我是Perl的新手。我需要在Perl中定义一个如下所示的数据结构:
city 1 -> street 1 - [ name , no of house , senior people ]
street 2 - [ name , no of house , senior people ]
city 2 -> street 1 - [ name , no of house , senior people ]
street 2 - [ name , no of house , senior people ]
我怎样才能实现这个目标?
答案 0 :(得分:5)
以下是使用哈希引用的另一个示例:
my $data = {
city1 => {
street1 => ['name', 'house no', 'senior people'],
street2 => ['name','house no','senior people'],
},
city2 => {
street1 => etc...
...
}
};
然后您可以通过以下方式访问数据:
$data->{'city1'}{'street1'}[0];
或者:
my @street_data = @{$data->{'city1'}{'street1'}};
print @street_data;
答案 1 :(得分:4)
我找到了答案,如
my %city ;
$city{$c_name}{$street} = [ $name , $no_house , $senior];
我可以用这种方式生成
答案 2 :(得分:1)
Perl数据结构手册perldsc可能会提供帮助。它有一些示例向您展示如何创建公共数据结构。
答案 3 :(得分:0)
my %city ;
如果你想推
push( @{ city{ $c_name } { $street } }, [ $name , $no_house , $senior] );
(OR)
push @{ city{ $c_name } { $street } }, [ $name , $no_house , $senior];
答案 4 :(得分:0)
您可以在this回答中阅读我的简短教程。简而言之,您可以将哈希引用到值中。
%hash = ( name => 'value' );
%hash_of_hash = ( name => \%hash );
#OR
$hash_of_hash{ name } = \%hash;
# NOTICE: {} for hash, and [] for array
%hash2 = ( of_hash => { of_array => [1,2,3] } );
# ---^ ---^
$hash2{ of_hash }{ of_array }[ 2 ]; # value is '3'
# ^-- lack of -> because declared by % and ()
# same but with hash reference
# NOTICE: { } when declare
# NOTICE: -> when access
$hash_ref = { of_hash => { of_array => [1,2,3] } };
# ---^
$hash_ref->{ of_hash }{ of_array }[ 2 ]; # value is '3'
# ---^