I have tables like below.
And I successed to extract T2 and I would like to join other tables and aggregate workflow.
First I would like to subtract all customers who has product = A
product customer
A A
B A
C B
B B
A C
B C
. .
. .
. .
Therefore T2is extracted from table like below.
product customer
A A
B A
A C
B C
By below code,I can get T2
WITH T2 AS (
SELECT t.*,
COUNT(CASE WHEN product = 'A' THEN 1 END) OVER (PARTITION BY customer) AS a_cnt
FROM yourTable t
)
SELECT product, customer
FROM T2
WHERE a_cnt > 0;
Further more
I would like to join T3 like
customer age
A 10
B 20
C 30
Therefore My desired result is like below
product customer age
A A 10
B A 10
A C 30
B C 30
Are there any sophisticated way to aggregate workflow?
If you have any opinion,please let me know.
Thanks
You can use EXISTS as follows:
SELECT T1.PRODUCT, T1.CUSTOMER, T3.AGE
FROM T1 T1 JOIN T3 T3 ON T1.CUSTOMER = T3.CUSTOMER
WHERE EXISTS ( SELECT 1
FROM T1 T
WHERE T.CUSTOMER = T1.CUSTOMER
AND T.PRODUCT = 'A')
Related
I have two datasets in Oracle Table1 and Table2.
When I run this:
SELECT A.ID, B.NUM_X
FROM TABLE1 A
LEFT JOIN TABLE2 B ON A.ID=B.ID
WHERE B.BOOK = 1
It returns this.
ID NUM_X
1 10
1 5
1 9
2 2
2 1
3 20
3 11
What I want are the DISTINCT ID where NUM_X is the MAX value, something like this:
ID NUM_x
1 10
2 2
3 20
You can use aggregation:
SELECT A.ID, MAX(B.NUM_X)
FROM TABLE1 A LEFT JOIN
TABLE2 B
ON A.ID = B.ID
WHERE B.BOOK = 1
GROUP BY A.ID;
If you wanted additional columns, I would recommend window functions:
SELECT A.ID, MAX(B.NUM_X)
FROM TABLE1 A LEFT JOIN
(SELECT B.*,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY NUM_X DESC) as seqnum
FROM TABLE2 B
) B
ON A.ID = B.ID AND B.seqnum = 1
WHERE B.BOOK = 1
GROUP BY A.ID;
I'd like to group by region where there are customerswho has type=a
region customer type score
A a a 1
A b b 2
A c a 3
B d c 4
B e d 5
C f a 6
C g c 7
Therefore after first step
region customer type score
A a a 1
A b b 2
A c a 3
C f a 6
C g c 7
And then I groupby in region
region sum(score)
A 6
C 13
also I'd like to extract customer whose type=a
region customer type
A a a
A c a
C f a
Then I'd like to merge above.
My desired result is like following
customer sum_in_region
a 6
c 6
f 13
Are there any way to achieve this?
My work is till the second step..
How can I proceed further?
SELECT t1.region,t1.customer, t1.type, t1.score
FROM yourTable t1
WHERE EXISTS (SELECT 1
FROM yourTable t2
WHERE t2.region = t1.region
AND t2.type = 'a');
Thanks
Join the table to a derived table that does your first two steps.
SELECT t3.customer,
x1.score
FROM yourtable t3
INNER JOIN (SELECT t1.region,
sum(score) score
FROM yourtable t1
WHERE EXISTS (SELECT *
FROM yourtable t2
WHERE t2.region = t1.region
AND t2.type = 'a')
GROUP BY t1.region) x1
ON x1.region = t3.region
WHERE t2.type = 'a';
You could use the windows functions to get your result; the first step filters for only rows where type is a, based on the region. The second step then gets the sum of scores, based again on the region, before selecting only customer and sum columns :
with filter_type_a as
(select region, customer, type, score
from
(select *,
sum(type=="a") over (partition by region) as counter
from your_table)
where counter > 0)
select customer, sum_region
from
(select customer, type,
sum(score) over (partition by region) as sum_region
from filter_type_a)
where type=="a";
You can use below query:
SQLFiddle
with country_tmp as
(SELECT t1.region,t1.customer, t1.type, t1.score
FROM country t1
WHERE EXISTS (SELECT 1
FROM country t2
WHERE t2.region = t1.region
AND t2.type = 'a'))
select y.customer, x.score from
(select a.region, sum(a.score) score from (
SELECT t1.region,t1.customer, t1.type, t1.score
FROM country_tmp t1) a
group by region) x , (SELECT t1.region,t1.customer, t1.type
FROM country_tmp t1
Where t1.type = 'a') y where x.region = y.region;
Need to join two table without having cross join between them.
The join condition need to be made on Tabl.month = Tab2.month
Input
Table1 Table2
Month ID Month ID
1 a 1 a
1 b 1 b
1 c 2 g
2 d 3 i
2 e 3 j
3 f 3 k
Output:
Month_Tab1 ID_Tab1 Month_Tab2 ID_Tab2
1 a 1 a
1 b 1 b
1 c Null Null
2 d 2 g
2 e Null Null
3 f 3 i
Null Null 3 j
Null Null 3 k
The above o/p is required, without cross join, have tried full outer but cross join is happening as the ID is duplicate in both Tables. Left/Right join also cannt be applicable as either of the table might have larger set of ID's.
You want a full join, but with row_number() to identify the matches:
select t1.month month_tab1, t1.id id_tab1, t2.month month_tab2, t2.id id_tab2
from (
select t.*, row_number() over(partition by month order by id) rn from table1 t
) t1
full join (
select t.*, row_number() over(partition by month order by id) rn from table2 t) t2
on t2.month = t1.month and t2.rn = t1.rn
You can use a full outer join:
select
a.month,
a.id,
b.month,
b.id
from (
select month, id,
row_number() over(partition by month order by id) as n
from table1
) a
full outer join (
select month, id,
row_number() over(partition by month order by id) as n
from table2
) b on b.month = a.month and b.n = a.n
order by coalesce(a.month, b.month), coalesce(a.n, b.n)
The question is:
Two tables (t1, t2)
Table t1:
SELLER | NON_SELLER
A B
A C
A D
B A
B C
B D
C A
C B
C D
D A
D B
D C
Table t2:
SELLER | COUPON | BAL
A 9 100
B 9 200
C 9 300
D 9 400
A 9.5 100
B 9.5 20
A 10 80
Using SELECT Statement to get this result:
SELLER| COUPON | SUM(BAL)
A 9 900
B 9 800
C 9 700
D 9 600
A 9.5 20
B 9.5 100
C 9.5 120
D 9.5 120
A 10 0 # !!!
B 10 80
C 10 80
D 10 80
For seller A SUM(BAL) means sum( B.BAL,C.BAL,D.BAL), for B, SUM(BAL)=SUM(A.BAL,C.BAL,D.BAL)...
Please find a way with good performance and don't use temporary table.
My solution:
Running this query will get the result but without the row "A 10 0":
select t1.seller, t2.coupon, sum(bal)
from t1, t2
where t1.non_seller = t2.seller
group by t1.seller, t2.coupon
order by t2.coupon
Please help ~~~~~~
If I understand you correctly, you're looking for data on all sellers and all coupons. So let's start with a cross join that generates a list of coupons and sellers:
select sellers.seller
, coupons.coupon
from (
select distinct seller
from Table2
) as sellers
cross join
(
select distinct coupon
from Table2
) as coupons
For each seller-coupon combination, you're looking for the sum they can buy from other sellers. This can be accomplished by a left join:
select sellers.seller
, coupons.coupon
, case when sum(t2.bal) is null then 0 else sum(t2.bal) end
from (
select distinct seller
from Table2
) as sellers
cross join
(
select distinct coupon
from Table2
) as coupons
left join
Table2 t2
on t2.seller <> sellers.seller
and t2.coupon = coupons.coupon
group by
sellers.seller
, coupons.coupon
The only function of the case statement is to replace a null sum with a 0.
The output matches the one in your answer. Note that this solution doesn't use Table1: the list of other sellers is produced by the t2.seller <> sellers.seller condition in the left join.
I get another way to this:
select t1.seller, t2.coupon, sum(bal)
from t1, t2
where t1.non_seller = t2.seller
group by t1.seller, t2.coupon
union
(select seller,coupon,0 from t2 group by coupon having count(seller) == 1);
And I don't know if it is better or worst than compare with #Andomar :
select sellers.seller
, coupons.coupon
, case when sum(t2.bal) is null then 0 else sum(t2.bal) end
from (
select distinct seller
from Table2
) as sellers
cross join
(
select distinct coupon
from Table2
) as coupons
left join
Table2 t2
on t2.seller <> sellers.seller
and t2.coupon = coupons.coupon
group by
sellers.seller
, coupons.coupon
I have 2 tables to join in a specific way. I think my query is right, but not sure.
select t1.userID, t3.Answer Unit, t5.Answer Demo
FROM
table1 t1
inner join (select * from table2) t3 ON t1.userID = t3.userID
inner join (select * from table2) t5 ON t1.userID = t5.userID
where
NOT EXISTS (SELECT * FROM table1 t2 WHERE t2.userID = t1.userID AND t2.date > t1.date)
and NOT EXISTS (SELECT * FROM table2 t4 WHERE t4.userID = t3.userID and t4.counter > t3.counter)
and NOT EXISTS (SELECT * FROM table2 t6 WHERE t6.userID = t5.userID and t6.counter > t5.counter)
and t1.date_submitted >'1/1/2009'
and t3.question = Unit
and t5.question = Demo
order by
t1.userID
From table1 I want distinct userID where date > 1/1/2009
table1
userID Date
1 1/2/2009
1 1/2/2009
2 1/2/2009
3 1/2/2009
4 1/1/2008
So The result I want from table1 should be this:
userID
1
2
3
I then want to join this on userID with table2, which looks like this:
table2
userID question answer counter
1 Unit A 1
1 Demo x 1
1 Prod 100 1
2 Unit B 1
2 Demo Y 1
3 Prod 100 1
4 Unit A 1
1 Unit B 2
1 Demo x 2
1 Prod 100 2
2 Unit B 2
2 Demo Z 2
3 Prod 100 2
4 Unit A 2
I want to join table1 with table2 with this result:
userID Unit Demo
1 B X
2 B Z
In other words,
select distinct userID from table2 where question = Unit for the highest counter
and then
select distinct userID from table2 where question = Demo for the highest counter.
I think what I've done is 3 self-joins then joined those 3 together.
Do you think it's right?
SELECT du.userID, unit.answer, demo.answer
FROM (
SELECT DISTINCT userID
FROM table1
WHERE date > '1/1/2009'
) du
LEFT JOIN
table2 unit
ON (userID, question, counter) IN
(
SELECT du.userID, 'Unit', MAX(counter)
FROM table2 td
WHERE userID = du.userID
AND question = 'Unit'
)
LEFT JOIN
table2 demo
ON (userID, question, counter) IN
(
SELECT du.userID, 'Demo', MAX(counter)
FROM table2 td
WHERE userID = du.userID
AND question = 'Demo'
)
Having an index on table2 (userID, question, counter) will greatly improve this query.
Since you mentioned SQL Server 2005, the following will be easier and more efficient:
SELECT du.userID,
(
SELECT TOP 1 answer
FROM table2 ti
WHERE ti.user = du.userID
AND ti.question = 'Unit'
ORDER BY
counter DESC
) AS unit_answer,
(
SELECT TOP 1 answer
FROM table2 ti
WHERE ti.user = du.userID
AND ti.question = 'Demo'
ORDER BY
counter DESC
) AS demo_answer
FROM (
SELECT DISTINCT userID
WHERE date > '1/1/2009'
FROM table1
) du
To aggregate:
SELECT answer, COUNT(*)
FROM (
SELECT DISTINCT userID
FROM table1
WHERE date > '1/1/2009'
) du
JOIN table2 t2
ON t2.userID = du.userID
AND t2.question = 'Unit'
GROUP BY
answer