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Message Posted: Tue, 28 Nov 2006 @ 22:15:01 GMT

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<-- Anonymously Posted: Tuesday, November 28, 2006 17:03 -->

We are trying to estimate volumetrics for a new application and I really need to understand how much physical disk space a Timestamp(0) datatype will consume. Dieter's comments pasted below leave me a little confused. I think what you are saying is that the columnlength column indicates the "logical" presentation of the timestamp value, e.g. Timestamp(0) = 19 but physical internal storage only consumes 10 bytes. What I don't understand is why other column datatypes like Smallint, Integer and Date have their "physical" column length displayed in the columnlength column Is it not safe to assume the TS value is physical as well?

I created a test table (shown below) with a few different datatypes for each column and then checked their columnlength values in DBC.Columns. It appears that the numeric datatypes display their "physical" length and the non-numeric datatypes display their "logical" length. Is my assumption correct? And if so, how do I determine/prove what it costs to store a Timestamp(0) datatype?

Thanks in advance for sharing your knowledge!

     ColumnName      Type                 Length   Format
     SMALLINTsamp    SMALLINT                2
     INTEGERsamp     INTEGER                 4
     DATEsamp        DATE                    4     YYYY-MM-DD
     TIMEHHMM        INTEGER                 4     99:99
     TIMEsamp        TIME                    8     HH:MI:SS
     TIME_WITH_ZONE  TIME WITH TIMEZONE     21     HH:MI:SS.S(6)Z
     BYTEINTsamp     BYTEINT                 1
     DECIMAL9_4      DECIMAL                 4     ------.9999
     DECIMAL10_0     DECIMAL                 8     ----------9.
     CHAR8           CHAR                    8     X(8)
     VARCHAR8        VARCHAR                 8     X(8)
     TIMESTAMPsix    TIMESTAMP              26     YYYY-MM-DDBHH:MI:SS.S(6)
     TIMESTAMPzero   TIMESTAMP              19     YYYY-MM-DDBHH:MI:SS TimeStamp(0)

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