我希望在从这个查询中删除周末方面得到一些帮助。我有多个可能在时间上重叠的未决任务的案例。此查询有效，但我希望删除周末。
DECLARE @pends table (id varchar(50), d1 datetime, d2 datetime);
INSERT @pends
--- Play below me --- This is the query that pulls the pend tasks.
SELECT
dc.fldReferenceNumber,
dt.fldCreatedTime,
dt.fldDoneDate
FROM ([Centerpoint_Reporting_Prod].[dbo].[dim_case] dc
INNER JOIN ([Centerpoint_Reporting_Prod].[dbo].[dim_task] dt
INNER JOIN ([Centerpoint_Reporting_Prod].[dbo].[fact_task] ft
INNER JOIN [Centerpoint_Reporting_Prod].[dbo].[dim_task_type] dtt
ON ft.fldTaskTypeIdTaskTypeKey = dtt.TaskTypeSK)
ON ft.TaskSK = dt.TaskSK)
ON ft.CaseKey = dc.CaseSK)
WHERE dtt.Name_enUS LIKE 'Pending%' AND dt.fldDoneDate <> 0 and dc.fldReferenceNumber = 'CDCR-4998513'
--- Play above me ---
SELECT id, SUM (n) AS hold_days
FROM (
SELECT DISTINCT
t1.id,
t1.d1 AS date,
-DATEDIFF(DAY, (SELECT MIN(d1) FROM @pends), t1.d1) AS n
FROM @pends t1
LEFT JOIN @pends t2
ON t2.ID = t1.ID
AND t2.d1 <> t1.d1
AND t1.d1 BETWEEN t2.d1 AND t2.d2
GROUP BY t1.ID, t1.d1, t1.d2
HAVING COUNT(t2.ID) = 0
UNION ALL
SELECT DISTINCT
t1.id,
t1.d2 AS date,
DATEDIFF(DAY, (SELECT MIN(d1) FROM @pends), t1.d2) AS n
FROM @pends t1
LEFT JOIN @pends t2
ON t2.ID = t1.ID
AND t2.d2 <> t1.d2
AND t1.d2 BETWEEN t2.d1 AND t2.d2
GROUP BY t1.ID, t1.d1, t1.d2
HAVING COUNT(t2.ID) = 0
) s
GROUP BY ID;