I need a solution for the below senario
I have a table temp with columns: a, b, c, d and the data looks like this:
TABLE TEMP
+---+----+----+----+
|a | b | c | d |
+===+====+====+====+
| 1 | 1 | 1 | m |
| 1 | 2 | 1 | d |
| 1 | 3 | 1 | w |
| 2 | 1 | 1 | m |
| 2 | 2 | 1 | d |
| 2 | 2 | 1 | w |
+---+----+----+----+
QUERY
SELECT CASE WHEN B=1 AND C=1 THEN D END as T1,
CASE WHEN B=2 AND C=1 THEN D END as T2,
CASE WHEN B=3 AND C=1 THEN D END as T3
FROM TEMP
WHERE A=1
The above query gives multiple rows with null values where value is not present
I need a result set with a single row that looks like this:
Expected Result
+------+-------+------+
| T1 | T2 | T3 |
+======+=======+======+
| m | d | w |
+------+-------+------+
Do like this (using CTE)
QUERY
WITH
CTE1 as (select top 1 d as T1 from temp where b=1 and c=1),
CTE2 as (select top 1 d as T2 from temp where b=2 and c=1),
CTE3 as (select top 1 d as T3 from temp where b=3 and c=1)
select CTE1.*, CTE2.*, CTE3.*
FROM CTE1 CROSS JOIN CTE2 CROSS JOIN CTE3
SQL fiddle
About the multiple CTE
Please let me whether it works!
Related
I have table like this
|-------------------------|
| A | B | C |
|-------------------------|
| 1 | 2 | 5 |
|-------------------------|
| 1 | 2 | 10 |
|-------------------------|
| 1 | 2 | 2 |
|-------------------------|
I need to delete all duplicated rows with equals A nad B value and lower C value
after running sql script i need to have only this row with top C Value for every equals A and B columns
|-------------------------|
| A | B | V |
|-------------------------|
| 1 | 2 | 10 |
|-------------------------|
One method is window functions:
select t.*
from (select t.*,
row_number() over (partition by a, b order by v desc) as seqnum
from t
) t
where seqnum = 1;
This returns the entire row, which can be handy if you want additional columns. If you really need just the three columns, then aggregation does what you want:
select a, b, max(v)
from t
group by a, b;
In standard SQL, you can keep only the maximum value using:
delete from t
where t.v < (select max(t2.v) from t t2 where t2.a = t.a and t2.b = t.b);
I have 4 columns a ,b ,c, d
sample data
a | b | c | d |
1 | 1 | 101 | 0
2 | 1 | 101 | 0
3 | 1 | 101 | 1
4 | 1 | 102 | 0
5 | 1 | 102 | 0
1 | 2 | 101 | 0
2 | 2 | 101 | 1
Write a SQL command such that it should return those rows where for every value of c in b, return rows with maximum a
i.e
Expect output
a | b | c | d |
3 | 1 | 101 | 1
5 | 1 | 102 | 0
2 | 2 | 101 | 1
You can use a correlated subquery:
select t.*
from t
where t.a = (select max(t2.a) from t t2 where t2.b = t.b and t2.c = t.c);
With an index on t(b, c, a), this often has the best performance.
An alternative is window functions:
select t.*
from (select t.*, row_number() over (partition by b, c order by a desc) as seqnum
from t
) t
where seqnum = 1;
You don't mention the database you are using. In PostgreSQL you can do:
select distinct on (b, c) a, b, c, d
from t
order by b, c, a desc
Yesterday, after the help of a SO user #
Iterate over the rows of a second table to return resultset
I was able to make a combination of rows with a selfjoin.
After some modifications, to adapt to my implementation, I faced a new challenge that I'm stuck: how to make an aggregate sum of a third column?
My issue is better explained in the image below:
Based on the code
SELECT
b1.table_a_id,
b1.label_x,
b2.label_y
FROM table_a a
INNER JOIN table_b b1
ON b1.table_a_id = a.table_a_id
INNER JOIN table_b b2
ON b2.table_a_id = b1.table_a_id AND
b2.label_y > b1.label_x
ORDER BY
b1.table_a_id,
b1.label_x,
b2.label_y;
I was able to acquire the combinations.
What should be the next step to get the cumulative sum based on a third column?
I couldn't think of a solution without using a second service, such as python with pandas, using a cumsum function.
To generate the expected resultset, you would need to join the table with itself with an inequality condition on the order column. Then, you can do a window sum:
select
t1.table_a_id,
t1.label_x,
t2.label_y,
sum(t2.value) over(
partition by t1.table_a_id, t1.label_x
order by t1."order", t2."order"
) agg_value
from
table_b t1
inner join table_b t2
on t1.table_a_id = t2.table_a_id
and t2."order" >= t1."order"
order by t1."order", t2."order"
Note: order is a reserved word, so it needs to be quoted; if you actual database column has a different name, you can remove the double quotes.
Demo on DB Fiddle:
TABLE_A_ID | LABEL_X | LABEL_Y | AGG_VALUE
---------: | :------ | :------ | --------:
1 | A | B | 1
1 | A | C | 3
1 | A | D | 6
1 | A | E | 10
1 | A | F | 15
1 | B | C | 2
1 | B | D | 5
1 | B | E | 9
1 | B | F | 14
1 | C | D | 3
1 | C | E | 7
1 | C | F | 12
1 | D | E | 4
1 | D | F | 9
1 | E | F | 5
You seem to want a cumulative sum:
SELECT b1.table_a_id, b1.label_x, b2.label_y,
SUM(b1.value) OVER (PARTITION BY b1.table_a_id, b1.label_x
ORDER BY b2.order
) as AGG_VALUE
I have the following table
session_id | page_viewed
1 | A
1 | B
1 | C
2 | B
2 | E
What I would like to do is a cross join of the page_viewed column with itself but where the cross join is done on the partitions from session_id. So, from the table above the query would return:
session_id | page_1 | page_2
1 | A | A
1 | A | B
1 | A | C
1 | B | A
1 | B | B
1 | B | C
1 | C | A
1 | C | B
1 | C | C
2 | B | B
2 | B | E
2 | E | B
2 | E | E
I have looked into window functions today trying to find a way around it but it seems join functions cannot be used. Can anyone help?
You may join giving only the session_id as the join criteria:
SELECT
t1.session_id,
t1.page_viewed AS page_1,
t2.page_viewed AS page_2
FROM yourTable t1
INNER JOIN yourTable t2
ON t1.session_id = t2.session_id;
-- ORDER BY clause optional, if you need it here
Demo
Hmmm . . . you seem to want a self-join:
select t1.session_id, t1.page_viewed as page_1, t2.page_viewed as page_2
from t t1 join
t t2
on t1.session_id = t2.session_id
order by t1.session_id, t1.page_viewed, t2.page_viewed;
I create the following table on http://sqlfiddle.com in PostgreSQL 9.3.1 mode:
CREATE TABLE t
(
id serial primary key,
m varchar(1),
d varchar(1),
c int
);
INSERT INTO t
(m, d, c)
VALUES
('A', '1', 101),
('A', '2', 102),
('A', '3', 103),
('B', '1', 104),
('B', '3', 105);
table:
| ID | M | D | C |
|----|---|---|-----|
| 1 | A | 1 | 101 |
| 2 | A | 2 | 102 |
| 3 | A | 3 | 103 |
| 4 | B | 1 | 104 |
| 5 | B | 3 | 105 |
From this I want to generate such a table:
| M | D | ID | C |
|---|---|--------|--------|
| A | 1 | 1 | 101 |
| A | 2 | 2 | 102 |
| A | 3 | 3 | 103 |
| B | 1 | 4 | 104 |
| B | 2 | (null) | (null) |
| B | 3 | 5 | 105 |
but with my current statement
select * from
(select * from
(select distinct m from t) as dummy1,
(select distinct d from t) as dummy2) as combi
full outer join
t
on combi.d = t.d and combi.m = t.m
I only get the following
| M | D | ID | C |
|---|---|--------|--------|
| A | 1 | 1 | 101 |
| B | 1 | 4 | 104 |
| A | 2 | 2 | 102 |
| A | 3 | 3 | 103 |
| B | 3 | 5 | 105 |
| B | 2 | (null) | (null) |
Attempts to order it by m,d fail so far:
select * from
(select * from
(select * from
(select * from
(select distinct m from t) as dummy1,
(select distinct d from t) as dummy2) as kombi
full outer join
t
on kombi.d = t.d and kombi.m = t.m) as result)
order by result.m
Error message:
ERROR: subquery in FROM must have an alias: select * from (select * from (select * from (select * from (select distinct m from t) as dummy1, (select distinct d from t) as dummy2) as kombi full outer join t on kombi.d = t.d and kombi.m = t.m) as result) order by result.m
It would be cool if somebody could point out to me what I am doing wrong and perhaps show the correct statement.
select * from
(select kombi.m, kombi.d, t.id, t.c from
(select * from
(select distinct m from t) as dummy1,
(select distinct d from t) as dummy2) as kombi
full outer join t
on kombi.d = t.d and kombi.m = t.m) as result
order by result.m, result.d
I think your problem is the order. You can solve this problem with the order by clause:
select * from
(select * from
(select distinct m from t) as dummy1,
(select distinct d from t) as dummy2) as combi
full outer join
t
on combi.d = t.d and combi.m = t.m
order by combi.m, combi.d
You need to specify which data you would like to order. In this case you get back the row from the combi table, so you need to say that.
http://sqlfiddle.com/#!15/ddc0e/17
You could also use column numbers instead of names to do the ordering.
select * from
(select * from
(select distinct m from t) as dummy1,
(select distinct d from t) as dummy2) as combi
full outer join
t
on combi.d = t.d and combi.m = t.m
order by 1,2;
| M | D | ID | C |
|---|---|--------|--------|
| A | 1 | 1 | 101 |
| A | 2 | 2 | 102 |
| A | 3 | 3 | 103 |
| B | 1 | 4 | 104 |
| B | 2 | (null) | (null) |
| B | 3 | 5 | 105 |
you just need a pivot table
the query is very simple
select classes.M, p.i as D, t.ID, t.C
from (select M, max(D) MaxValue from t group by m) classes
inner join pivot p
on p.i =< classes.MaxValue
left join t
on t.M = classes.M
and t.D = p.i
pivot table is a dummy table some how
CREATE TABLE Pivot (
i INT,
PRIMARY KEY(i)
)
populate is some how
CREATE TABLE Foo(
i CHAR(1)
)
INSERT INTO Foo VALUES('0')
INSERT INTO Foo VALUES('1')
INSERT INTO Foo VALUES('2')
INSERT INTO Foo VALUES('3')
INSERT INTO Foo VALUES('4')
INSERT INTO Foo VALUES('5')
INSERT INTO Foo VALUES('6')
INSERT INTO Foo VALUES('7')
INSERT INTO Foo VALUES('8')
INSERT INTO Foo VALUES('9')
Using the 10 rows in the Foo table, you can easily populate the Pivot table with 1,000 rows. To get 1,000 rows from 10 rows, join Foo to itself three times to create a Cartesian product:
INSERT INTO Pivot
SELECT f1.i+f2.i+f3.i
FROM Foo f1, Foo F2, Foo f3
you can read about that in Transac-SQL Cookbook by Jonathan Gennick, Ales Spetic
You just need to order by the final column definitions. t.m and t.d. SO your final SQL would be...
SELECT *
FROM (SELECT *
FROM (SELECT DISTINCT m FROM t) AS dummy1,
(SELECT DISTINCT d FROM t) AS dummy2) AS combi
FULL OUTER JOIN t
ON combi.d = t.d
AND combi.m = t.m
ORDER BY t.m,
t.d;
Also for query optimization perspective, it is better to now have many layers of sub queries.
I think you need another correlation name - dummy3? - after 'as result )' before the order by.