Oracle SQL - Comparing Rows - sql

Question

I have a problem I'm working on with Oracle SQL that goes something like this.
TABLE
PurchaseID CustID Location
----1------------1-----------A
----2------------1-----------A
----3------------2-----------A
----4------------2-----------B
----5------------2-----------A
----6------------3-----------B
----7------------3-----------B
I'm interested in querying the Table to return all instances where the same customer makes a purchase in different locations. So, for the table above, I would want:
OUTPUT
PurchaseID CustID Location
----3------------2-----------A
----4------------2-----------B
----5------------2-----------A
Any ideas on how to accomplish this? I haven't been able to think of how to do it, and most of my ideas seem like they would be pretty clunky. The database I'm using has 1MM+ records, so I don't want it to run too slowly.
Any help would be appreciated. Thanks!

Answer 1

SELECT *
FROM YourTable T
WHERE CustId IN (SELECT CustId
FROM YourTable
GROUP BY CustId
HAVING MIN(Location) <> MAX(Location))

Answer 2

You should be able to use something similar to the following:
select purchaseid, custid, location
from yourtable
where custid in (select custid
from yourtable
group by custid
having count(distinct location) >1);
See SQL Fiddle with Demo.
The subquery in the WHERE clause is returning all custids that have a total number of distinct locations that are greater than 1.

Answer 3

In English:
Select a row if another row exists with the same customer and a different location.
In SQL:
SELECT *
FROM atable t
WHERE EXISTS (
SELECT *
FROM atable
WHERE CustID = t.CustID
AND Location <> t.Location
);

Answer 4

Here's one approach using a sub-query
SELECT T1.PurchaseID
,T1.CustID
,T1.Location
FROM YourTable T1
INNER JOIN
(SELECT T2.CustID
,COUNT (DISTINCT T2.Location )
FROM YourTable T1
GROUP BY
T2.CustID
HAVING COUNT (DISTINCT T2.Location )>1
) SQ
ON SQ.CustID = T1.CustID

Answer 5

This should only require one full table scan.
create table test (PurchaseID number, CustID number, Location varchar2(1));
insert into test values (1,1,'A');
insert into test values (2,1,'A');
insert into test values (3,2,'A');
insert into test values (4,2,'B');
insert into test values (5,2,'A');
insert into test values (6,3,'B');
insert into test values (7,3,'A');
with repeatCustDiffLocations as (
select PurchaseID, custid, location, dense_rank () over (partition by custid order by location) r
from test)
select b.*
from repeatCustDiffLocations a, repeatCustDiffLocations b
where a.r > 1
and a.custid = b.custid;

Answer 6

This makes most sense to me as I was trying to return the rows with the same values throughout the table, specifically for two columns as shown in this stackoverflow answer here.
The answer to your problem in this format is:
SELECT DISTINCT a.*
FROM TEST a
INNER JOIN TEST b
ON a.CUSTOMERID = b.CUSTOMERID AND
a.LOCATION <> b.LOCATION;
However, the solution to a problem such as mine with two columns having matching values in multiple rows (2 in this instance, would yield no results because all PurchaseID's are unique):
SELECT DISTINCT a.*
FROM TEST a
INNER JOIN TEST b
ON a.CUSTOMERID = b.CUSTOMERID AND
a.PURCHASEID = b.PURCHASEID AND
a.LOCATION <> b.LOCATION;
Although, this wouldn't return the correct results based on the what needs to be queried, it shows that the query logic works
SELECT DISTINCT a.*
FROM TEST a
INNER JOIN TEST b
ON a.CUSTOMERID = b.CUSTOMERID AND
a.PURCHASEID <> b.PURCHASEID AND
a.LOCATION = b.LOCATION;
If anyone wants to try in Oracle here is the table and values to insert:
CREATE TABLE TEST (
PurchaseID integer,
CustomerID integer,
Location varchar(1));
INSERT ALL
INTO TEST VALUES (1, 1, 'A')
INTO TEST VALUES (2, 1, 'A')
INTO TEST VALUES (3, 2, 'A')
INTO TEST VALUES (4, 2, 'B')
INTO TEST VALUES (5, 2, 'A')
INTO TEST VALUES (6, 3, 'B')
INTO TEST VALUES (7, 3, 'B')
SELECT * FROM DUAL;