看了一眼自关联,没搞懂,试了一下也没成功。
over方式一下结果就出来了,好用。
/* 需求:累计求和六种算法效率比较 作者:felix 日期:2020-06-23 */ --第一步,准备测试数据 --IF OBJECT_ID(N'dbo.t') IS NOT NULL -- DROP TABLE dbo.t; --GO --CREATE TABLE dbo.t --( -- i BIGINT IDENTITY(1, 1) PRIMARY KEY, -- d MONEY --)www.devze.com; --INSERT t -- d --) --SELECT TOP 31465 -- ROUND(10000 * RAND(CHECKSUM(NEWID())), 2) --FROM sys.all_objects AS a -- CROSS JOIN sys.all_objects; ----第二步,创建记录时间的表格 --IF OBJECT_ID(N'dbo.record_time') IS NOT NULL -- DROP TABLE dbo.record_time; --CREATE TABLE dbo.record_time -- i INT IDENTITY PRIMARY KEY, -- 算法 NVARCHAR(10), -- bt DATETIME2,--开始时间 -- et DATETIME2,--结束时间 -- idiff AS DATEDIFF(ms, bt, et)--所用的毫秒数 --第一种方法,自连接法,sql server 2008以上版本测试通过,157255661.40 SET STATISTICS TIME OFF; SET STATISTICS IO OFF; GO DECLARE @bt DATETIME2 = GETDATE(); SELECT a.i, a.d, SUM(b.d) AS total_sum FROM dbo.t AS a INNER JOIN dbo.t AS b ON b.i <= a.i GROUP BY a.i, a.d; DECLARE @et DATETIME2 = GETDATE(); INSERT INTO dbo.record_time ( 算法, bt, et ) VALUES ('自连接', @bt, @et); --ORDER BY a.i; ; --第二种方法,递归,sql server 2008以上版本测试通过,157255661.40 WITH cte_total_sum AS (SELECT i, d, d AS total_sum FROM dbo.t WHERE i = 1 UNION ALL SELECT s.i, s.d, p.total_sum + s.d AS total_sum FROM dbo.t开发者_PostgreSQL AS s INNER JOIN cte_total_sum AS p ON s.i - 1 = p.i) SELECT * FROM cte_total_sum OPTION (MAXRECURSION 0); ('递归', @bt, @et); --第三种方法,over 子句,sql server 2012测试通过,sql server 2008不支持,157255661.40 SELECT i, d, SUM(d) OVER (ORDER BY i) AS total_sum FROM dbo.t; ('over子句', @bt, @et); --第四种,相关子查询,sql server 2008以上版本测试通过,156625045.22 SELECT outquery.i, outquery.d, ( SELECT SUM(innerq.d) FROM dbo.t AS ihttp://www.devze.comnnerq WHERE innerq.i <= outquery.i ) AS ct --内部查询 FROM dbo.t AS outquery; ('相关子查询', @bt, @et); --ORDER BY outquery.i; --外部查询 --游标方法,有两种方法可以实现,一种是临时表更新,一种是变量叠加更新,157255661.40 --先增加一个存储累计和的列 --第5种,游标_临时表更新 --ALTER TABLE dbo.t ADD total_d MONEY DEFAULT (0);--只运行一次 DECLARE @t TABLE --定义表变量,存储累计求和临时结果 i INT PRhttp://www.devze.comIMARY KEY IDENTITY, d MONEY, total_d MONEY ); DECLARE @i INT = 0, @d MONEY = 0, @total_d MONEY = 0; DECLARE c1 CURSOR FOR SELECT i, d FROM dbo.t ORDER BY i; OPEN c1; FETCH c1 INTO @i, @d; WHILE @@FETCH_STATUS = 0 BEGIN SET @total_d += @d; INSERT INTO @t ( d, total_d ) VALUES (@d, @total_d); FETCH c1 INTO @i, @d; END; CLOSE c1; D编程客栈EALLOCATE c1; UPDATE dbo.t SET total_d = b.total_d INNER JOIN @t AS b ON a.i = b.i; ('游标_临时表更新', @bt, @et); --第6种,游标_变量叠加更新 DECLARE c1 CURSOR FOR SELECT i, d FROM dbo.t; --ORDER BY i;编程客栈 UPDATE dbo.t SET total_d = @total_d WHERE i = @i; ('游标_变量叠加更新', @bt, @et); --执行时间 over子句<游标临时表更新<游标变量叠加更新<自连接<相关子查询<递归查询
补充:下面看下SQL server 累加求和
SQL server 累加求和
1.
SELECT SalesOrderID, ProductID, OrderQty ,SUM(OrderQty) OVER(PARTITION BY SalesOrderID) AS Total ,AVG(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Avg" ,COUNT(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Count" ,MIN(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Min" ,MAX(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Max" FROM Sales.SalesOrderDetail WHERE SalesOrderID IN(43659,43664);
2.
select SchSno,convert(varchar(10),a.Dates,120) Dates, sum(Amt_avail) over(partition by SchSno order by convert(varchar(10),a.Dates,120)) as PeriodPreAmt from jr_creditUserAcct a
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