数据库时间结构是
年 smallint
月 smallint
日 smallint
时 smallint
数值 int
怎样输出 与前一天的数据差
结果类似
年 月 日 时 数值 24小时差值
2014 1 1 1 99
2014 1 2 1 92 7
2014 1 3 1 92 0
2014 1 3 1 96 4
2014 1 3 1 99 3
------解决方案--------------------
这样？
----------------------------------------------------------------
-- Author :DBA_HuangZJ(發糞塗牆)
-- Date :2014-08-19 11:02:23
-- Version:
-- Microsoft SQL Server 2012 - 11.0.5058.0 (X64)
-- May 14 2014 18:34:29
-- Copyright (c) Microsoft Corporation
-- Enterprise Edition: Core-based Licensing (64-bit) on Windows NT 6.3 <X64> (Build 9600: ) (Hypervisor)
--
----------------------------------------------------------------
--> 测试数据：[huang]
if object_id('[huang]') is not null drop table [huang]
go
create table [huang]([年] int,[月] int,[日] int)
insert [huang]
select 2014,1,1 union all
select 2014,1,2 union all
select 2014,1,3 union all
select 2014,1,3 union all
select 2014,1,3
--------------开始查询--------------------------
select CAST(年 AS CHAR(4))+'-'+RIGHT('00'+CAST(月 AS VARCHAR(2)),2)+'-'+RIGHT('00'+CAST([日] AS VARCHAR(2)),2)[年月日]
from [huang]
----------------结果----------------------------
/*
年月日
--------------
2014-01-01
2014-01-02
2014-01-03
2014-01-03
2014-01-03
*/