Question

您好我正在为SSRS中的报告开发数据集我有一个查询，它给出了积压中的总请求数：

SELECT 
COUNT(*) as NB
FROM p_rqt WITH (NOLOCK) 
INNER JOIN p_cpy WITH (NOLOCK) ON p_cpy.CpyInCde = p_rqt.OrigCpyInCde 
WHERE 
    CpyTypInCde IN (27, 31) 
    AND p_rqt.RqtNatInCde IN (74, 75, 76) 
    AND HeadRqtInCde = 0 
    AND p_rqt.OrigCpyInCde LIKE CASE WHEN @Client = 0 THEN '%' ELSE @Client END
    AND ((RcvDte < DATEADD(day, 1, @DateDeb) AND RqtEndDte IS NULL)   OR 
(RcvDte < DATEADD(day, 1, @DateDeb) AND RqtEndDte > DATEADD(day, 1, @DateDeb)))

我想要检索每天剩余的总金额。我尝试了很多这样的事情：

SELECT CONVERT(date,rcvdte,103), count(*) as nb
FROM p_rqt p WITH (NOLOCK) 
INNER JOIN p_cpy WITH (NOLOCK) ON p_cpy.CpyInCde = p.OrigCpyInCde 
WHERE 
    CpyTypInCde IN (27, 31) 
    AND p.RqtNatInCde IN (74, 75, 76) 
    AND HeadRqtInCde = 0 
    AND ((RcvDte < DATEADD(day, 1, '20170901') AND RqtEndDte IS NULL)   OR (RcvDte < DATEADD(day, 1, '20170901') AND RqtEndDte > DATEADD(day, 1, '20170901')))
    group by CONVERT(date,rcvdte,103)
    order by CONVERT(date,rcvdte,103)

我尝试了内连接子查询，Sum和其他东西但我能做的就是每天增加记录数量我想要这样的东西：

date:               NB:
01/01/2017        1950
02/01/2017        1954               (+4 items)
03/01/2017        1945               (-9 items)

谢谢

Answer 1

使用LAG：

WITH cte AS (
    SELECT
        CONVERT(date, rcvdte, 103) AS date,
        COUNT(*) AS nb
    FROM p_rqt p WITH (NOLOCK) 
    INNER JOIN p_cpy WITH (NOLOCK)
        ON p_cpy.CpyInCde = p.OrigCpyInCde 
    WHERE 
        CpyTypInCde IN (27, 31) AND
        p.RqtNatInCde IN (74, 75, 76) AND
        HeadRqtInCde = 0 AND
        ((RcvDte < DATEADD(day, 1, '20170901') AND RqtEndDte IS NULL)   OR (RcvDte < DATEADD(day, 1, '20170901') AND RqtEndDte > DATEADD(day, 1, '20170901')))
    GROUP BY CONVERT(date, rcvdte, 103)
    ORDER BY CONVERT(date, rcvdte, 103)
)

SELECT
    t1.date,
    (SELECT SUM(t2.nb) FROM cte t2 WHERE t2.date <= t1.date) AS nb,
    CASE WHEN t1.nb - LAG(t1.nb, 1, t1.nb) OVER (ORDER BY t1.date) > 0
         THEN '(+' + (t1.nb - LAG(t1.nb, 1, t1.nb) OVER (ORDER BY t1.date)) + ' items)'
         ELSE '('  + (t1.nb - LAG(t1.nb, 1, t1.nb) OVER (ORDER BY t1.date)) + ' items)'
    END AS difference
FROM cte t1
ORDER BY t1.date;

Answer 2

所以我找到了一个解决方案，但它真的很慢，我仍然会发布答案

            DECLARE @Tb TABLE (  Colonne1 Datetime, Colonne2 INT )
DECLARE @Debut Datetime = '01/09/2017'
WHILE @Debut < '13/09/2017'
BEGIN
   DECLARE @Compteur int = (
         SELECT 
                COUNT(1) NB 
                FROM p_rqt WITH (NOLOCK) 
                INNER JOIN p_cpy WITH (NOLOCK) ON p_cpy.CpyInCde = p_rqt.OrigCpyInCde 
                WHERE 
                       CpyTypInCde IN (27, 31) 
                       AND p_rqt.RqtNatInCde IN (74, 75, 76) 
                       AND HeadRqtInCde = 0 
                       AND p_rqt.OrigCpyInCde LIKE '%' 
                       AND (
                              (RcvDte < @Debut AND RqtEndDte IS NULL)   
                              OR 
                              (RcvDte < @Debut AND RqtEndDte > @Debut)
                       )
   )
   INSERT INTO @Tb (Colonne1, Colonne2) VALUES (@Debut, @Compteur)
   SET @Debut = DATEADD(day, 1, @Debut)
   IF @Debut > '13/09/2017'
    BREAK
   ELSE
  CONTINUE
  END
  SELECT * FROM @Tb

SQL Server 2014选择每天的总计

2 个答案: