Question

我有一个包含一个开关的sql语句，我想在ms-access-2010上执行。

sql：

SELECT 
TABLE_1.ID, 
SWITCH
(
    (TABLE_3.ID IN (7,8) AND TABLE_1.SOME_COLUMN = 2), (SELECT MAX(TABLE_4.ID) FROM TABLE_4 WHERE  TABLE_4.TABLE_2_ID = TABLE_2.ID),
    (SELECT MIN(TABLE_4.SOME_COLUMN) FROM TABLE_4 WHERE TABLE_4.TBL_2_ID = TABLE_1.TBL_2_ID)
)AS SOME_ALIAS 
FROM 
    ((TABLE_1 
        LEFT JOIN TABLE_2 ON TABLE_1.TABLE_2_ID = TABLE_2.ID)
        LEFT JOIN TABLE_3 ON TABLE_2.TABLE_3_ID = TABLE_3.ID)

在我的结果中，“ some_column”列始终为#Function!

如果我不使用切换语句，而是将两种情况下的子选择添加为普通选择字段，如下所示：

SELECT 
TABLE_1.ID, 
(SELECT MAX(TABLE_4.ID) FROM TABLE_4 WHERE  TABLE_4.TABLE_2_ID = TABLE_2.ID) AS SOME_COLUMN_1,
(SELECT MIN(TABLE_4.SOME_COLUMN) FROM TABLE_4 WHERE TABLE_4.TBL_2_ID = TABLE_1.TBL_2_ID) AS SOME_COLUMN_2
FROM 
    ((TABLE_1 
        LEFT JOIN TABLE_2 ON TABLE_1.TABLE_2_ID = TABLE_2.ID)
        LEFT JOIN TABLE_3 ON TABLE_2.TABLE_3_ID = TABLE_3.ID)

我没有任何问题（在SOME_COLUMN_1和SOME_COLUMN_2内只有两行没有值）。

是什么导致#FUNCTION!的结果在这里？

Answer 1

Switch语句需要偶数个参数。您仅传递了3个参数，因此会导致错误。

您可以使用IIF，也可以添加其他比较参数（如果第一个参数不正确，如果要返回最后一个参数，则可能只是True作为第三个参数）。

SELECT 
TABLE_1.ID, 
IIF
(
    (TABLE_3.ID IN (7,8) AND TABLE_1.SOME_COLUMN = 2), (SELECT MAX(TABLE_4.ID) FROM TABLE_4 WHERE  TABLE_4.TABLE_2_ID = TABLE_2.ID),
    (SELECT MIN(TABLE_4.SOME_COLUMN) FROM TABLE_4 WHERE TABLE_4.TBL_2_ID = TABLE_1.TBL_2_ID)
)AS SOME_ALIAS 
FROM 
    ((TABLE_1 
        LEFT JOIN TABLE_2 ON TABLE_1.TABLE_2_ID = TABLE_2.ID)
        LEFT JOIN TABLE_3 ON TABLE_2.TABLE_3_ID = TABLE_3.ID)

ms-access-2010返回#func [感叹号]

1 个答案: