Decode JSON (PHP)

Question

如何选择"total_items"one_answers"average_price"的数量？

JSON:

   {
  "status" : "success",
  "data" : {
    "items" : [
      {
        "market_hash_name" : "AK-47 | Redline (Field-Tested)",
        "total_items" : 698,
        "lowest_price" : "3.90",
        "highest_price" : "300.00",
        "cumulative_price" : "4669.62",
        "recent_sales_info" : {
          "hours" : "17.94",
          "average_price" : "4.23"
        }
      }
    ]
  }
}

这是我的PHP脚本：

$link = 'skin.json';
    $string = file_get_contents($link);
    $obj = json_decode($string, TRUE);
$name = $obj['items']['market_hash_name'];
$itmes = $name['total_items'];
$itmes = $name['average_price'];

如何保存变量中的"total_items"one_answers"average_price"的数量？非常感谢。当做Enge

Answer 1

就像Andreas写的那样，或者：

// first, new variable (shorter lines/less depth afterwards)
$article = $obj['data']['items'][0];
// then 
$total_items = $article['total_items'];
$avg_price = $article['recent_sales_info']['average_price'];

Enge，注意"items"：[是JSON对象中的一个开放数组，这是零。你缺少…

Answer 2

试试这个：

$total_items = $obj['data']['items'][0]['total_items'];
$avg_price = $obj['data']['items'][0]['recent_sales_info']['average_price'];