您可以使用以下解决方案：
SELECT
a.StoreID,
a.weekdayname,
b.isOpen
FROM
(
SELECT *
FROM
(
SELECT DISTINCT StoreID
FROM MyTable
) aa
CROSS JOIN
(
SELECT 1 AS weekdaynum, 'Sunday' AS weekdayname UNION ALL
SELECT 2, 'Monday' UNION ALL
SELECT 3, 'Tuesday' UNION ALL
SELECT 4, 'Wednesday' UNION ALL
SELECT 5, 'Thursday' UNION ALL
SELECT 6, 'Friday' UNION ALL
SELECT 7, 'Saturday'
) bb
) a
LEFT JOIN
MyTable b ON a.StoreID = b.StoreID AND
a.weekdaynum = b.[DayOfWeek]
WHERE
a.StoreID = @I_StoreID
ORDER BY
a.StoreID, a.weekdaynum
在这里，我们手动选择所有工作日名称CROSS JOIN，每个名称各不相同StoreID。然后，将选择的结果包装在FROM子句中，并LEFT JOIN在条件StoreID与星期几都匹配的情况下返回主表。如果不是，则该isOpen字段将为NULL，但StoreID仍会显示和相应的工作日。
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