Question

JSON：

{"location":{"name":"Tirana","region":"Tirane","country":"Albania","lat":41.33,"lon":19.82,"tz_id":"Europe/Tirane","localtime_epoch":1484543668,"localtime":"2017-01-16 5:14"},"current":{"last_updated_epoch":1484543668,"last_updated":"2017-01-16 05:14","temp_c":4.0,"temp_f":39.2,"is_day":0,"condition":{"text":"Overcast","icon":"//cdn.apixu.com/weather/64x64/night/122.png","code":1009},"wind_mph":6.9,"wind_kph":11.2,"wind_degree":150,"wind_dir":"SSE","pressure_mb":1009.0,"pressure_in":30.3,"precip_mm":0.0,"precip_in":0.0,"humidity":60,"cloud":0,"feelslike_c":1.2,"feelslike_f":34.2}}

PHP：

$response = file_get_contents('http://api.apixu.com/v1/current.json?key=a54959ce2b294134bda34330171601&q=Paris');
var_dump(json_decode($response)); 
echo $response->location[0]->name;

API致电：http://api.apixu.com/v1/current.json?key=a54959ce2b294134bda34330171601&q=Tirana

Answer 1

使用json_decode将json字符串解析为数组，然后使用索引访问它。

<强> the demo

$response = file_get_contents('http://api.apixu.com/v1/current.json?key=a54959ce2b294134bda34330171601&q=Paris');
$array = json_decode($response, true); 
echo $array['location']['name'];

Answer 2

请尝试这样做。您可以使用json格式获取内容。使用第二个参数json_decode使用true（）转换为数组。

<?php
$json = '{"location":{"name":"Tirana","region":"Tirane","country":"Albania","lat":41.33,"lon":19.82,"tz_id":"Europe/Tirane","localtime_epoch":1484543668,"localtime":"2017-01-16 5:14"},"current":{"last_updated_epoch":1484543668,"last_updated":"2017-01-16 05:14","temp_c":4.0,"temp_f":39.2,"is_day":0,"condition":{"text":"Overcast","icon":"//cdn.apixu.com/weather/64x64/night/122.png","code":1009},"wind_mph":6.9,"wind_kph":11.2,"wind_degree":150,"wind_dir":"SSE","pressure_mb":1009.0,"pressure_in":30.3,"precip_mm":0.0,"precip_in":0.0,"humidity":60,"cloud":0,"feelslike_c":1.2,"feelslike_f":34.2}}
';
$array = json_decode($json,true);
//print_r($array);
$location = $array['location'];
echo $location['name'];

?>

如何使用PHP访问嵌套JSON中的名称值？

2 个答案: