求教二条SQL语句，十分感谢
第一个SQL语句
T1
编号 日期 级别
1001 2012-08-01 1
1001 2012-08-02 1
1001 2012-08-04 1
1001 2012-08-05 1
1002 2012-08-03 2
1002 2012-08-04 2
1002 2012-08-05 2
1002 2012-08-07 2
。。。

结果
编号 日期 天数 级别
1001 2012-08-01 1 1
1001 2012-08-02 2 1
1001 2012-08-04 4 1
1001 2012-08-05 5 1
1002 2012-08-03 1 2
1002 2012-08-04 2 2
1002 2012-08-05 3 2
1002 2012-08-07 5 2
天数的意思是，根据相同编号的最早的那天开始计算，最早的那天为第一天，依次往下排，如果有天数间隔的，天数也随之进行间隔。

[解决办法]

SQL code

------------------------------ Author  :TravyLee(努力工作中！！！)-- Date    :2012-08-08 13:08:24-- Version:--      Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (X64) --    Apr  2 2010 15:48:46 --    Copyright (c) Microsoft Corporation--    Enterprise Edition (64-bit) on Windows NT 6.1 <X64> (Build 7600: ) (Hypervisor)--------------------------------> 测试数据：[T1]if object_id('[T1]') is not null drop table [T1]go create table [T1]([编号] int,[日期] datetime,[级别] int)insert [T1]select 1001,'2012-08-01',1 union allselect 1001,'2012-08-02',1 union allselect 1001,'2012-08-04',1 union allselect 1001,'2012-08-05',1 union allselect 1002,'2012-08-03',2 union allselect 1002,'2012-08-04',2 union allselect 1002,'2012-08-05',2 union allselect 1002,'2012-08-07',2goselect *,DATEDIFF(DD,(select min([日期]) from T1 b where a.[编号]=b.[编号]),[日期])+1 as 天数from T1 a/*编号    日期    级别    天数------------------------------------------------1001    2012-08-01 00:00:00.000    1    11001    2012-08-02 00:00:00.000    1    21001    2012-08-04 00:00:00.000    1    41001    2012-08-05 00:00:00.000    1    51002    2012-08-03 00:00:00.000    2    11002    2012-08-04 00:00:00.000    2    21002    2012-08-05 00:00:00.000    2    31002    2012-08-07 00:00:00.000    2    5*/
[解决办法]
SQL code
--> 测试数据：[T1]if object_id('[T1]') is not null drop table [T1]go create table [T1]([编号] int,[日期] datetime,[级别] int)insert [T1]select 1001,'2012-08-01',1 union allselect 1001,'2012-08-02',1 union allselect 1001,'2012-08-04',1 union allselect 1001,'2012-08-05',1 union allselect 1002,'2012-08-03',2 union allselect 1002,'2012-08-04',2 union allselect 1002,'2012-08-05',2 union allselect 1002,'2012-08-07',2go--> 测试数据：[T2]if object_id('[T2]') is not null drop table [T2]go create table [T2]([级别] int,[起始天数1] int,[终止天数1] int,[金额1] int,[起始天数2] int,[终止天数2] int,[金额2] int)insert [T2]select 1,1,2,500,3,10,600 union allselect 2,1,2,700,3,10,800gowith tas(select *,DATEDIFF(DD,(select min([日期]) from T1 b where a.[编号]=b.[编号]),[日期])+1 as 天数from T1 a)select a.*,case when a.天数 between b.[起始天数1] and b.终止天数1 then [金额1] else b.金额2 end as 金额from t a inner join T2 bon a.级别=b.级别go/*编号    日期    级别    天数    金额-----------------------1001    2012-08-01 00:00:00.000    1    1    5001001    2012-08-02 00:00:00.000    1    2    5001001    2012-08-04 00:00:00.000    1    4    6001001    2012-08-05 00:00:00.000    1    5    6001002    2012-08-03 00:00:00.000    2    1    7001002    2012-08-04 00:00:00.000    2    2    7001002    2012-08-05 00:00:00.000    2    3    8001002    2012-08-07 00:00:00.000    2    5    800*/