PARTITION BY segregate sets, this enables you to be able to work(ROW_NUMBER(),COUNT(),SUM(),etc) on related set independently. In your query, the related set comprised of rows with similar cdt.country_code, cdt.account, cdt.currency. When you partition on those columns and you apply ROW_NUMBER on them. Those other columns on those combination/set will receive sequential number from ROW_NUMBER But … Read more
You can try the following: * Updated ** — Test Data DECLARE @YourTable TABLE(Product INT, Timestamp DATETIME, Price NUMERIC(16,4)) INSERT INTO @YourTable SELECT 5678, ‘20080101 12:00:00’, 12.34 UNION ALL SELECT 5678, ‘20080101 12:01:00’, NULL UNION ALL SELECT 5678, ‘20080101 12:02:00’, NULL UNION ALL SELECT 5678, ‘20080101 12:03:00’, 23.45 UNION ALL SELECT 5678, ‘20080101 12:04:00’, NULL … Read more
SELECT MIN(MinuteBar) AS MinuteBar5, Opening, MAX(High) AS High, MIN(Low) AS Low, Closing, Interval FROM ( SELECT FIRST_VALUE([Open]) OVER (PARTITION BY DATEDIFF(MINUTE, ‘2015-01-01 00:00:00’, MinuteBar) / 5 ORDER BY MinuteBar) AS Opening, FIRST_VALUE([Close]) OVER (PARTITION BY DATEDIFF(MINUTE, ‘2015-01-01 00:00:00’, MinuteBar) / 5 ORDER BY MinuteBar DESC) AS Closing, DATEDIFF(MINUTE, ‘2015-01-01 00:00:00’, MinuteBar) / 5 AS Interval, … Read more
To get around this issue, wrap your select statement in a CTE, and then you can query against the CTE and use the windowed function’s results in the where clause. WITH MyCte AS ( select employee_id, RowNum = row_number() OVER ( order by employee_id ) from V_EMPLOYEE ORDER BY Employee_ID ) SELECT employee_id FROM MyCte … Read more
Basically, you need a window function. That’s a standard feature nowadays. In addition to genuine window functions, you can use any aggregate function as window function in Postgres by appending an OVER clause. The special difficulty here is to get partitions and sort order right: SELECT ea_month, id, amount, ea_year, circle_id , sum(amount) OVER (PARTITION … Read more
Use this: SELECT x.id, x.position, x.name FROM (SELECT t.id, t.name, @rownum := @rownum + 1 AS position FROM TABLE t JOIN (SELECT @rownum := 0) r ORDER BY t.name) x WHERE x.name=”Beta” …to get a unique position value. This: SELECT t.id, (SELECT COUNT(*) FROM TABLE x WHERE x.name <= t.name) AS position, t.name FROM TABLE … Read more