How do i determine what the next value is of the Primary Key Auto increment?
I have been looking in the system databases but i couldnt find any answers.
I am guessing you mean "the next value to be assigned to an identity column", and your identity column also happens to be used as a primary key. Otherwise, there is no concept in Sybase ASE that would qualify as "Primary Key Auto increment".
Assuming the above, you can get the next value to be assigned by using the function next_identity('table_name'). Note that the result is in varchar format so you may need to convert it back to a number.
Related
I want to pass a value for the row key as a parameter just like the partition key. But the UI only gives me an option to use unique identifier or a source column. Actually I need to use this same entity somewhere else. How will I query this entity if the row key is going to be a random value?
Based on the official statement , partition key can be set custom value yet row key only could be set column name from source or GUID default value.
I think it's because of guaranteed uniqueness constraint restrictions. So, if you want to control row key value , you could add row key into your source data.
Hope it helps you.
I have an existing sybase ase table which is using IDENTITY as its primary key. Now i need to recreate this table but i want to start the PK from next value of IDENTITY PK in prod env. e.g. If currently PK = 231 then after re-creating i want it to start from 232 onwards or any other INTEGER value > 231.
In oracle its easy to configure a sequence number and we can give start with but in sybase ase as we dont have sequence available so i tried using newid() function but it gives binary(16) values whereas i want integer values.
Can anyone suggest something ?
I am planning to use something like mentioned below and i think it will resolve my problem. Let me know if anyone has a better solution.
select abs(hextoint(newid()))
Any thoughts on this solution ? Can this ever generate the same number which it generated already?
select next_identity('tablename') will return the identity value of the next insert for a table with an identity column so you know which ID will be allocated next.
Select ##identity immediately after an insert will return the ID which was just given to the row inserted.
However you need to be careful as identity columns are not the same as sequences and should not be relied upon if you want a sequence with no gaps because you will get a gap (albeit small sometimes) if the database crashes or is shutdown with nowait. For these a number fountain/insert trigger type generation of IDs is a better option. Using 'identity insert' is only really for when you want to bulk-load a whole table - you should not be setting that with every insert or you will defeat the whole purpose of an identity column, which is fast generation of new key values.
there's some Cassandra operation to verify if a Column Family contains a key? We don't need any row data, only key existence or not.
Best Regards
If you're using Java then create a SliceQuery for the rowKey and set begin/end values equal to the specific column key you're looking for. If there is a column with the specific key then the following expression will be true:
sliceQuery.execute().get().getColumns().size() > 0
One quick way of doing it is to ask for the column count for the row, if it's positive the row exists. Because of tombstones there's a gray area around "does not exist". You can remove all columns for a row, but asking for data for the row may result in an empty set of columns instead of null (this depends a lot on which driver you're using). You should consider rows that don't have columns as non-existent, and therefore asking for the column count is probably the best way to determine if a row exists or not.
There's some more information about this in the Cassandra FAQ under "range ghosts".
I'm having problems with using the time_uuid type as a key in my columnfamily. I want to store my records, and have them ordered by when they were inserted, and then I figured that the time_uuid is a good way to go. This is how I've set up my column family:
sys.create_column_family("keyspace", "records", comparator_type=TIME_UUID_TYPE)
When I try to insert, I do this:
q=pycassa.ColumnFamily(pycassa.connect("keyspace"), "records")
myKey=pycassa.util.convert_time_to_uuid(datetime.datetime.utcnow())
q.insert(myKey,{'somedata':'comevalue'})
However, when I insert data, I always get an error:
Argument for a v1 UUID column name or value was neither a UUID, a datetime, or a number.
If I change the comparator_type to UTF8_TYPE, it works, but the order of the items when returned are not as they should be. What am I doing wrong?
The problem is that in your data model, you are using the time as a row key. Although this is possible, you won't get a meaningful ordering unless you also use the ByteOrderedPartitioner.
For this reason, most people insert time-ordered data using the time as a column name, not a row key. In this model, your insert statement would look like:
q.insert(someKey, {datetime.datetime.utcnow(): 'somevalue'})
where someKey is a key that relates to the entire time series that you're inserting (for example, a username). (Note that you don't have to convert the time to UUID, pycassa does it for you.) To store something more than a single value, use a supercolumn or a composite key.
If you really want to store the time in your row keys, then you need to specify key_validation_class, not comparator_type. comparator_type sets the type of the column names, while key_validation_class sets the type of the row keys.
sys.create_column_family("keyspace", "records", key_validation_class=TIME_UUID_TYPE)
Remember the rows will not be sorted unless you also use the ByteOrderedPartitioner.
The comparator for a column family is used for ordering the columns within each row. You are seeing that error because 'somedata' is valid utf-8 but not a valid uuid.
The ordering of the rows stored in cassandra is determined by the partitioner. Most likely you are using RandomPartitioner which distributes load evenly across your cluster but does not allow for meaningful range queries (the rows will be returned in a random order.)
http://wiki.apache.org/cassandra/FAQ#range_rp
I'm using C# and Microsoft.Jet.OLEDB.4.0 provider to insert rows into an Access mdb.
Yes, I know Access sucks. It's a huge legacy app, and everything else works OK.
The table has an autonumber column. I insert the rows, but the autonumber column is set to zero.
I Googled the question and read all the articles I could find on this subject. One suggested inserting -1 for the autonumber column, but this didn't work. None of the other suggestions I could find worked.
I am using OleDbParameter's, not concatenating a big SQL text string.
I've tried the insert with and without a transaction. No difference.
How do I get this insert to work (i.e. set the autonumber column contents correctly)?
Thanks very much in advance,
Adam Leffert
In Access it is possible to INSERT an explicit value into an IDENTITY (a.k.a. Automnumber) column. If you (or your middleware) is writing the value zero to the IDENTITY column and there is no unique constraint on the IDENTITY column then that might explain it.
Just to be clear you should be using the syntax
INSERT INTO (<column list>) ...
and the column list should omit the IDENTITY column. Jet SQL will allow you to omit the entire column list but then implicitly include the IDENTITY column. Therefore you should use the INSERT INTO () syntax to explicitly omit the IDENTITY column.
In Access/Jet, you can write explicit values to the IDENTITY column, in which case the value will obviously not be auto-generated. Therefore, ensure both you and your middleware (ADO.NET etc) are not explicitly writing a zero value to the IDENTITY column.
BTW just for the IDENTITY column in the below table will auto-generate the value zero every second INSERT:
CREATE Table Test1
(
ID INTEGER IDENTITY(0, -2147483648) NOT NULL,
data_col INTEGER
);
When doing the insert, you need to be sure that you are NOT specifying a value for the AutoNumber column. Just like in SQL Server you don't insert a value for an identity column.