My problem is that I would like to select some records which appears in a row.
For example we have table like this:
x
x
x
y
y
x
x
y
Query should give answer like this:
x 3
y 2
x 2
y 1
SQL tables represent unordered sets. Your question only makes sense if there is a column that specifies the ordering. If so, you can use the difference-of-row-numbers to determine the groups and then aggregate:
select col1, count(*)
from (select t.*,
row_number() over (order by <ordering col>) as seqnum,
row_number() over (partition by col1 order by <ordering col>) as seqnum_2
from t
) t
group by col1, (seqnum - seqnum_2)
I made a SQL Fiddle
http://sqlfiddle.com/#!18/f8900/5
CREATE TABLE [dbo].[SomeTable](
[data] [nchar](1) NULL,
[id] [int] IDENTITY(1,1) NOT NULL
);
INSERT INTO SomeTable
([data])
VALUES
('x'),
('x'),
('x'),
('y'),
('y'),
('x'),
('x'),
('y')
;
select * from SomeTable;
WITH SomeTable_CTE (Data, total, BaseId, NextId)
AS
(
SELECT
Data,
1 as total,
Id as BaseId,
Id+1 as NextId
FROM SomeTable
where not exists(
Select * from SomeTable Previous
where Previous.Id+1 = SomeTable.Id
and Previous.Data = SomeTable.Data)
UNION ALL
select SomeTable_CTE.Data, SomeTable_CTE.total+1, SomeTable_CTE.BaseId as BaseId, SomeTable.Id+1 as NextId
from SomeTable_CTE inner join SomeTable on
SomeTable.Data = SomeTable_CTE.Data
and
SomeTable.Id = SomeTable_CTE.NextId
)
SELECT Data, max(total) as total
FROM SomeTable_CTE
group by Data, BaseId
order by BaseId
The elephant in the room is the missing column(s) to establish the order of rows.
SELECT col1, count(*)
FROM (
SELECT col1, order_column
, row_number() OVER (ORDER BY order_column)
- row_number() OVER (PARTITION BY col1 ORDER BY order_column) AS grp
FROM tbl
) t
GROUP BY col1, grp
ORDER BY min(order_column);
To exclude partitions with only a single row, add a HAVING clause:
SELECT col1, count(*)
FROM (
SELECT col1, order_column
, row_number() OVER (ORDER BY order_column)
- row_number() OVER (PARTITION BY col1 ORDER BY order_column) AS grp
FROM tbl
) t
GROUP BY col1, grp
HAVING count(*) > 1
ORDER BY min(order_column);
db<>fiddle here
Add a final ORDER BY to maintain original order (and a meaningful result). You may want to add a column like min(order_column) as well.
Related:
Find the longest streak of perfect scores per player
Select longest continuous sequence
Group by repeating attribute
Given the following table:
Column A Column B
east red
west blue
east green
I want to find out the of column values of each column and how many times each value is present in the table. Given the output above the result should look like:
A values A value counts B values B value counts
east 2 red 1
west 1 blue 1
green 1
This is achievable by running SELECT colX, count(colX) From Table GROUP BY colX for each column. This is not a scalable solution if there is a complex WHERE condition since it needs to be executed for each query.
An alternative is to execute the complex where query once and compute the aggregations in the server code. But is there a single SQL query that can compute that?
You can use window function :
select cola, count(*) over (partition by cola) as a_count,
colb, count(*) over (partition by colb) as b_count
This will count for both columns (a & b) with their values display.
You can use subqueries to aggregate, then union all and aggregate again to combine the results:
select max(a) as a, max(a_cnt) as a_cnt, max(b) as b, max(b_cnt) as b
from ((select a, count(*) as a_cnt, null as b, null as b_cnt,
row_number() over (order by count(*) desc) as seqnum
from t
group by a
) union all
(select null, null, b, count(*),
row_number() over (order by count(*) desc) as seqnum
from t
group by b
)
) ab
group by seqnum
order by seqnum;
If you are using Oracle you can use user_tab_cols to generate the SQL for all columns in your table
SELECT 'SELECT '
|| Listagg(column_name
||',count(1) over (partition by '
||column_name
||') as '
||column_name
||'_cnt', ',')
within GROUP (ORDER BY column_id)
||' FROM '
||'TEST_DATA'
FROM user_tab_cols
WHERE table_name = 'TEST_DATA'
Sample output is below
SELECT ID,count(1) over (partition by ID) as ID_cnt,VALUE,count(1) over (partition by
VALUE) as VALUE_cnt FROM TEST_DATA
I have ROW_NUMBER() OVER (ORDER BY NULL) rnum in a sql statement to give me row numbers. is there a way to attach the max rnum to the same dataset going out?
what I want is the row_number() which I get, but I also want the MAXIMUM rownumber of the total return on each record.
SELECT
ROW_NUMBER() OVER (ORDER BY NULL) rnum,
C.ID, C.FIELD1, C."NAME", C.FIELD2, C.FIELD3
FROM SCHEMA.TABLE
WHERE (C.IS_CRNT = 1)
), MAX_NUM as (
SELECT DATA.ID, max(rnum) as maxrnum from DATA GROUP BY DATA.COMPONENT_ID
) select maxrnum, DATA.* from DATA JOIN MAX_NUM on DATA.COMPONENT_ID = MAX_NUM.COMPONENT_ID
DESIRED RESULT (ASSUMING 15 records):
1 15 DATA
2 15 DATA
3 15 DATA
etc...
I think you want count(*) as a window function:
SELECT ROW_NUMBER() OVER (ORDER BY NULL) as rnum,
COUNT(*) OVER () as cnt,
C.ID, C.FIELD1, C."NAME", C.FIELD2, C.FIELD3
FROM SCHEMA.TABLE
WHERE C.IS_CRNT = 1
Based on my assumptions in your dataset, this is the approach I would take:
WITH CTE AS (
select C.ID, C.FIELD1, C."NAME", C.FIELD2, C.FIELD3, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY ID)
FROM SCHEMA.TABLE WHERE (C.IS_CRNT = 1))
SELECT *, (select count(*) from cte) "count" from cte;
I have data like below -
Year,winning_country
2001,IND
2002,IND
2003,IND
2004,AUS
2005,AUS
2006,SA
2007,SA
2008,SA
2009,IND
2010,IND
2011,IND
2012,IND
2013,AUS
2014,AUS
2015,SA
2016,NZ
2017,SL
2018,IND
The question here is to find out the longest streak of wins for each country and desired output will be like below -
Country,no_of_wins
IND,4
AUS,2
SA,3
SL,1
NZ,1
Can someone help here.
This is a gaps and islands problem, but the simplest method is to subtract a sequence from the year. So, to get all the sequences:
select country, count(*) as streak,
min(year) as from_year, max(year) as to_year
from (select year, country,
row_number() over (partition by country order by year) as seqnum
from t
) t
group by country, (year - seqnum);
To get the longest per country, aggregate again or use window functions:
select country, streak
from (select country, count(*) as streak,
min(year) as from_year, max(year) as to_year,
row_number() over (partition by country order by count(*) desc) as seqnum_2
from (select year, country,
row_number() over (partition by country order by year) as seqnum
from t
) t
group by country, (year - seqnum)
) cy
where seqnum_2 = 1;
I prefer using row_number() to get the longest streak because it allows you to also get the years when it occurred.
Looks like an gaps-and-islands problem.
The SQL below calculates some ranking based on 2 row_number.
Then it's just a matter of grouping.
SELECT q2.Country, MAX(q2.no_of_wins) AS no_of_wins
FROM
(
SELECT q1.winning_country as Country,
COUNT(*) AS no_of_wins
FROM
(
SELECT t.Year, t.winning_country,
(ROW_NUMBER() OVER (ORDER BY t.Year ASC) -
ROW_NUMBER() OVER (PARTITION BY t.winning_country ORDER BY t.Year)) AS rnk
FROM yourtable t
) q1
GROUP BY q1.winning_country, q1.rnk
) q2
GROUP BY q2.Country
ORDER BY MAX(q2.no_of_wins) DESC
If Redshift supports analytic function, below would be the query.
with t1 as
(
select 2001 as year,'IND' as cntry from dual union
select 2002,'IND' from dual union
select 2003,'IND' from dual union
select 2004,'AUS' from dual union
select 2005,'AUS' from dual union
select 2006,'SA' from dual union
select 2007,'SA' from dual union
select 2008,'SA' from dual union
select 2009,'IND' from dual union
select 2010,'IND' from dual union
select 2011,'IND' from dual union
select 2012,'IND' from dual union
select 2013,'AUS' from dual union
select 2014,'AUS' from dual union
select 2015,'SA' from dual union
select 2016,'NZ' from dual union
select 2017,'SL' from dual union
select 2018,'IND' from dual) ,
t2 as (select year, cntry, year - row_number() over (partition by cntry order by year) as grpBy from t1 order by cntry),
t3 as (select cntry, count(grpBy) as consWins from t2 group by cntry, grpBy),
res as (select cntry, consWins, row_number() over (partition by cntry order by consWins desc) as rnk from t3)
select cntry, consWins from res where rnk=1;
Hope this helps.
Here is a solution that leverages the use of Redshift Python UDF's
There may be simpler ways to achieve the same but this is a good example of how to create a simple UDF.
create table temp_c (competition_year int ,winning_country varchar(4));
insert into temp_c (competition_year, winning_country)
values
(2001,'IND'),
(2002,'IND'),
(2003,'IND'),
(2004,'AUS'),
(2005,'AUS'),
(2006,'SA'),
(2007,'SA'),
(2008,'SA'),
(2009,'IND'),
(2010,'IND'),
(2011,'IND'),
(2012,'IND'),
(2013,'AUS'),
(2014,'AUS'),
(2015,'SA'),
(2016,'NZ'),
(2017,'SL'),
(2018,'IND')
;
create or replace function find_longest_streak(InputStr varChar)
returns integer
stable
as $$
MaxStreak=0
ThisStreak=0
ThisYearStr=''
LastYear=0
for ThisYearStr in InputStr.split(','):
if int(ThisYearStr) == LastYear + 1:
ThisStreak+=1
else:
if ThisStreak > MaxStreak:
MaxStreak=ThisStreak
ThisStreak=1
LastYear=int(ThisYearStr)
return max(MaxStreak,1)
$$ language plpythonu;
select winning_country,
find_longest_streak(listagg(competition_year,',') within group (order by competition_year))
from temp_c
group by winning_country
order by 2 desc
;
How about something like...
SELECT
winning_country,
COUNT(*)
GROUP BY winning_country
HAVING MAX(year) - MIN(year) = COUNT(year) - 1
This assumes no duplicate entries.
Creating a session abstraction do the trick:
WITH winning_changes AS (
SELECT *,
CASE WHEN LAG(winning_country) OVER (ORDER BY year) <> winning_country THEN 1 ELSE 0 END AS same_winner
FROM winners
),
sequences AS (
SELECT *,
SUM(same_winner) OVER (ORDER BY year) AS winning_session
FROM winning_changes
),
streaks AS (
SELECT winning_country AS country,
winning_session,
COUNT(*) streak
FROM sequences
GROUP BY 1,2
)
SELECT country,
MAX(streak) AS no_of_wins
FROM streaks
GROUP BY 1;
For the below table
ID DATE FIELD1 FIELD2 FIELD4
123456 01.07.2014 00:00:00 EMPLOYER GROUPS TMC SELECT CARE HMO
123789 01.07.2017 00:00:00 EMPLOYER GROUPS MQC SELECT CARE HMO
How to select only one row that have max(date)? i.e. 01.07.2017
select *
from theTable
where Date = (select max(date) from theTable)
Should do it. Add a top 1 if multiple rows have the same date.
In Oracle 12+, you can use:
select t.*
from t
order by date desc
fetch first 1 row only;
In earlier versions, you an use a subquery:
select t.*
from (select t.*
from t
order by date desc
) t
where rownum = 1;
If you need more than one record with the same maximum date:
select t.*
from t
where t.date = (select max(t2.date) from t);
For tsql / SQL Server you can use the below
DECLARE #max_date datetime;
SELECT #max_date = max(DATE) from table_name;
SELECT TOP 1 * FROM table_name WHERE DATE = #max_date;
The 'TOP 1' makes sure you only receive 1 row.
Which row this is will arbitrary though, unless you add an 'ORDER BY' statement in your query.
Alternatively
dense_rank() function may be used as :
select dt, ID
from
(
select dense_rank() over (order by "Date" desc) as dr,
"Date" as dt,
ID
from tab
)
where dr = 1;
or if you're sure about there's no tie(for the column "Date"), even row_number() function may be used :
select dt, ID
from
(
select row_number() over (order by "Date" desc) as rn,
"Date" as dt,
ID
from tab
)
where rn = 1;
SQL Fiddle Demo