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Thread: RDBMS Basic question

  1. #1
    Join Date
    Nov 2001
    Location
    Singapore
    Posts
    182

    RDBMS Basic question

    1)how to create multiple tables using single command?
    2)what is boys codds normal form?
    3) What is null? which option is correct?
    ans
    a.undefined
    b.unassigned
    c.assigned zero
    d.ununderstandable
    e.blank space
    f. None of the above

    4)in which normal form functional dependendencies
    are considerd?
    5)what is fourth normal form ?
    6)what type of relationship(parent/child) exists
    in hierarchial model ?
    7)what is returned for %isopen in implicit cursor?
    8)which one doesnot consider null?
    a.count
    b.max
    c.min
    d.avg
    Last edited by ganga; 11-09-2002 at 03:20 PM.
    J Gangadhar

  2. #2
    Join Date
    May 2002
    Posts
    2,645
    1) Can't. You can run a script that has multiple create table statements in it, but there is not a command that says "create the following tables..."

    2) Boyce-Codd normal form is how relational databases are structured with respect to how tables are designed. You can look this up on the Internet, look in any database design book, etc. The goal for most designs is third normal form (3NF). Google returned (only) 3,460 hits in .12 seconds, so I am sure that with a little effort, you can find a link that will help you.

    3) Nulls are nulls, they are not equal to anything, nor are they equal to each other (a null value does not equal another null value). The concept of a null is understandable, the value of a null is not zero, and an attribute can be assigned a null value.

    4-6) When you read up on BCNF, you'll be able to answer these questions.

    7. http://otn.oracle.com/docs/products/...ms48.htm#36258

    8.
    SQL> select comm from emp;

    COMM
    ----------

    500

    1400




    0





    13 rows selected.

    SQL> select max(comm) from emp;

    MAX(COMM)
    ----------
    1400

    SQL> select min(comm) from emp;

    MIN(COMM)
    ----------
    0

    SQL> select avg(comm) from emp;

    AVG(COMM)
    ----------
    633.333333

    SQL> select count(comm) from emp;

    COUNT(COMM)
    -----------
    3
    Last edited by stecal; 11-09-2002 at 03:35 PM.

  3. #3
    Join Date
    Nov 2001
    Location
    Singapore
    Posts
    182
    Thankz for giving hints

    Ganga
    J Gangadhar

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