Re: How to determine if a table has a column that is autoincrement(Identity)?

From: Miha Markic [MVP C#] (miha)
Date: 06/03/04 

Date: Thu, 3 Jun 2004 07:58:19 +0200

Hi Joe,

What database are we talking about? 

-- 
Miha Markic [MVP C#] - RightHand .NET consulting & development
miha at rthand com
www.rthand.com
"Joe" <anon03@comcast.net> wrote in message
news:fc5825fe.0406021922.ec5b4c8@posting.google.com...
> Hi All,
>
>   I am trying to find out how to determine if a table that I am
>   working with contains a column that is an identity column (ie,
>   has autoincrement set to true). I am working in C# and I'm
>   already using the GetOleDbSchemaTable method to get the schema
>   for the table I am interested in. However, this DataTable that
>   I get back does give me the names of the columns but it does
>   not tell me if a column is an identity column, or at the very
>   least I don't know where to look for it in the Datatable. This
>   is the call I am using:
>
>   DataTable dtCols = conn.GetOleDbSchemaTable(
>                     OleDbSchemaGuid.Columns,
>                     new Object[]{null, null, sTableName, null});
>
>   Now I've seen posts and visted websites that say you can get
>   the autoincrement/identity flag by accessing the Columns data
>   member as such:
>
>   dtCols.Columns[i].AutoIncrement
>
>   Now, although this flag is there, it seems to be associated
>   with some meta-data column. In fact, all of the Columns that
>   are in the DataTable(dtCols, the table that I get from the
>   call above) seem to be associated with the table's meta-data
>   (like, TABLE_CATALOG, TABLE_SCHEMA, TABLE_NAME). Its the Rows
>   data member that contains information on the columns of the
>   table but I can't find any reference to the autoincrement or
>   identity flag. I was hoping it would be in this DataTable
>   since I could make one function call and get all I need.
>
>   Can someone please tell me how I can programatically get this
>   flag or tell me where to find it? Thanks in advance.
>
> Sincerely,
>   Joe

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