo"; print "Inside : $date\n"; } print "Outside 2 : $date\n";` [download] Incidentally, you can make your `$date` setting very compact by using \|\| or the "ternary operator" (see perlop): `my $date = $ARGV[0] \|\| yesterday(); sub yesterday { my ($yr,$mo,$day) = (localtime( time - 86400 ))[5,4,3]; return sprintf "%04d%02d%02d",1900+$yr,1+$mo,$day; }` [download] if `$ARGV[0]` evaluates to false ( because it doesn't exist, or was 0 ), then `$date` gets set to what the `yesterday` sub returns. HTH. /msg me if you need anything explained. `perl -e 'print "How sweet does a rose smell? "; chomp ($n = <STDIN>); +$rose = "smells sweet to degree $n"; other_name = rose; print "$oth +er_name\n"'` [download]	[reply] [d/l] [select]
Re: Copying a value from an element in the ARGV array to a variable... by davorg (Chancellor) on Sep 17, 2001 at 19:40 UTC
You need to declare the variable $date outside of either of the blocks created by the if/else statement. `my $date; if (@ARGV) { $date = 'something'; } else { $date = 'something else'; }` [download] -- <http://www.dave.org.uk> Perl Training in the UK <http://www.iterative-software.com>	[reply] [d/l]
Re: Copying a value from an element in the ARGV array to a variable... by mr.nick (Chaplain) on Sep 17, 2001 at 20:02 UTC
Just to build on arturo's comment of using the ternary \|\| operator: if you have (or plan to have) multiple arguments, then something along these lines might work better for you: `my $date = @ARGV ? shift @ARGV : yesterday();` [download] That way you can easily inline other arguments: `my $arg1 = @ARGV ? shift @ARGV : 'default value'; my $date = @ARGV ? shift @ARGV : yesterday(); my $arg2 = @ARGV ? shift @ARGV : 'other value';` [download] mr.nick ...	[reply] [d/l] [select]