PHP Shopify集成-变量删除

PHP Shopify集成-变量删除,php,shopify,Php,Shopify,我正在开发一个集成到shopify的PHP应用程序。基本上,“商店”和应用程序之间必须同步产品信息。在产品内部,我们有变体(类似于子产品)。Shopify用来发送带有json数据的Webhook来报告这些更改。每次我更改/添加/删除变体时,shopify都会发送一个“产品更新”webhook,该webhook只更改json内容。这是一个例子: { ... "variants": [{ "id": 279656846, "position": 1,

我正在开发一个集成到shopify的PHP应用程序。基本上,“商店”和应用程序之间必须同步产品信息。在产品内部,我们有变体(类似于子产品)。Shopify用来发送带有json数据的Webhook来报告这些更改。每次我更改/添加/删除变体时,shopify都会发送一个“产品更新”webhook,该webhook只更改json内容。这是一个例子:

{
...
    "variants": [{
        "id": 279656846,
        "position": 1,
        "price": "80.00",
        "product_id": 123022448,
        "sku": "1000",
        "option1": "30 cm",
        "inventory_quantity": 10
    },
    {
        "id": 291321287,
        "position": 2,
        "price": "15.00",
        "product_id": 123022448,
        "sku": "1003",
        "option1": "15 cm",
        "inventory_quantity": 23
    }],
...
}
如果我创建了一个新的变体,它会向我发送一个当前状态的“产品更新”,并且新的变体是json格式的。类似地,如果我删除,它只向我发送当前状态的“产品更新”,但没有json中已删除的变量

我创建了以下代码,可以正确处理更改/添加案例:

foreach ($jsonArr['variants'] as $rows) {

    $variant = $rows['option1'];
    $sku = $rows['sku'];
    $salesPrice = $rows['price'];
    $stockQty = $rows['inventory_quantity'];

    $idVar = $rows['id'];

    $dupchk = mysql_query("SELECT * FROM `variants` WHERE `idVar`='$idVar'",$con) or die (mysql_error());
    $num_rows = mysql_num_rows($dupchk);

    if ($num_rows > 0) {

        $sql = "UPDATE `variants` SET `variant`='$variant',`sku`='$sku',`salesPrice`='$salesPrice',`stockQty`='$stockQty' WHERE `idVar`='$idVar'";

        if (!mysql_query($sql,$con)) {
            die('Error: ' . mysql_error());
        }

    }

    else {

        $sql = "INSERT INTO `variants`(`idVariant`, `idProduct`, `variant`, `sku`, `salesPrice`, `stockQty`, `comments`, `idVar`) VALUES ('','$idProduct','$variant','$sku','$salesPrice','$stockQty','','$idVar')";

        if (!mysql_query($sql,$con)) {
            die('Error: ' . mysql_error());
        }

    }

}

问题是此代码不处理delete variant情况。我曾尝试过这样做,但直到现在,我只创建了一个无法工作的“大混乱”代码。如果您有任何建议,请告知如何明智地处理此问题。

为了解决此问题,我使用了以下代码:

//Variable to count variant update or create
$var_count = 0;

foreach ($jsonArr['variants'] as $rows) {

    $variant = $rows['option1'];
    $sku = $rows['sku'];
    $salesPrice = $rows['price'];
    $stockQty = $rows['inventory_quantity'];

    $idVar = $rows['id'];

    $dupchk = mysql_query("SELECT * FROM `variants` WHERE `idVar`='$idVar'",$con) or die (mysql_error());
    $num_rows = mysql_num_rows($dupchk);

    if ($num_rows > 0) {

        $sql = "UPDATE `variants` SET `variant`='$variant',`sku`='$sku',`salesPrice`='$salesPrice',`stockQty`='$stockQty' WHERE `idVar`='$idVar'";
        $var_count++;


        if (!mysql_query($sql,$con)) {
            die('Error: ' . mysql_error());
        }

    }

    else {

        $sql = "INSERT INTO `variants`(`idVariant`, `idProduct`, `variant`, `sku`, `salesPrice`, `stockQty`, `comments`, `idVar`) VALUES ('','$idProduct','$variant','$sku','$salesPrice','$stockQty','','$idVar')";
        $var_count++;

        if (!mysql_query($sql,$con)) {
            die('Error: ' . mysql_error());
        }

    }

}

//Start checking to erase variant if needed
$result = mysql_query("SELECT `idVar` FROM products NATURAL JOIN variants WHERE `idShopify`='$idShopify'",$con);
$num_rows = mysql_num_rows($result);

if ($num_rows>$var_count) {

    while($row = mysql_fetch_array($result))
        {

            $clear = 0;

            foreach ($jsonArr['variants'] as $rows) {

                if ($rows['id']==$row['idVar']) {
                    $clear++;
                }

            }

            if ($clear==0) {

                $idVar = $row['idVar'];
                $sql = "DELETE FROM `variants` WHERE `idVar`='$idVar'";

                if (!mysql_query($sql,$con)) {
                    die('Error: ' . mysql_error());
                }

            }

        }

}
这不是优雅,而是做工。请随意提出代码改进建议