所以我有这个php文件和下面的查询

$query = "select Customer.CustID, LastName,FirstName,DOB,CustPhone, 
Max(TransDate) as LatestTransaction from Customer, History where 
Customer.AuthKey='A111' and Customer.AuthKey=History.AuthKey and 
Customer.CustID=History.CustID  Group by CustID;";

$result = mysqli_query($con,$query);

$rows = array();
while($r = mysqli_fetch_array($result)) {
$rows[] = $r;
}
  echo json_encode($rows);

  mysqli_close($con);

然后我在这里做这个c#来得到结果并把它们放到datagridview中

WebClient wc = new WebClient();
            var json = wc.DownloadString(URL);
            List<Customer> customers = 
            JsonConvert.DeserializeObject<List<Customer>>(json);
            CustomerDataViewGrid.DataSource = customers;

这是没有问题的，但我想采取

Customer.AuthKey = 'A111'

从上面的php中选择query，并将c代码中的值传递给它

$aKey = $_POST['Authkey'];
Customer.AuthKey = '$aKey'

我曾尝试在同一个按钮点击事件中执行此操作，但我不太了解如何发送此字符串，然后立即获得下载字符串。当我硬编码customer.authkey时，它会返回结果，但当我尝试发送此字符串时，它不会返回任何内容。我怎样才能在同一事件中做到这一点？
谢谢