将JSON解析为变量或MySQL

Question

很抱歉我不是PHP技术人员：）

我从JSON解析数据时遇到问题。 这是转储后的JSON

$response = curl_exec($ch);
$data = json_decode($response, true);
$c = count($data['products']);
var_dump($data);

array(4) {
  ["products"]=>
  array(23347) {
    [0]=>
    array(25) {
      ["name"]=>
      string(15) "DOOKY Strieška"
      ["description"]=>
      string(0) ""
      ["count"]=>
      float(0)
      ["measureType"]=>
      string(2) "ks"
      ["price"]=>
      float(20.83)
      ["vat"]=>
      float(20)
      ["stockCardNumber"]=>
      string(6) "000004"
      ["typeId"]=>
      int(1)
      ["warehouseCode"]=>
      string(3) "MOL"
      ["timestamp"]=>
      string(19) "2019-05-09T10:38:30"
      ["foreignName"]=>
      string(0) ""
      ["customText"]=>
      string(12) "silver stars"
      ["warrantyPeriod"]=>
      float(0)
      ["weightPerUnit"]=>
      float(0)
      ["volumePerUnit"]=>
      float(0)
      ["numberingSequenceCode"]=>
      string(2) "SK"
      ["ean"]=>
      string(13) "5038278982969"
      ["stn"]=>
      string(0) ""
      ["jkpov"]=>
      string(6) "126367"
      ["din"]=>
      string(0) ""
      ["bn"]=>
      string(0) ""
      ["plu"]=>
      int(0)
      ["warrantyPeriodTypeId"]=>
      int(0)
      ["priceWithVat"]=>
      float(25)
      ["eshop"]=>
      NULL
    }

我需要获取每个EAN，JKPOV，COUNT个变量，然后插入到MySQL

我有这段代码，但是它不能正常工作，因为没有向我写入所有数据。但是在var_dump($data);中，我有这个丢失的数据...

$i=0;
for ($i=0; $i <= $c; $i++) {
$ean = $data['products'][$i]['ean'];
$count = $data['products'][$i]['count'];
$jkpov = $data['products'][$i]['jkpov'];
conf::$db->query("insert ignore into sklad set `ean`='$ean', `jkpov`='$jkpov', `count`='$count'");
$i++; }