Comparison of the 2 columns from the same table

Mr President.

I have a table xyz that has 10 million rows

who has 3 columns

ProfileId personid fpersonid


I find on all of the profileids that do not match personid = fpersonid

is there a way better and faster outside self-join?


Select a. profileid xyz a, b of xyz where a.personid! = b.fpersonid

Your SQL is incorrect.

Just use... . Select xyz profileid where personid! = fpersonid

Tags: Database

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    A21849B1020792 02/01/2015

    A12349 B1020792 01/01/2015

    A12345B1234567 01/03/2015      

    A43567B1234567 01/04/2015      


    Here's the query I used:

    substr (a.id, 7) select sc, max (a.create_date) cd1

    of table a, table b

    where a.id! = b.id

    and substr (a.id, 7) = substr (b.id, 7)

    and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

    and a.id ('A12349B1020792', 'A21849B1020792','A12345B1234567 ','A43567B1234567')

    Group of substr (a.id, 7);

    Result:

    ---------

    sc                    cd1

    ---                    ------

    B1020792 02/01/2015

    B1234567 04/01/2015

    There is one thing that I'm not able to do with my request, that is, be able to get real IDs list and not list of substrings. If I added a.id column in the select clause and the group by clause, the query includes both by substr (a.id, 7) and a.id and as a result, the query returns four lines as below:

    substr (a.id, 7) select sc, max (a.create_date) cd1, a.id ID

    of table a, table b

    where a.id! = b.id

    and substr (a.id, 7) = substr (b.id, 7)

    and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

    and a.id ('A12349B1020792', 'A21849B1020792','A12345B1234567 ','A43567B1234567')

    Group by substr (a.id, 7), a.id.

    Result:

    ---------

    sc                    cd1                      ID

    ---                    ------                   ---------

    01/01/2015 A12349B1020792 B1020792

    B1020792 01/02/2015 A21849B1020792

    B1234567 01/03/2015 A12345B1234567

    B1234567 01/04/2015 A43567B1234567


    My goal is to retrieve a list of all the unique identifiers that have the later dates.


    ID:

    ---

    A21849B1020792

    A43567B1234567


    Thanks for your time in advance!

    Hello

    user11951344 wrote:

    I have a table with ID (14 digit string values) starting with "A". Ex: A21849B1020792. There may be a different ID with the same substring "B1020792" in the same table, example: A12349B1020792. If this is the case, the ID of the last creation date must be returned in the result. In other words, comparison is done on the same table. "In the example provided, say A12349B1020792 updated 1 January 2015 ' and A21849B1020792 a date February 1, 2015", as A21849B1020792 has the most recent date, the result should contain only the A21849B1020792.

    ID create_date

    ---                              ------------------

    A21849B1020792 02/01/2015

    A12349B1020792 01/01/2015

    A12345B1234567 03/01/2015

    A43567B1234567 01/04/2015

    Here's the query I used:

    substr (a.id, 7) select sc, max (a.create_date) cd1

    of table a, table b

    where a.id! = b.id

    and substr (a.id, 7) = substr (b.id, 7)

    and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

    and a.id ('A21849B1020792', 'A12345B1234567', 'A12349B1020792', 'A43567B1234567')

    Group of substr (a.id, 7);

    Result:

    ---------

    sc                    cd1

    ---                    ------

    B1020792 02/01/2015

    B1234567 01/04/2015

    There is one thing that I am not able to do with my request, that is, the ability to retrieve the list of IDs real and not a list of substrings. If I added a.id column in the select clause and the group by clause, the query groups according to the two substr (a.id, 7) and a.id and as a result, the query returns four lines as below:

    substr (a.id, 7) select sc, max (a.create_date) cd1, a.id ID

    of table a, table b

    where a.id! = b.id

    and substr (a.id, 7) = substr (b.id, 7)

    and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

    and a.id ('A21849B1020792', 'A12345B1234567', 'A12349B1020792', 'A43567B1234567')

    Group by substr (a.id, 7), a.id.

    Result:

    ---------

    sc                    cd1                      ID

    ---                    ------                   ---------

    B1020792 01/01/2015 A12349B1020792

    B1020792 02/01/2015 A21849B1020792

    B1234567 03/01/2015 A12345B1234567

    B1234567 01/04/2015 A43567B1234567

    My goal is to retrieve a list of all the unique identifiers that have the later dates.

    ID:

    ---

    A21849B1020792

    A43567B1234567

    Thanks for your time in advance!

    If it makes sense to treat the first 6 characters of the ID separately from the rest of the id in this problem, maybe it makes sense to store those two parts of the id in two different columns.  Relational databases work best when each column of each row contains 1 single piece of data (at most).  It is so fundamental to the design of table that he called the first normal form.

    Given that the two parts are stored in column 1, so you can something like this Request Top - N:

    WITH got_r_num AS

    (

    SELECT r.id

    r.create_date AS cd1

    , RANK () OVER (PARTITION OF SUBSTR (r.id, 7))

    ORDER BY r.create_date DESC

    ) AS r_num

    FROM table_x g - g for data values

    JOIN table_x r - r for related values

    ON SUBSTR (r.id, 7) = SUBSTR (g.id, 7)

    WHERE g.id IN ('A12349B1020792'

    , "A21849B1020792".

    , "A12345B1234567".

    , "A43567B1234567".

    )

    - AND g.id LIKE 'A %' - if necessary.   The above condition ensures already g.id start with "A".

    AND r.id LIKE 'a % '.

    )

    SELECT id, create_date

    OF got_r_num

    WHERE r_num = 1

    ;

    If you would care to post CREATE TABLE and INSERT instructions for the sample data, and then I could test this.

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