Question

我有一个MSSQL 2000数据库，其中包含我需要执行查询的表。它有一个'STEP'列和列'DATE1'，'DATE2'......'DATE7'。当一个特定的脚本运行时，它会更新'DATE $ step'列并将'STEP'递增一个（必要时包裹）。

我正在尝试创建一个查询，该查询返回最后一个日期的行，例如如果'STEP'= 4，'DATE3'比X天还要早，但我对如何进行此查询感到有些困惑。

Answer 1

这样的东西？

SELECT
  [DateColumn]=CASE STEP WHEN 1 THEN DATE1 
WHEN 2 THEN DATE2
WHEN 3 THEN DATE3
ELSE DATE4 END
FROM <Table>

Answer 2

select Step, [Date]
from 
  (select 
    Step,
    case Step
      when 1 then Date1
      when 2 then Date2
      when 3 then Date3
      when 4 then Date4
      when 5 then Date5
      when 6 then Date6
      when 7 then Date7
    end as [Date]
  from YourTable) as T
where datediff(d, T.[Date], getdate()) > @XDays

Answer 3

 Select MyTable.* from MyTable INNER JOIN
     (Select tableID, CASE STEP
                WHEN 7 THEN DATE6   -- The STEP column increments are done 
                WHEN 6 THEN DATE5   -- *after* updating the date, therefore
                WHEN 5 THEN DATE4   -- the last update date is in Date[STEP-1]
                WHEN 4 THEN DATE3
                WHEN 3 THEN DATE2
                WHEN 2 THEN DATE1
                WHEN 1 THEN DATE7   -- Rollover (special case)
             END AS LastDate
     From MyTable
     -- DATEDIFF Returns the count (signed integer) of the specified 
     --          datepart boundaries crossed. I therefore use seconds
     --          to get predicable results regardless of execution time.
     --          (86400 seconds in a day)
     WHERE DATEDIFF('s',LastDate,getdate()) > 
                    (86400 * @DaysParameter)) as dateResult           
     ON MyTable.tableID = dateResult.tableID

使用列的值在SQL中选择另一列

3 个答案: