I'm using TransactionScope
to do some batch insert and updates. Problem is, I'm getting timeout exceptions on a 30 min long operation even when I set the timeout of the TransactionScope
to one hour.
Also after the exception it inserts seemingly random amount of the records of the batch. For example the last operation had 12440 inserts and after the timeout there were 7673 records inserted to the table.
The timeout of the SqlConnection
and SqlCommand
are both set to int.MaxValue
.
What am I doing wrong?
Here's my code:
using (TransactionScope transaction = new TransactionScope(TransactionScopeOption.Required, TimeSpan.FromHours(1)))
{
try
{
using (db = new DB())
{
//operations here
}
}
catch (Exception ex)
{
throw new Exception("DB Error:\r\n\r\n" + ex.Message);
}
transaction.Complete();
} // <--- Exception here: Transaction aborted (Inner exception: Timeout)
Is your transaction failing after 10 minutes? If so, you are probably hitting the Transaction Manager Maximum Timeout which is set in the machine.config. If I recall correctly, if you try to set a timeout greater than the maximum value then your setting will be ignored. Try upping the value in machine.config and see if that helps your issue.
In terms of random commits do you set Transaction Binding=Explicit Unbind
on your connection string? The default value is Transaction Binding=Implicit Unbind
. From MSDN:
Implicit Unbind causes the connection to detach from the transaction when it ends. After detaching, additional requests on the connection are performed in autocommit mode. The System.Transactions.Transaction.Current property is not checked when executing requests while the transaction is active. After the transaction has ended, additional requests are performed in autocommit mode.
Basically, when the transaction times out all inserts up to that point will be rolled back but any additional inserts done using the same connection will be done in autocommit mode where every insert statement will be immediately committed. That does sound similar to the scenario you are seeing (but it's hard to know without seeing the full code/repro).