CodeIgniter：第一个查询成功运行，第二个未能更改数据库。 相同的值，不同的表

Question

$ customer_number是从xml文件中获取的，由于我添加了新代码，因此需要对其进行清理以确保它确实是一个INT（我只是想缩小原因）。

将其清除后存储在上述变量名中（如果我回显此变量，则会显示一个数字）。 然后将其放置在两个数组中，以分别针对exp_store_orders和exp_members表进行两个单独的查询。

第一个查询执行顺利。 使用exp_store_order表中的$ customer_number值可以很好地更新account_number。

但是，第二个较小的查询根本无法更新exp_members表中的account_number字段。

如果我将$ updateData2数组中的变量替换为一个简单的数字（例如“ 1”），则第二个查询将执行应做的事情。 然后，使用exp_members更新account_number字段。

我并没有真正知道发生了什么。我不知道如何将它清楚地当作INT并在第一个array \\ first查询中工作，但在第二个数组中却绝对不做任何操作（没有错误！）。

任何帮助将不胜感激。

//searches for and grabs all xml files within the directory. Parses all elements and stores values in variables.
foreach(glob("/random/directory/listing/*xml") as $filename) {
    $xml= simplexml_load_file($filename);

    $order_id = $xml->order_id; 
    $hash = $xml->hash; 
    $timestamp = $xml->timestamp;
    $date = $xml->date;
    $time = $xml->time;
    $customer_number = $xml->customer_number;
    $customer_number = filter_var($customer_number, FILTER_SANITIZE_NUMBER_INT);
    $member_id = $xml->member_id;
    $shipping_account = $xml->shipping_account;
    $shipping_carrier = $xml->shipping_carrier;
    $shipping_tracking = $xml->shipping_tracking;
    $response = $xml->response;
    $response = filter_var($response, FILTER_SANITIZE_STRING);
    $status_update = time();



   //places variables within arrays to be fed into queries.
   $updateData = array(
       'response' => "$response",
       'order_custom2' => "$shipping_account",
       'order_custom3' => "$shipping_carrier",
       'shipping_tracking' => "$shipping_tracking",
       'order_completed_date' => "$timestamp",
       'order_status_updated' => "$status_update",
       'order_status_name' => "Shipping",
       'account_number' => "$customer_number"
   );

   $updateData2 = array(
       'account_number' => "$customer_number"
   );

   //update database with values from XML file, specifically the exp_store_orders and exp_members tables.
   $qry = $this->EE->db->where('id = '.$order_id.'')
                        ->update('exp_store_orders', $updateData); 

   $qry2 = $this->EE->db->where('member_id = '.$member_id.'')
                        ->update('exp_members', $updateData2); 

   //remove files from directory.
   //unlink($filename);
 }

编辑：当我更改代码并执行以下操作：

$updateData2 = array(
       'AS400_account_number' => $xml->customer_number //$customer_number
   );

我收到以下错误：

错误号：1064

您的SQL语法有误； 检查与您的MySQL服务器版本相对应的手册以获取正确的语法，以在第1行的“ WHERE member_id = 1”附近使用

更新exp_members SET AS400_account_number = WHERE member_id = 1

Answer 1

   $updateData2 = array(
       'account_number' => $customer_number
   );

删除$ customer_number周围的引号