如何解析JSON数组和检索PHP字符串变量？

好的。。。到目前为止，以下工作已经完成：

// Open connection
$ch = curl_init();

// Make the curl call
curl_setopt($ch, CURLOPT_USERPWD, $username . ":" . $password);
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-type: application/json'));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
$head = curl_exec($ch);
$http_code = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);


$data = json_decode($head, true);

foreach($data['events'] as $key => $item) {

    // $event_id = $data['events'][$key]['eventId'];
    $event_id = $item['eventId'];

    // $name = $data['events'][$key]['name'];
    $name = $item['name'];

    // $desc = $data['events'][$key]['description'];
    $desc = $item['description'];



    echo "ID: $event_id<br> 
          NAME: $name<br>
          DESCRIPTION: $desc<br>
          <br>";
} 

我想使用数据作为PHP字符串变量，这些变量可以在“primaryImage”、“categories”、“location”等元素中找到。

应该是这样的：

echo $item['primaryImage']->smallImageUrl;

---

不幸的是，这不起作用。在这种情况下，它应该显示警告。收到任何警告？或者，再次尝试将其放入json_解码中

echo $item['primaryImage']->smallImageUrl;