日期:2014-05-17 浏览次数:20752 次
select cast(ltrim(@i/60)+'.'+ltrim(@i%60) as decimal(18,2))
------解决方案--------------------
借4楼数据一用
IF NOT OBJECT_ID('[tb]') IS NULL
DROP TABLE [tb]
GO
CREATE TABLE [tb]([col] INT)
INSERT [tb]
SELECT 60 UNION ALL
SELECT 70 UNION ALL
SELECT 119 UNION ALL
SELECT 120 UNION ALL
SELECT 1800 UNION ALL
SELECT 18000
GO
select cast(col/60+(col%60)/100.0 as dec(18,2))
from tb
/**
---------------------------------------
1.00
1.10
1.59
2.00
30.00
300.00
(6 行受影响)
**/
------解决方案--------------------
CAST(COL*1./60 AS DEC(18,2))
------解决方案--------------------
--------------------SQL Server数据格式化工具-------------------
---------------------------
-- DESIGNER :happycell188(喜喜)
-- QQ :584738179
-- Development Tool :Microsoft Visual C++ 6.0 C Language
-- FUNCTION :CONVERT DATA TO T-SQL
---------------------------
-- Microsoft SQL Server 2005
-- Developer Edition on Microsoft Windows XP [版本 5.1.2600]
---------------------------
---------------------------
use test
go
if object_id('test.dbo.tb') is not null drop table tb
-- 创建数据表
create table tb
(
cnt int
)
go
--插入测试数据
insert into tb select 60
union all select 70
union all select 119
union all select 120
union all select 1800
union all select 18000
go
--代码实现
select str(ltrim(cnt/60)+'.'+ltrim(cnt%60),10,2)cnt from tb
/*测试结果
cnt
---------------------
1.00
1.10
1.59
2.00
30.00
300.00
(6 行受影响)
*/