Question

所以我要从如下所示的JSON格式的URL中获取数据。

[
{
"id": 1,
"name": "Leanne Graham",
"username": "Bret",
"email": "Sincere@april.biz",
"address": {
  "street": "Kulas Light",
  "suite": "Apt. 556",
  "city": "Gwenborough",
  "zipcode": "92998-3874",
  "geo": {
    "lat": "-37.3159",
    "lng": "81.1496"
  }
},
"phone": "1-770-736-8031 x56442",
"website": "hildegard.org",
"company": {
  "name": "Romaguera-Crona",
  "catchPhrase": "Multi-layered client-server neural-net",
  "bs": "harness real-time e-markets"
}
},
// And So on upto 10 id places.

现在，我要使用cURL php在我的网站中获取数据。

我的php代码如下：

$url = "https://jsonplaceholder.typicode.com/users" ;

$ch = curl_init() ;
curl_setopt($ch, CURLOPT_URL, $url) ;
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$result = curl_exec($ch);
curl_close($ch) ;

$id = 1;
$data = json_decode($result) ;
foreach ($data as $mydata) {
  if ($mydata["id"] == $id) {
 echo $mydata["address"]["street3"] ;
 break ;
}
}

但是我得到的是一个错误：

不能将stdClass类型的对象用作数组

Answer 1

second param为TRUE时，返回的对象将转换为关联数组。

$data = json_decode($result,true) ;
foreach ($data as $mydata) {
  if ($mydata["id"] == $id) {
      echo $mydata["address"]["street"] ;//changes `street3` is not there
      break ;
  }
}

默认情况下，json_decode将返回您的对象，如果您想使用[]语法，则需要关联数组，并且仅为此需要传递第二个参数true，请参阅此 json_decode用于进一步查询

使用curl PHP从嵌套的JSON数据中获取数据

1 个答案: