如何使用PHP访问嵌套JSON中的名称值？

JSON:

PHP:

---

API调用：

使用json_decode将json字符串解析为数组，然后使用索引访问它

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

---

尝试这样做。您将获得

json

格式的内容。使用

json_decode

（）和第二个参数

true

将其转换为数组

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

json\u decode（$response）捕获一个局部变量，然后解析这个json，怎么样？位置是一个一维数组，而不是多维数组。所以删除

[0]

->

echo$response->location->name；

只是一个想法：即使基本服务是免费的，也不应该提供包含有效API密钥的链接。但是，您应该将该密钥交换为新的密钥。（而且，由于密钥嵌入URL中，您将来应该使用HTTPS）

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

<?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'];

?>