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SQL Server Forum / DB Engine / SQL Server CE / December 2007

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Unspecified error when inserting/updating record using SqlCeResultSet on Sql Compact

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Jeffry van de Vuurst - 14 Nov 2007 22:13 GMT
Hi,

Inside a single transaction I'm inserting/updating multiple records into
multiple tables, like this:
table 1 record 1
table 2 record 1
table 3 record 1
table 1 record 2
table 2 record 1
table 1 record 3
table 2 record 3
table 3 record 3

Now I'm getting an unspecified error on a certain table:

Unspecified error [ 3,-1300704038,-1834095882,activitypointerBase,x lock (x
blocks),PAG (idx): 1078 ]

From msdn I see that:

PAG (idx) means a lock on an index page.
x lock means an exclusive lock:
Indicates a data modification, such as an insert, an update, or a deletion.
Ensures that multiple updates cannot be made to the same resource at the
same time. (I assume that multiple updates within the SAME transaction can
be made, only multiple updates from different transaction cannot be made,
right?)

I cannot find any reference to this error message and don't know what the
numbers mean. Maybe it relates to data that can be found in the
sys.lock_information table like explained here,
http://technet.microsoft.com/en-us/library/ms172932.aspx, but I'm not sure.

Furthermore, the sys.lock_information table is empty. I haven't been able to
reproduce the problem myself. I only received an error log and the database
to investigate it.

So, does anybody have an idea what this error message means and what I can
do to troubleshoot this?

Thanks,

Signature

Jeffry van de Vuurst
CWR Mobility BV
www.cwrmobility.com/weblog
--

Simon Hart [MVP] - 02 Dec 2007 10:10 GMT
All I can guess it that you are accidentally updating the same record in that
transaction by mistake. Or you have another transaction open somewhere that
has not been rolled or commited.
Signature

Simon Hart
Visual Developer - Device Application Development MVP
http://simonrhart.blogspot.com

> Hi,
>
[quoted text clipped - 37 lines]
>
> Thanks,
Jeffry van de Vuurst - 03 Dec 2007 14:17 GMT
Hi Simon,

Indeed, it turned out that I started a transaction outside a try/catch. Then
something else failed, also outside the try/catch, which caused the
transaction to hang around. Then in the next loop, the previous transaction
was still running. I just had to improve my error handling a bit.

Thanks,

Signature

Jeffry van de Vuurst
CWR Mobility BV
www.cwrmobility.com/weblog
--

> All I can guess it that you are accidentally updating the same record in
> that
[quoted text clipped - 50 lines]
>>
>> Thanks,
 
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