PHP如何访问特定的json数据

debugcn 发表于 Dev

卡梅尔西

再会，

我直接从 API 端点获取我的数据。我能够获得id但我如何才能"type":"participant",具体访问skillTier？

这是我的代码，但它抛出一个错误： NOTICE: UNDEFINED INDEX: SKILLTIER IN C:\XAMPP\HTDOCS\VGAPP\INDEX.PHP ON LINE 219

if(is_array($finaljson) || is_object($finaljson)){
if(!empty($finaljson['included']) && $finaljson['included'] !== ""){
    foreach ($finaljson['included'] as $key => $value) {
        if(!empty($value['attributes']['name']) && $value['attributes']['name'] === $username){
            $userid = $value['id'];
            $skillTier = $value['skillTier'];
        }
    }

这是json数据：

https://gist.github.com/kamaelxiii/18ef0c1600330b98717b96db00532d6a

米克马克库萨

减少一个大的 json / 数组以定位和隔离您想要的数据是一项耐心的工作。您只需要逐步遍历结构，将每个级别的数据打印到屏幕上，并确定包含您想要的内容的键。

代码：

$array = json_decode($json, true);
foreach ($array['included'] as $included) {  // iterate all children of "included"
    if ($included['type'] === 'participant') {  // only access "participant" subset data
        $result[$included['id']] = $included['attributes']['stats']['skillTier'];  // access the skillTier data which is beneath "stats" beneath "attributes"
    }
}
echo "<pre>";
var_export($result);
echo "</pre>";

输出：

array (
  '4cfe4736-f47b-11e6-b633-0242ac110030' => 16,
  '4cfe473f-f47b-11e6-b633-0242ac110030' => 10,
  '4cfe4727-f47b-11e6-b633-0242ac110030' => 27,
  'd4003883-f423-11e6-865e-0242ac110008' => 23,
  'd4003869-f423-11e6-865e-0242ac110008' => 24,
  'd400387a-f423-11e6-865e-0242ac110008' => 23,
  '4cfe471f-f47b-11e6-b633-0242ac110030' => 27,
  'd400386d-f423-11e6-865e-0242ac110008' => 24,
  'd4003887-f423-11e6-865e-0242ac110008' => 23,
  '4cfe4724-f47b-11e6-b633-0242ac110030' => 24,
  'd4003864-f423-11e6-865e-0242ac110008' => 18,
  '4cfe473b-f47b-11e6-b633-0242ac110030' => 17,
)

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