On 29 Jun 2001 09:26:55 -0400, Kim Green wrote:
> What's the proper syntax to indicate that a variable is optional? The script
> that I have created works great when I pass in a variable, but the script
> need to execute the SQL even if I don't pass in a variable.
>
> Thanks,
> Kim
>
>
you might try
$var = $ARGV[0] || DEFAULT;
where DEFAULT is some reasonable value. The || operator returns the
first true value; if $ARGV[0] has not been set on the command line then
it will contain undef which evaluates to false in boolean context so the
default gets returned to the assignment operator.
IMPORTANT: this construct will fail if the argument is the number 0.
In cases where you expect 0 to be a valid value it may be better to use
this construct:
$var = defined($ARGV[0]) ? $ARGV[0] : DEFAULT;
The ?: operator (the only trinary operator I am aware of) returns the
second argument if the first argument is true, otherwise it returns the
third argument.
--
Today is Setting Orange, the 34th day of Confusion in the YOLD 3167
Hail Eris!
