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Re: Replacing a list with one variable

by Ea (Hermit)
on Nov 21, 2013 at 12:15 UTC ( #1063725=note: print w/replies, xml ) Need Help??

in reply to Replacing a list with one variable

You might also like the qw() operator which splits a string on whitespace to produce a list. So
my @bookIDList = qw(abc ghj zxc bnm qwe rty iop sdf ert); my $param = = { IDs => [ @bookIDList ],
or another way to make a reference to a list
my $param = = { IDs => \@bookIDList,
or maybe cut out the assignment line
my $param = = { IDs => [ qw(abc ghj zxc bnm qwe rty iop sdf ert) ], sort => "AUTHOR", maxResults => 100 };
I find using the qw() to be quicker and cleaner for me.

As always, there's more than one way to do it and the best way is the one that makes the code easier to write and read. Express your style :) Modern Perl will give you more ideas.

The other 3 ways to write this are left as an exercise to the reader

Sometimes I can think of 6 impossible LDAP attributes before breakfast.

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