Question

我试图使用简单的天气API，但无论如何，它无论如何都不会将它识别为对象。这是我的代码：

$url = "https://www.amdoren.com/api/weather.php?api_key=za8LEJ8F9mcHK8SvLxdM98rM9mNFjW&lat=40.7127837&lon=-74.0059413";
$curl = curl_init($url);
$curl_response = curl_exec($curl);
$jsonobj = json_decode($curl_response);
$msg = "Temperature in ".$city."will be: ". $jsonobj->forecast->max_c;

这是我试图通过$jsonojb->forecast->max_c：

访问的数据

{
 "error" : 0,
 "error_message" : "-",
 "forecast":[ 
 {"date":"2016-12-02",
 "avg_c":8,
 "min_c":5,
 "max_c":11,
 "avg_f":46,
 "min_f":41,
 "max_f":52,
  (...)

但它不起作用。我做错了什么人？

Answer 1

预测是一个数组，所以你必须这样使用：

$forecast = $jsonobj->forecast;
$forecast[0]->max_c;

Answer 2

您可以尝试以下代码：

$url = "https://www.amdoren.com/api/weather.php?api_key=za8LEJ8F9mcHK8SvLxdM98rM9mNFjW&lat=40.7127837&lon=-74.0059413";
$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HTTPHEADER, array('Content-Type: application/json'));
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);

$curl_response = curl_exec($curl);
$jsonobj = json_decode($curl_response);

$msg = "Temperature in ". $city . " will be: ". $jsonobj->forecast[0]->max_c;
echo $msg;

希望它有所帮助！

Json_decode抛出＆＃34;试图获得非对象的属性＆＃34;

2 个答案: