Question

我正在将一个项目从MySQL迁移到MS SQL。这与我以前的习惯有很大不同，因为我是从1个MySQL表转为MS SQL视图和表。我希望可以坚持使用MySQL，但这是我无法控制的。

我需要插入数据，否则将数据更新到sql表中。这是我的尝试和错误。请帮助：

$serverName = "myserver";
$connectionInfo = array( "Database"=>"sqldatabasename", "UID"=>"myusername", "PWD"=>"youguysarethebest" );
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn === false ) {
 die( print_r( sqlsrv_errors(), true));
}

$sql = "INSERT INTO CallSettings (Order_Number, ODetail_id, Serial_Id, AssignedPerson, Status, Comments, Customer_IPAddress, customer_assistance_name, customer_assistance_number, customer_assistance_email,networksettings, addressbook, accountcodes, installedprintdrivers) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?,     ?, ?) ELSE UPDATE CallSettings SET (Order_Number= (?), ODetail_id= (?), Serial_Id= (?), AssignedPerson= (?), Status= (?), Comments= (?), Customer_IPAddress= (?), customer_assistance_name= (?), customer_assistance_number=(?), customer_assistance_email= (?), networksettings= (?), addressbook= (?), accountcodes= (?),     installedprintdrivers= (?))";

$params = array($Order_Number, $ODetail_id, $Serial_Id, $AssignedPerson, $Status, $Comments,         $Customer_IPAddress, $customer_assistance_name, $customer_assistance_number, $customer_assistance_email,         $networksettings, $addressbook, $accountcodes, $installedprintdrivers);

$stmt = sqlsrv_query( $conn, $sql, $params);
if( $stmt === false ) {
 die( print_r( sqlsrv_errors(), true));
}
sqlsrv_close($conn);

错误

Array ( [0] => Array ( [0] => 07002 [SQLSTATE] => 07002 [1] => 0 [code] => 0 [2] => [Microsoft][SQL Server Native Client 11.0]COUNT field incorrect or syntax error [message] => [Microsoft][SQL Server Native Client 11.0]COUNT field incorrect or syntax error ) )

Answer 1

错误消息是因为传递的参数比预期的要少（14个而不是28个）。

您需要查询中所有?标记的参数。您仅传递前14个元素的参数（更新部分还有14个参数，即使它们与第一部分相同），因此您需要在查询调用中重复参数数组（参数列表的2倍））以传递所有值。

$stmt = sqlsrv_query( $conn, $sql, array_merge($params,$params));

PHP SQLSRV插入其他更新

1 个答案: