如何找到日期

當天使用SQL Server 2000如何找到日期

表1

ID Date Value1 Value2 

001 01/01/2012 100 0 
001 02/01/2012 200 200 
... 
... 
001 31/01/2012 250 0 
002 01/01/2012 050 100 
002 02/01/2012 100 0 
... 
002 31/01/2012 075 
....

我想通過ID顯示按日期明智的（日期列）組值（值列）

輸出像

ID 01/01/2012 02/01/2012 ... 31/01/2012 

001 100 200 .... 250 
002 050 100 .... 075 
.....

查詢

DECLARE @loop int,@date Date,@sql nvarchar(4000) 
DECLARE @TempTable TABLE 
(
    [Date] DATE 
) 
INSERT INTO @TempTable SELECT DISTINCT [Date] FROM table1 ORDER BY [Date] 
    SET @sql = '' 
    SET @loop = 1 
    WHILE (@loop<=31) 
    BEGIN 
     IF EXISTS(SELECT * FROM @TempTable WHERE DAY([Date])[email protected]) 
     BEGIN 
      SET @date = (SELECT [Date] FROM @TempTable WHERE DAY([Date])[email protected]) 
      SET @sql = @sql+ ',MAX(CASE CONVERT(nvarchar(10),[Date],103) WHEN '''+CONVERT(nvarchar(10),@date,103)+''' THEN [Value1] END) AS [DATE'+CONVERT(nvarchar(2),@loop)+']' 
     END 
     ELSE 
      SET @sql = @sql+ ', NULL AS [DATE'+CONVERT(nvarchar(2),@loop)+']' 
     SET @loop = @loop+1 
    END 

EXEC('SELECT ID'[email protected]+' FROM table1 GROUP BY ID')

上面的查詢工作，從上述查詢我想找到的星期天，如果是星期天則日期欄應該顯示從值2列中的值，否則值1列應顯示

如何做到這一點。

需要查詢幫助

來源

2012-05-23 JetJack

DATEPART function給你一週的整數一天。星期天 - 星期六成爲1 - 7

試圖通過這一個

SET @sql = @sql+ ',MAX(CASE CONVERT(nvarchar(10),[Date],103) WHEN '''+CONVERT(nvarchar(10),@date,103)+''' THEN CASE DATEPART(weekday,[Date])WHEN 1 THEN [Value1] ELSE [Value2] END END) AS [DATE'+CONVERT(nvarchar(2),@loop)+']'

改變線對總計算每個ID

SELECT ID, 
SUM(CASE WHEN DATEPART(weekday,[Date]) <> 1 THEN [Value1] END) AS Value1, 
SUM(CASE DATEPART(weekday,[Date]) WHEN 1 THEN [Value2] END) AS Value2, 
FROM table1 GROUP BY ID

來源

2012-05-23 12:17:55

謝謝你，如何總結值1和值2的，從我希望做一個總和（值）和sum（值2）輸出。 sum（value1）表示除星期日剩餘的所有日子，sum（value2）只有星期日，如何做到這一點，PLZ幫助我.... – JetJack

它可能會變成你不需要迭代，也不動態腳本。考慮以下幾點：

SELECT 
    ID, 
    Day1 = MAX(CASE D WHEN 1 THEN V END), 
    Day2 = MAX(CASE D WHEN 2 THEN V END), 
    Day3 = MAX(CASE D WHEN 3 THEN V END), 
    Day4 = MAX(CASE D WHEN 4 THEN V END), 
    Day5 = MAX(CASE D WHEN 5 THEN V END), 
    Day6 = MAX(CASE D WHEN 6 THEN V END), 
    Day7 = MAX(CASE D WHEN 7 THEN V END), 
    Day8 = MAX(CASE D WHEN 8 THEN V END), 
    Day9 = MAX(CASE D WHEN 9 THEN V END), 
    Day10 = MAX(CASE D WHEN 10 THEN V END), 
    Day11 = MAX(CASE D WHEN 11 THEN V END), 
    Day12 = MAX(CASE D WHEN 12 THEN V END), 
    Day13 = MAX(CASE D WHEN 13 THEN V END), 
    Day14 = MAX(CASE D WHEN 14 THEN V END), 
    Day15 = MAX(CASE D WHEN 15 THEN V END), 
    Day16 = MAX(CASE D WHEN 16 THEN V END), 
    Day17 = MAX(CASE D WHEN 17 THEN V END), 
    Day18 = MAX(CASE D WHEN 18 THEN V END), 
    Day19 = MAX(CASE D WHEN 19 THEN V END), 
    Day20 = MAX(CASE D WHEN 20 THEN V END), 
    Day21 = MAX(CASE D WHEN 21 THEN V END), 
    Day22 = MAX(CASE D WHEN 22 THEN V END), 
    Day23 = MAX(CASE D WHEN 23 THEN V END), 
    Day24 = MAX(CASE D WHEN 24 THEN V END), 
    Day25 = MAX(CASE D WHEN 25 THEN V END), 
    Day26 = MAX(CASE D WHEN 26 THEN V END), 
    Day27 = MAX(CASE D WHEN 27 THEN V END), 
    Day28 = MAX(CASE D WHEN 28 THEN V END), 
    Day29 = MAX(CASE D WHEN 29 THEN V END), 
    Day30 = MAX(CASE D WHEN 30 THEN V END), 
    Day31 = MAX(CASE D WHEN 31 THEN V END) 
FROM (
    SELECT 
    ID, 
    V = CASE DATENAME(WEEKDAY, Date) WHEN 'Sunday' THEN Value2 ELSE Value1 END, 
    D = DAY(Date) 
    FROM Table1 
    WHERE … /* perhaps, a condition to retrieve a specific month is due here */ 
) s 
GROUP BY ID 
;

來源

2012-05-24 07:44:43

如何找到日期

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