I have a table as below
Id | PriceCardId | Days
1 | 1 | 2
2 | 1 | 4
3 | 1 | 5
4 | 1 | 6
5 | 1 | 3
6 | 2 | 5
7 | 2 | 3
8 | 3 | 6
How to write SQL query to get all PriceCardId has Day contain
List<int> days
Example days = [2,4,5,6], with data as above result is 1
Thanks!
I think you want:
select pricecardid
from t
where day in (2, 4, 5, 6)
group by pricecardid
having count(*) = 4; -- the number of days you are looking for
This assumes no duplicates in your table. If there are duplicates, use having count(distinct day) instead of count(*).
Note: You can phrase this as:
with d as (
select v.dy
from values ( (2), (4), (5), (6) ) v(dy)
)
select pricecardid
from t
where day in (select d.dy from d)
group by pricecardid
having count(*) = (select count(*) from d);
Try this:
SELECT PriceCardId
FROM [My_Table]
WHERE [Day] IN(2,4,5,6)
GROUP BY PriceCardId
HAVING COUNT(DISTINCT [Day])=4
Related
My table has account_id and device_id. One account_id could have multiple device_ids and vice versa. I am trying to count the depth of each connected many-to-many relationship.
Ex:
account_id | device_id
1 | 10
1 | 11
1 | 12
2 | 10
3 | 11
3 | 13
3 | 14
4 | 15
5 | 15
6 | 16
How do I construct a query that knows to combine accounts 1-3 together, 4-5 together, and leave 6 by itself? All 7 entries of accounts 1-3 should be grouped together because they all touched the same account_id or device_id at some point. I am trying to group them together and output the count.
Account 1 was used on device's 10, 11, 12. Those devices used other accounts too so we want to include them in the group. They used additional accounts 2 and 3. But account 3 was further used by 2 more devices so we will include them as well. The expansion of the group brings in any other account or device that also "touched" an account or device already in the group.
A diagram is shown below:
You can use a recursive cte:
with recursive t(account_id, device_id) as (
select 1, 10 union all
select 1, 11 union all
select 1, 12 union all
select 2, 10 union all
select 3, 11 union all
select 3, 13 union all
select 3, 14 union all
select 4, 15 union all
select 5, 15 union all
select 6, 16
),
a as (
select distinct t.account_id as a, t2.account_id as a2
from t join
t t2
on t2.device_id = t.device_id and t.account_id >= t2.account_id
),
cte as (
select a.a, a.a2 as mina
from a
union all
select a.a, cte.a
from cte join
a
on a.a2 = cte.a and a.a > cte.a
)
select grp, array_agg(a)
from (select a, min(mina) as grp
from cte
group by a
) a
group by grp;
Here is a SQL Fiddle.
You can GROUP BY the device_id and then aggregate together the account_id into a Postgres array. Here is an example query, although I'm not sure what your actual table name is.
SELECT
device_id,
array_agg(account_id) as account_ids
FROM account_device --or whatever your table name is
GROUP BY device_id;
Results - hope it's what you're looking for:
16 | {6}
15 | {4,5}
13 | {3}
11 | {1,3}
14 | {3}
12 | {1}
10 | {1,2}
-- \i tmp.sql
CREATE TABLE graph(
account_id integer NOT NULL --references accounts(id)
, device_id integer not NULL --references(devices(id)
,PRIMARY KEY(account_id, device_id)
);
INSERT INTO graph (account_id, device_id)VALUES
(1,10) ,(1,11) ,(1,12)
,(2,10)
,(3,11) ,(3,13) ,(3,14)
,(4,15)
,(5,15)
,(6,16)
;
-- SELECT* FROM graph ;
-- Find the (3) group leaders
WITH seed AS (
SELECT row_number() OVER () AS cluster_id -- give them a number
, g.account_id
, g.device_id
FROM graph g
WHERE NOT EXISTS (SELECT*
FROM graph nx
WHERE (nx.account_id = g.account_id OR nx.device_id = g.device_id)
AND nx.ctid < g.ctid
)
)
-- SELECT * FROM seed;
;
WITH recursive omg AS (
--use the above CTE in a sub-CTE
WITH seed AS (
SELECT row_number()OVER () AS cluster_id
, g.account_id
, g.device_id
, g.ctid AS wtf --we need an (ordered!) canonical id for the tuples
-- ,just to identify and exclude them
FROM graph g
WHERE NOT EXISTS (SELECT*
FROM graph nx
WHERE (nx.account_id = g.account_id OR nx.device_id = g.device_id) AND nx.ctid < g.ctid
)
)
SELECT s.cluster_id
, s.account_id
, s.device_id
, s.wtf
FROM seed s
UNION ALL
SELECT o.cluster_id
, g.account_id
, g.device_id
, g.ctid AS wtf
FROM omg o
JOIN graph g ON (g.account_id = o.account_id OR g.device_id = o.device_id)
-- AND (g.account_id > o.account_id OR g.device_id > o.device_id)
AND g.ctid > o.wtf
-- we still need to exclude duplicates here
-- (which could occur if there are cycles in the graph)
-- , this could be done using an array
)
SELECT *
FROM omg
ORDER BY cluster_id, account_id,device_id
;
Results:
DROP SCHEMA
CREATE SCHEMA
SET
CREATE TABLE
INSERT 0 10
cluster_id | account_id | device_id
------------+------------+-----------
1 | 1 | 10
2 | 4 | 15
3 | 6 | 16
(3 rows)
cluster_id | account_id | device_id | wtf
------------+------------+-----------+--------
1 | 1 | 10 | (0,1)
1 | 1 | 11 | (0,2)
1 | 1 | 12 | (0,3)
1 | 1 | 12 | (0,3)
1 | 2 | 10 | (0,4)
1 | 3 | 11 | (0,5)
1 | 3 | 13 | (0,6)
1 | 3 | 14 | (0,7)
1 | 3 | 14 | (0,7)
2 | 4 | 15 | (0,8)
2 | 5 | 15 | (0,9)
3 | 6 | 16 | (0,10)
(12 rows)
Newer version (I added an Id column to the table)
-- for convenience :set of all adjacent nodes.
CREATE TEMP VIEW pair AS
SELECT one.id AS one
, two.id AS two
FROM graph one
JOIN graph two ON (one.account_id = two.account_id OR one.device_id = two.device_id)
AND one.id <> two.id
;
WITH RECURSIVE flood AS (
SELECT g.id, g.id AS parent_id
, 0 AS lev
, ARRAY[g.id]AS arr
FROM graph g
UNION ALL
SELECT c.id, p.parent_id AS parent_id
, 1+p.lev AS lev
, p.arr || ARRAY[c.id] AS arr
FROM graph c
JOIN flood p ON EXISTS (
SELECT * FROM pair WHERE p.id = pair.one AND c.id = pair.two)
AND p.parent_id < c.id
AND NOT p.arr #> ARRAY[c.id] -- avoid cycles/loops
)
SELECT g.*, a.parent_id
, dense_rank() over (ORDER by a.parent_id)AS group_id
FROM graph g
JOIN (SELECT id, MIN(parent_id) AS parent_id
FROM flood
GROUP BY id
) a
ON g.id = a.id
ORDER BY a.parent_id, g.id
;
New results:
CREATE VIEW
id | account_id | device_id | parent_id | group_id
----+------------+-----------+-----------+----------
1 | 1 | 10 | 1 | 1
2 | 1 | 11 | 1 | 1
3 | 1 | 12 | 1 | 1
4 | 2 | 10 | 1 | 1
5 | 3 | 11 | 1 | 1
6 | 3 | 13 | 1 | 1
7 | 3 | 14 | 1 | 1
8 | 4 | 15 | 8 | 2
9 | 5 | 15 | 8 | 2
10 | 6 | 16 | 10 | 3
(10 rows)
I've a table with 2 columns in SQL
+------+--------+
| WEEK | OUTPUT |
+------+--------+
| 1 | 10 |
| 2 | 20 |
| 3 | 30 |
| 4 | 40 |
| 5 | 50 |
| 6 | 50 |
+------+--------+
How do I calculate to sum up output for 2 weeks before (ex : on week 3, it will sum up the output for week 3, 2 and 1), I've seen many tutorials to do moving average but they are using date, in my case i want to use (int), is that possible ?.
Thanks !.
I think you want something like this :
SELECT *,
(SELECT Sum(output)
FROM table1 b
WHERE b.week IN( a.week, a.week - 1, a.week - 2 )) AS SUM
FROM table1 a
OR
In clause can be converted to between a.week-2 and a.week.
sql fiddle
You can use a self-join. The idea is to put you table beside itself with a condition that brings matching rows in a single row:
SELECT * FROM [output] o1
INNER JOIN [output] o2 ON o1.Week between o2.Week and o2.Week + 2
this select will produce this output:
o1.Week o1.Output o2.Week o2.Output
--------------------------------------------
1 10 1 10
2 20 1 10
2 20 2 20
3 30 1 10
3 30 2 20
3 30 3 30
4 40 2 20
4 40 3 30
4 40 4 40
and so on. Note that for weeks 1 and 2 there aren't previous weeks available.
Now you should just group the data by o1.Week and get the SUM:
SELECT o1.Week, SUM(o2.Output)
FROM [output] o1
INNER JOIN [output] o2 ON o1.Week between o2.Week and o2.Week + 2
GROUP BY o1.Week
If week is continuous, you can simply use Window function
SELECT [Week], [Output],
SUM([Output]) OVER (ORDER BY [Week] ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
FROM dbo.SomeTable
Range is more accurate for your calculation, but it not implemented in SQL Server yet. Other database engines may support
SELECT [Week], [Output],
SUM([Output]) OVER (ORDER BY [Week] RANGE BETWEEN 2 PRECEDING AND CURRENT ROW)
FROM dbo.SomeTable
Try this:
SELECT SUM(t1.output) / 3
FROM yourtable t1
WHERE t1.week <=
(select t2.week from yourtable t2 where t2.week - t1.week > 0 and t2.week - t1.week <= 2)
You are not written your sqlserver, if it is sqlserver2012 or above , then the simple example is
declare #table table(wk int,outpt int )
insert into #table values (1,10)
,(2,20)
,(3,30)
,(4,40)
,(5,50)
,(6,60)
select *,SUM(outpt) over(partition by id order by id rows between unbounded preceding and current row ) dd
from (
select * , 1 id
from #table
where wk < 5
) a
I want to create an additional column which calculates the value of a row from count column with its predecessor row from the sum column. Below is the query. I tried using ROLLUP but it does not serve the purpose.
select to_char(register_date,'YYYY-MM') as "registered_in_month"
,count(*) as Total_count
from CMSS.USERS_PROFILE a
where a.pcms_db != '*'
group by (to_char(register_date,'YYYY-MM'))
order by to_char(register_date,'YYYY-MM')
This is what i get
registered_in_month TOTAL_COUNT
-------------------------------------
2005-01 1
2005-02 3
2005-04 8
2005-06 4
But what I would like to display is below, including the months which have count as 0
registered_in_month TOTAL_COUNT SUM
------------------------------------------
2005-01 1 1
2005-02 3 4
2005-03 0 4
2005-04 8 12
2005-05 0 12
2005-06 4 16
To include missing months in your result, first you need to have complete list of months. To do that you should find the earliest and latest month and then use heirarchial
query to generate the complete list.
SQL Fiddle
with x(min_date, max_date) as (
select min(trunc(register_date,'month')),
max(trunc(register_date,'month'))
from users_profile
)
select add_months(min_date,level-1)
from x
connect by add_months(min_date,level-1) <= max_date;
Once you have all the months, you can outer join it to your table. To get the cumulative sum, simply add up the count using SUM as analytical function.
with x(min_date, max_date) as (
select min(trunc(register_date,'month')),
max(trunc(register_date,'month'))
from users_profile
),
y(all_months) as (
select add_months(min_date,level-1)
from x
connect by add_months(min_date,level-1) <= max_date
)
select to_char(a.all_months,'yyyy-mm') registered_in_month,
count(b.register_date) total_count,
sum(count(b.register_date)) over (order by a.all_months) "sum"
from y a left outer join users_profile b
on a.all_months = trunc(b.register_date,'month')
group by a.all_months
order by a.all_months;
Output:
| REGISTERED_IN_MONTH | TOTAL_COUNT | SUM |
|---------------------|-------------|-----|
| 2005-01 | 1 | 1 |
| 2005-02 | 3 | 4 |
| 2005-03 | 0 | 4 |
| 2005-04 | 8 | 12 |
| 2005-05 | 0 | 12 |
| 2005-06 | 4 | 16 |
Let's say I have the following table:
create temp table test (id serial, number integer);
insert into test (number)
values (5), (4), (3), (2), (1), (0);
If I sort by number descending, I get:
select * from test order by number desc;
id | number
---+--------
1 | 5
2 | 4
3 | 3
4 | 2
5 | 1
6 | 0
If I sort by number ascending, I get:
select * from test order by number asc;
6 | 0
5 | 1
4 | 2
3 | 3
2 | 4
1 | 5
How do I stripe the order so that it alternates between ascending and descending per row?
for example:
6 | 0 or 1 | 5
1 | 5 6 | 0
5 | 1 2 | 4
2 | 4 5 | 1
4 | 2 3 | 3
3 | 3 4 | 2
Update
WITH x AS (
SELECT *
, row_number() OVER (ORDER BY number) rn_up
, row_number() OVER (ORDER BY number DESC) rn_down
FROM test
)
SELECT id, number
FROM x
ORDER BY LEAST(rn_up, rn_down), number;
Or:
...
ORDER BY LEAST(rn_up, rn_down), number DESC;
to start with the bigger number.
I had two CTE at first, but one is enough - simpler and faster.
Or like this (similar to the already given answer but slightly shorter :)
WITH x AS (
SELECT *, row_number() OVER (ORDER BY number) rn, count(*) over () as c
FROM test
)
SELECT id, number
FROM x
ORDER BY ABS((c + 1.5) / 2 - rn) DESC;
If the reverse order is needed then it should be
ORDER BY ABS((c + 0.5) / 2 - rn) DESC;
I'm trying to find out the average number of times a value appears in a column, group it based on another column and then perform a calculation on it.
I have 3 tables a little like this
DVD
ID | NAME
1 | 1
2 | 1
3 | 2
4 | 3
COPY
ID | DVDID
1 | 1
2 | 1
3 | 2
4 | 3
5 | 1
LOAN
ID | DVDID | COPYID
1 | 1 | 1
2 | 1 | 2
3 | 2 | 3
4 | 3 | 4
5 | 1 | 5
6 | 1 | 5
7 | 1 | 5
8 | 1 | 2
etc
Basically, I'm trying to find all the copy ids that appear in the loan table LESS times than the average number of times for all copies of that DVD.
So in the example above, copy 5 of dvd 1 appears 3 times, copy 2 twice and copy 1 once so the average for that DVD is 2. I want to list all the copies of that (and each other) dvd that appear less than that number in the Loan table.
I hope that makes a bit more sense...
Thanks
Similar to dotjoe's solution, but using an analytic function to avoid the extra join. May be more or less efficient.
with
loan_copy_total as
(
select dvdid, copyid, count(*) as cnt
from loan
group by dvdid, copyid
),
loan_copy_avg as
(
select dvdid, copyid, cnt, avg(cnt) over (partition by dvdid) as copy_avg
from loan_copy_total
)
select *
from loan_copy_avg lca
where cnt <= copy_avg;
This should work in Oracle:
create view dvd_count_view
select dvdid, count(1) as howmanytimes
from loans
group by dvdid;
select avg(howmanytimes) from dvd_count_view;
Untested...
with
loan_copy_total as
(
select dvdid, copyid, count(*) as cnt
from loan
group by dvdid, copyid
),
loan_copy_avg as
(
select dvdid, avg(cnt) as copy_avg
from loan_copy_total
group by dvdid
)
select lct.*, lca.copy_avg
from loan_copy_avg lca
inner join loan_copy_total lct on lca.dvdid = lct.dvdid
and lct.cnt <= lca.copy_avg;