复杂的PHP数组逻辑

我基本上需要找到一种方法来查看订购最多的产品，然后返回前5个“post_id”的数组

这是包含产品详细信息的不同订单的数组：

Array
(
    [1] => Array
        (
            [post_id] => 1
            [post_ident] => macbook_pro
            [post_name] => Macbook Pro
            [post_parent_ident] => rings
            [post_cat_ident] => default
            [post_module_ident] => store
            [post_price] => 999.00
            [post_currency] => EUR
            [item_link] => index.php?module=store&show=post&category=default&parent=rings&post=macbook_pro
            [qty] => 1
        )

)

Array
(
    [1] => Array
        (
            [post_id] => 1
            [post_ident] => macbook_pro
            [post_name] => Macbook Pro
            [post_parent_ident] => rings
            [post_cat_ident] => default
            [post_module_ident] => store
            [post_price] => 999.00
            [post_currency] => EUR
            [item_link] => index.php?module=store&show=post&category=default&parent=rings&post=macbook_pro
            [qty] => 1
        )

)

Array
(
    [1] => Array
        (
            [post_id] => 1
            [post_ident] => macbook_pro
            [post_name] => Macbook Pro
            [post_parent_ident] => rings
            [post_cat_ident] => default
            [post_module_ident] => store
            [post_price] => 999.00
            [post_currency] => EUR
            [item_link] => index.php?module=store&show=post&category=default&parent=rings&post=macbook_pro
            [qty] => 1
        )

)

如何确定阵列中的哪些产品最多

---

例如，post_id为1的产品在阵列中存在3次。我如何计算它，然后将post_id作为数组中的第一项返回？

该数组是否来自SQL？否，产品详细信息已序列化，这是取消序列化后的输出。该数组是否来自SQL？否，产品详细信息已序列化，这是取消序列化后的输出。

$result = array();
foreach($array as $value)
{
   $postId = $value[1]['post_id'];
   if(isset($result[$postId])){
      $result[$postId]++;     // increment count of post id if already exists in result
   }
   else{
      $result[$postId] = 1;    // start count for post id
   }  
}

$keys = array_keys(asort($result));   // sort the array and find all the keys
echo $keys[count($keys)-1];           // last key will be the answer