Question

所以我有一段JSON数据。记住：我正在展示作品！

{
  "status" : "success",
  "prices" : [
    {
      "market_hash_name" : "AK-47 | Aquamarine Revenge (Battle-Scarred)",
      "price" : "15",
      "created_at" : 1460708937
    },

这是我的PHP，它应该得到价格，但它不起作用。可能是什么问题？ 'price'=>$prices->prices[0]->market_hash_name[$row['market_hash_name']]->price);

Answer 1

因为您在 market_hash_name 节点下没有价格。这应该可以解决问题：

//the second parameter (true) says that the function should return an array
$data = json_decode($jsonString, true); 
echo $data['prices'][0]['price'];

重要说明：这只是使用给定json字符串部分的示例。您必须自己索引节点。：）

Answer 2

查看答案：

在线链接：3v4l.org

主JSON数据太大，无法在任何Web服务器中运行沙箱或3v41。

有json的部分。

$market_hash_name = 'AK-47 | Aquamarine Revenge (Battle-Scarred)';

$json = '{
  "status" : "success",
  "prices" : [
    {
      "market_hash_name" : "AK-47 | Aquamarine Revenge (Battle-Scarred)",
      "price" : "15",
      "created_at" : 1460708937
    },
    {
      "market_hash_name" : "AK-47 | Aquamarine Revenge (Factory New)",
      "price" : "41.4",
      "created_at" : 1460708934
    }
    ]
}';

$arr = json_decode($json, true);

foreach($arr['prices'] as $key => $val){
    if($val['market_hash_name'] == $market_hash_name){
        echo $val['price']; //15
    }   
}

Answer 3

您需要使用$row['market_hash_name']迭代价格以找到匹配的 market_hash_name 。然后，您可以从该匹配对象中提取价格。

我尽量让你的代码尽可能接近。

$sql = $db->query('SELECT * FROM `items`');
$prices2 = file_get_contents('cache/SuiPrices.json');
$prices2 = json_decode($prices);
while ($row = $sql->fetch()) {
    $price = null;
    foreach($prices2->prices as $data) {
        if($row['market_hash_name'] == $data->market_hash_name) {
            $price = (float) $data->price;

            break;
        }
    }

    $array['items'][] = array(
        'img' => 'http://steamcommunity-a.akamaihd.net/economy/image/'.$row['img'].'/200fx200',
         'name' => $row['market_hash_name'],
         'assetid' => $row['assetid'],
         'price' => $price
    }
}

exit(json_encode($array));

PHP - 读取JSON数据无法正常工作？

3 个答案: