Question

我希望有人可以帮我解决一个相当直接的问题，但让我感到难过。

我写了以下查询

SELECT a.post_id, 
       a.date_from, 
       a.date_to, 
       b.description, 
       b.parttime_pct                       AS budget_hours, 
       Sum(a.parttime_pct)                  AS contract_hours, 
       b.parttime_pct - Sum(a.parttime_pct) AS remainder 
FROM   aprresourcepost a 
       LEFT OUTER JOIN aprpost b 
                    ON(( a.client = b.client 
                         AND a.post_id = b.post_id )) 
       INNER JOIN ahsresources c 
               ON(( c.client = a.client 
                    AND c.resource_id = a.resource_id )) 
       LEFT OUTER JOIN agladdress d 
                    ON(( 'C0' = d.attribute_id 
                         AND c.client = d.client 
                         AND c.resource_id = d.dim_value 
                         AND d.address_type = '1' )) 
       LEFT OUTER JOIN agldimvalue e 
                    ON(( e.client = b.client 
                         AND e.dim_value = b.dim_value )) 
WHERE  a.client = 'CI' 
       AND a.post_id = '14013' 
GROUP  BY a.post_id, 
          b.description, 
          b.parttime_pct, 
          a.parttime_pct, 
          a.date_from, 
          a.date_to

返回以下内容

post_id   date_from   date_to     descr   budget_hours    contract_hours    remainder

14013     2014-10-01  2099-12-31  Reg     140.00          105.00             35.00           
14013     2014-12-20  2099-12-31  Reg     140.00          140.00            105.00

我想要展示的内容（如果可能）是

post_id   date_from   date_to     descr   budget_hours    contract_hours    remainder

14013     2014-10-01  2099-12-31  Reg     140.00          105.00             35.00           
14013     2014-12-20  2099-12-31  Reg     140.00          140.00              0.00

这基本上表明，截至12月20日之前的任何日期，此角色的建议小时数（员工）仍有35小时，但截至12月20日，该职位已100％填补，但是我想要包括日期和日期数据是根据位置和日期和日期相加，我想我有点需要滚动总数

我知道如果我从select和group中删除日期，它将显示140.00 140.00但是我需要一种方法来分隔日期。

非常感谢任何帮助。

谢谢你们

Answer 1

没有数据我发现这很难测试。但这是我最好的一击。使用CTE和ROW_NUMBER提供虚拟行id（需要2005年以后），我们自己加入前一行来累积余数：

WITH qry AS 
(
  SELECT ROW_NUMBER() OVER (ORDER BY date_from) AS row,
         a.post_id, 
         a.date_from, 
         a.date_to, 
         b.description, 
         b.parttime_pct                       AS budget_hours, 
         Sum(a.parttime_pct)                  AS contract_hours, 
         b.parttime_pct - Sum(a.parttime_pct) AS remainder 
  FROM   aprresourcepost a 
         LEFT OUTER JOIN aprpost b 
                      ON(( a.client = b.client 
                           AND a.post_id = b.post_id )) 
         INNER JOIN ahsresources c 
                 ON(( c.client = a.client 
                      AND c.resource_id = a.resource_id )) 
         LEFT OUTER JOIN agladdress d 
                      ON(( 'C0' = d.attribute_id 
                           AND c.client = d.client 
                           AND c.resource_id = d.dim_value 
                           AND d.address_type = '1' )) 
         LEFT OUTER JOIN agldimvalue e 
                      ON(( e.client = b.client 
                           AND e.dim_value = b.dim_value )) 
  WHERE  a.client = 'CI' 
         AND a.post_id = '14013' 
  GROUP  BY a.post_id, 
            b.description, 
            b.parttime_pct, 
            a.parttime_pct, 
            a.date_from, 
            a.date_to 
)
SELECT
  qry2.post_id,
  qry2.date_from,
  qry2.date_to,
  qry2.description,
  qry2.budget_hours,
  qry2.contract_hours,
  qry2.budget_hours - (qry2.remainder + ISNULL(qry1.remainder, 0)) AS Cumulative_Remainder
FROM
  qry qry2
  LEFT JOIN qry qry1 ON qry1.row = qry2.row - 1

Answer 2

您需要做的就是将查询作为子查询并在外部查询中查找remainder

select *,
budget_hours- contract_hours AS remainder 
from(
SELECT a.post_id, 
       a.date_from, 
       a.date_to, 
       b.description, 
       b.parttime_pct                       AS budget_hours, 
       Sum(a.parttime_pct)                  AS contract_hours
FROM   aprresourcepost a 
       LEFT OUTER JOIN aprpost b 
                    ON(( a.client = b.client 
                         AND a.post_id = b.post_id )) 
       INNER JOIN ahsresources c 
               ON(( c.client = a.client 
                    AND c.resource_id = a.resource_id )) 
       LEFT OUTER JOIN agladdress d 
                    ON(( 'C0' = d.attribute_id 
                         AND c.client = d.client 
                         AND c.resource_id = d.dim_value 
                         AND d.address_type = '1' )) 
       LEFT OUTER JOIN agldimvalue e 
                    ON(( e.client = b.client 
                         AND e.dim_value = b.dim_value )) 
WHERE  a.client = 'CI' 
       AND a.post_id = '14013' 
GROUP  BY a.post_id, 
          b.description, 
          b.parttime_pct, 
          a.parttime_pct, 
          a.date_from, 
          a.date_to ) A

Answer 3

你是否有理由通过a.parttime_pct进行分组？如果从group by子句中删除它会发生什么？它不会在select语句中作为列返回，因此您不需要按它进行分组。

SELECT a.post_id, 
   a.date_from, 
   a.date_to, 
   b.description, 
   b.parttime_pct                       AS budget_hours, 
   Sum(a.parttime_pct)                  AS contract_hours, 
   b.parttime_pct - Sum(a.parttime_pct) AS remainder 
FROM   aprresourcepost a 
   LEFT OUTER JOIN aprpost b 
                ON(( a.client = b.client 
                     AND a.post_id = b.post_id )) 
   INNER JOIN ahsresources c 
           ON(( c.client = a.client 
                AND c.resource_id = a.resource_id )) 
   LEFT OUTER JOIN agladdress d 
                ON(( 'C0' = d.attribute_id 
                     AND c.client = d.client 
                     AND c.resource_id = d.dim_value 
                     AND d.address_type = '1' )) 
   LEFT OUTER JOIN agldimvalue e 
                ON(( e.client = b.client 
                     AND e.dim_value = b.dim_value )) 
WHERE  a.client = 'CI' 
   AND a.post_id = '14013' 
GROUP  BY a.post_id, 
      b.description, 
      b.parttime_pct, 
      a.date_from, 
      a.date_to

按位置滚动日期总计的SQL总和

3 个答案: