获取对象或数组对象php中的值答案

【问题标题】：Get value in object or array object php获取对象或数组对象php中的值
【发布时间】：2020-08-07 05:26:35
【问题描述】：

如何在 php 中从这个对象和这个对象数组中获取值名称？以及如何检查是对象还是数组对象然后获取值名称？

"director": {
"@type": "Person",
"url": "/name/nm0001104/",
"name": "Frank Darabont"}



"director": [
{
  "@type": "Person",
  "url": "/name/nm5156926/",
  "name": "Devon Downs"
},
{
  "@type": "Person",
  "url": "/name/nm2632302/",
  "name": "Kenny Gage"
}],

【问题讨论】：

标签： php arrays json object arrayobject

【解决方案1】：

$json = '{
    "director": [
          {
            "@type": "Person",
            "url": "/name/nm5156926/",
            "name": "Devon Downs"
          },
          {
            "@type": "Person",
            "url": "/name/nm2632302/",
            "name": "Kenny Gage"
          }
  ]
  }';

  $names = [];
  $data = json_decode($json, true);

  if (true === isset($data['name'])) {
      $name[] = $data['name'];
      return $names;
  }
  foreach ($data as $director) {
    foreach($director as $itme){
      if (true === isset($itme['name'])) {
          $names[] = $itme['name'];
      }
    }

  }
  return $names;

输出：

Array
(
    [0] => Devon Downs
    [1] => Kenny Gage
)

【讨论】：

【解决方案2】：

使用json_decode。

$names = [];
$data = json_decode($json, true);
if (true === isset($data['name'])) {
    $name[] = $data['name'];
    return $names;
}
foreach ($data as $director) {
    if (true === isset($director['name'])) {
        $names[] = $director['name'];
    }
}
return $names;

【讨论】：