Are all the columns completely filled? ie all the "ename" are 10 chars long
and all the "job" fields are 9 chars. When you use the varchar2 field it
only stores the length _used_.
Jim
-----Original Message-----
To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
Date: Friday, March 23, 2001 9:21 AM
Hi Ron,
Let's take a example:
SQL> desc emp
Name Null? Type BYTES
---------------------- -------- ------ ------
EMPNO NOT NULL NUMBER(4) 21
ENAME VARCHAR2(10) 10
JOB VARCHAR2(9) 9
MGR NUMBER(4) 21
HIREDATE DATE 7
SAL NUMBER(7,2) 21
COMM NUMBER(7,2) 21
DEPTNO NUMBER(2) 21
TOTAL=3+(SUM OF COLUMN SIZES)+(# OF COLUMNS < 250)+(3* # OF
COLUMNS > 250)
So according to ur formula
TOTAL=3+131+8=142 BYTES
Is this OK 'coz after analyzing the table with COMPUTE I got AVG_ROW_LEN 40.
Can there be such a HUGE difference ?
Pl. comment.
Thanks,
- Harsh
-----Original Message-----
Sent: Friday, March 23, 2001 3:16 PM
To: Multiple recipients of list ORACLE-L
max(row size) is the sum of all of the columns sizes in the
table plus overhead.
The following is the column sizes and the sum formula to
calculate the size of a row in a data block.
NUMBER =21
CHAR = COLUMN SIZE
VARCHAR = COLUMN SIZE
DATE = 7
TOTAL=3+(SUM OF COLUMN SIZES)+(# OF COLUMNS < 250)+(3* # OF
COLUMNS > 250)
This is part of the calculations used to determine the storage
needed for a table.
OR: you could analyze you table and use the average row size is
you need a general calculation of actual usage.
ROR m*?*m
>>> [EMAIL PROTECTED] 03/22/01 08:45PM >>>
Dear Listers,
I've download a presentation talking about Calculating PCT_USED .
the formula is quite simple :
PCT_USED + PCT_FREE < 100
PCT_USED = PCT_FREE + 1 row size
but I have no idea to calculate the size of 1 row in a table .
is there anyone who knows to calculate it ?
or any alternative formula to calculate PCT_USED ?
Thanks a lot in advance : )
=bambang=
<> Bambang Setiawan <>
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