假設Geo::Coder::Google
數據轉儲的這種結構--- dd $location;
。如何訪問Perl結構中某個鍵的特定值?
address_components => [
{
long_name => "Blackheath Avenue",
short_name => "Blackheath Ave",
types => ["route"],
},
{
long_name => "Greater London",
short_name => "Gt Lon",
types => ["administrative_area_level_2", "political"],
},
{
long_name => "United Kingdom",
short_name => "GB",
types => ["country", "political"],
},
{
long_name => "SE10 8XJ",
short_name => "SE10 8XJ",
types => ["postal_code"],
},
{ long_name => "London", short_name => "London", types => ["postal_town"] },
],
formatted_address => "Blackheath Avenue, London SE10 8XJ, UK",
geometry => {
bounds => {
northeast => { lat => 51.4770228, lng => 0.0005404 },
southwest => { lat => 51.4762273, lng => -0.0001147 },
},
location => { lat => 51.4766277, lng => 0.0002212 },
location_type => "APPROXIMATE",
viewport => {
northeast => { lat => 51.4779740302915, lng => 0.00156183029150203 },
southwest => { lat => 51.4752760697085, lng => -0.00113613029150203 },
},
},
types => ["route"],
}
一個例子電話:
my $long_name = &get_field_for_location("long_name", $location);
繼子返回第一個long_name
(在這個例子---類型=路線):
sub get_field_for_location($$) {
my $field = shift;
my $location = shift;
my $address = $location->{address_components};
return $_->{$field} for @$address;
}
如何訪問的另一個long_name
類型?即如何修改此子項以訪問給定類型條目的$field
?
我不明白這個問題。 – simbabque
這不是我打算只得到第一個條目。爲了節省時間,我問如何應對這種結構。 – mnemonic