4

Below is an example of my customer table. There some records having multiple values in BIRTHDAY DATE (by mistake or so). I only want to select those records that have same values for LASTNAME, MIDDLENAME, FIRSTNAME, SSN but different BIRTHDAY:

Member table

LASTNAME   MIDDLENAME   FIRSTNAME    SSN      BIRTHDAY
Jones      M            Carol        1234     17-DEC-45
Jones      M            Carol        1234     17-DEC-45
Jones      M            Carol        4425     20-APR-70
Black      S            Ted          5555     15-MAY-57
Roberts    T            Cole         1412     14-MAY-57
Roberts    T            Cole         1412     20-OCT-57
Roberts    S            Cole         1412     15-MAY-57

I would like the result to be:

LASTNAME   MIDDLEANME  FIRSTNAME    SSN      BIRTHDAY
Roberts    T           Cole         1412     14-MAY-57
Roberts    T           Cole         1412     20-OCT-57

Notice that there were few accounts with same SSN or full name in the table, they are not selected since they don't have everything same. Also Jones M. Carol with 1234 as SSN is not selected either since she does not have different Birthday for two different account.

This is my SQL query that I have so far and this is not working well necessarily.

SELECT x.FIRST_NM, x.MDL_NM, x.LAST_NM, x.SSN, x.BRTH_DT
FROM Member_table x
WHERE EXISTS
(
    SELECT FIRST_NM, MDL_NM, LAST_NM, SSN, COUNT(*)
    from Member_table
    WHERE CURRENT_RECORD_IN = 'Y'
    group by FIRST_NM, MDL_NM, LAST_NM, SSN
    having count(distinct BRTH_DT) > 1
)
ORDER BY FIRST_NM ASC, LAST_NM ASC, MDL_NM ASC, SSN ASC;

Any advice for this query?

  • 1
    What should happen if there are 2 rows with same birthday and a 3rd row with different birthday? – ypercubeᵀᴹ Jun 28 '18 at 13:30
4

This also works on Oracle.

       CREATE TABLE MEMBER 
(
    LASTNAME    VARCHAR2(7),
    MIDDLENAME   CHAR(1),
    FIRSTNAME    VARCHAR2(5),
    SSN          INT,
    BIRTHDAY     VARCHAR2(9)
);

Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Jones','M','Carol',1234,'17-DEC-45');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Jones','M','Carol',1234,'17-DEC-45');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Jones','M','Carol',4425,'20-APR-70');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Black','S','Ted',5555,'15-MAY-57');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Roberts','T','Cole',1412,'14-MAY-57');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Roberts','T','Cole',1412,'20-OCT-57');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('Roberts','S','Cole',1412,'15-MAY-57');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('James','N','Rob',7890,'18-JUN-58');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('James','N','Rob',7890,'15-JUN-58');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('James','N','Rob',7890,'20-MAR-56');
Insert into MEMBER (LASTNAME,MIDDLENAME,FIRSTNAME,SSN,BIRTHDAY) values ('James','N','Rob',7890,'14-APR-55');

SELECT      DISTINCT a.* 
 FROM        member a,member b
 WHERE       a.lastname=b.lastname
       AND   a.middlename=b.middlename
       AND   a.firstname=b.firstname
       AND   a.ssn=b.ssn
       AND   a.birthday != b.birthday

  ORDER BY  a.lastname,a.middlename,a.firstname,a.ssn,a.birthday;

output

James   N   Rob   7890  14-APR-55
James   N   Rob   7890  15-JUN-58
James   N   Rob   7890  18-JUN-58
James   N   Rob   7890  20-MAR-56
Roberts T   Cole  1412  14-MAY-57
Roberts T   Cole  1412  20-OCT-57
  • 1
    This would work but may produce multiple results (if for example there are 4 rows with same everything and different birthday, all 4 will appear in the result 3 times) – ypercubeᵀᴹ Jun 28 '18 at 13:26
  • You're right ,fixed the code with "distinct".I never thought of that possibility but it was nice of you to point out. – Sam Jun 28 '18 at 14:33
3

Here's an example of using EXISTS and a correlated subquery. I tested on SQL Server, but will probably work on other RDBMS's.

drop table if exists table1
CREATE TABLE Table1
    (LASTNAME varchar(7), MIDDLENAME varchar(1), FIRSTNAME varchar(5), SSN int, BIRTHDAY varchar(9))
;

INSERT INTO Table1
    (LASTNAME, MIDDLENAME, FIRSTNAME, SSN, BIRTHDAY)
VALUES
    ('Jones', 'M', 'Carol', 1234, '17-DEC-45'),
    ('Jones', 'M', 'Carol', 1234, '17-DEC-45'),
    ('Jones', 'M', 'Carol', 4425, '20-APR-70'),
    ('Black', 'S', 'Ted', 5555, '15-MAY-57'),
    ('Roberts', 'T', 'Cole', 1412, '14-MAY-57'),
    ('Roberts', 'T', 'Cole', 1412, '20-OCT-57'),
    ('Roberts', 'S', 'Cole', 1412, '15-MAY-57')
;


SELECT *
FROM table1 t1
WHERE EXISTS (
        SELECT *
        FROM table1
        WHERE LASTNAME = t1.LASTNAME
            AND MIDDLENAME = t1.MIDDLENAME
            AND FIRSTNAME = t1.FIRSTNAME
            AND SSN = t1.SSN
            AND BIRTHDAY <> t1.BIRTHDAY
        )

| LASTNAME | MIDDLENAME | FIRSTNAME | SSN  | BIRTHDAY  |
|----------|------------|-----------|------|-----------|
| Roberts  | T          | Cole      | 1412 | 14-MAY-57 |
| Roberts  | T          | Cole      | 1412 | 20-OCT-57 |
  • 1
    It might be rewritten using an IINNER JOIN if you need the list of all pairs SELECT * FROM table1 AS t1 INNER JOIN table1 AS t2 ON (t2.LASTNAME = t1.LASTNAME AND t2.MIDDLENAME = t1.MIDDLENAME AND t2.FIRSTNAME = t1.FIRSTNAME AND t2.SSN = t1.SSN AND t2.BIRTHDAY <> t1.BIRTHDAY) Otherwise, SELECT DISTINCT t1.* FROM... can be used to get only t1 rows (not the t1 t2 pairs) – Xenos Jun 28 '18 at 13:45
2

Using aggregation, this is simple:

select *
from member_table
where (firstname, middle_name, last_name, ssn) in (
  select firstname, middle_name, last_name, ssn
  from member_table
  group by firstname, middle_name, last_name, ssn
  having min(birthday) <> max(birthday)
);

The outer query is only necessary if you need the actual birthdays (as in your sample output), otherwise the inner query suffices.

Note that this works for all datatypes; min() and max() might not be returning the minimum and maximum values, respectively, but in this particular case that doesn't matter -- as long as they are different.

1

Note that there's nothing to tie your WHERE EXISTS subquery to the outer query; if there's any records in your table where the first four values are identical, but the birth date is different, then all rows of your table qualify.

The simplest solution would be to use the WHERE EXISTS subquery as a derived table, and joining it to Member_table:

SELECT DISTINCT x.FIRST_NM, x.MDL_NM, x.LAST_NM, x.SSN, x.BRTH_DT
FROM Member_table x
       INNER JOIN (
                   SELECT FIRST_NM, MDL_NM, LAST_NM, SSN, COUNT(*)
                   from Member_table
                   WHERE CURRENT_RECORD_IN = 'Y'
                   group by FIRST_NM, MDL_NM, LAST_NM, SSN
                   having count(distinct BRTH_DT) > 1
                  ) bd ON (    x.FIRST_NM = bd.FIRST_NM
                           AND x.MDL_NM = bd.MDL_NM
                           AND x.LAST_NM = bd.LAST_NM
                           AND x.SSN = bd.SSN
                          )
ORDER BY FIRST_NM ASC, LAST_NM ASC, MDL_NM ASC, SSN ASC;

So:

  • Your subquery returns the name and SSN of all rows where different birth dates exist for those same name and SSN values.
  • By joining that to Member_table on the name and SSN columns, you ensure that you're only grabbing rows where multiple birth dates exist.
  • I also added DISTINCT to the main query'sSELECT` list, so that you only get 1 copy of each row in your output.

Untested, as your specific DBMS was unstated.

1

If your Oracle version supports window function:

select LASTNAME, MIDDLENAME, FIRSTNAME, SSN, BIRTHDAY 
from (
    select LASTNAME, MIDDLENAME, FIRSTNAME, SSN, BIRTHDAY 
         , first_value(BIRTHDAY) over (partition by LASTNAME, MIDDLENAME, FIRSTNAME, SSN order by BIRTHDAY) as fst
         , first_value(BIRTHDAY) over (partition by LASTNAME, MIDDLENAME, FIRSTNAME, SSN order by BIRTHDAY desc) as lst
    from Table1
) t
where fst <> lst;

Note that fst have asc order and lst have desc order. An alternative is to use last_value for the latter, but then we would have to declare the window frame:

last_value(BIRTHDAY) over (
    partition by LASTNAME, MIDDLENAME, FIRSTNAME, SSN 
    order by BIRTHDAY 
    rows between current row and unbounded following
) as lst

This is not necessary for first_value since the default is:

range between unbounded preceding and current row

DB<>Fiddle

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