Question

我正在尝试在查询中合并以下逻辑，类似于2个vlookup，如果一个不可用，请使用下一个。

eventStart.setYear(date.getYear()); Logger.log(eventStart);

因此它应该先查询D列（单元2），然后如果没有查询，则查询C列（值）

到目前为止，这是我尝试过的=IFERROR(VLOOKUP(D2,'Segmentation'!$A:$C,3,0),IFERROR(VLOOKUP(C2,'Segmentation'!$A:$C,3,0),""))连接：

OR

也尝试过：

SELECT DISTINCT A.[unit], B.[Hierarchy] FROM [Mapping$] as A

inner join [Segmentation$] as B

ON (A.[unit2] = B.[Number] OR
      A.[Value] = B.[Number])

WHERE B.[Hierarchy] <> ''

Order By A.[unit] asc

细分表中的样本数据：

SELECT DISTINCT A.[unit], B.[Hierarchy] FROM [Mapping$] as A

inner join [Segmentation$] as B

ON Nz(A.[unit2], A.[Value])=B.[Number]

WHERE B.[Hierarchy] <> ''

Order By A.[unit] asc

映射表：

+--------------+-------------+-----------+
| Number       |  Name       | Hierarchy |
+--------------+-------------+-----------+
|        11773 | Stack       | Overflow  |
+--------------+-------------+-----------+

预期结果：

+-------+-----------+-------+-------+-----------+---------+
| unit  | variable1 | Value | unit2 | variable2 | value2  |
+-------+-----------+-------+-------+-----------+---------+
| 11773 | lvl0      | 80000 | 11773 | lvl0_nm   | thing 1 |
| 11773 | lvl1      | 89000 | 11773 | lvl1_nm   | thing 2 |
| 11773 | lvl2      | 89022 | 11773 | lvl2_nm   | thing 3 |
| 11773 | lvl3      | 89756 | 11773 | lvl3_nm   | thing 4 |
| 11773 | lvl4      | 89833 | 11773 | lvl4_nm   | thing 5 |
| 11773 | lvl5      | 11773 | 11773 | lvl5_nm   |         |
| 11773 | lvl6      |       | 11773 | lvl6_nm   |         |
| 11773 | lvl7      |       | 11773 | lvl7_nm   |         |
| 11773 | lvl8      |       | 11773 | lvl8_nm   |         |
| 11773 | lvl9      |       | 11773 | lvl9_nm   |         |
| 11773 | lvl10     |       | 11773 | lvl10_nm  |         |
| 11773 | lvl11     |       | 11773 | lvl11_nm  |         |
| 11773 | lvl12     |       | 11773 | lvl12_nm  |         |
| 11773 | lvl13     |       | 11773 | lvl13_nm  |         |
| 11773 | lvl14     |       | 11773 | lvl14_nm  |         |
| 11773 | lvl15     |       | 11773 | lvl15_nm  |         |
| 11773 | lvl16     |       | 11773 | lvl16_nm  |         |
| 11773 | lvl17     |       | 11773 | lvl17_nm  |         |
| 11773 | lvl18     |       | 11773 | lvl18_nm  |         |
+-------+-----------+-------+-------+-----------+---------+

第一个vlookup失败的示例：

+-------+-----------+-------+-------+-----------+---------+-----------+
| unit  | variable1 | Value | unit2 | variable2 | value2  | Hierarchy |
+-------+-----------+-------+-------+-----------+---------+-----------+
| 11773 | lvl0      | 80000 | 11773 | lvl0_nm   | thing 1 | Overflow  |
| 11773 | lvl1      | 89000 | 11773 | lvl1_nm   | thing 2 | Overflow  |
| 11773 | lvl2      | 89022 | 11773 | lvl2_nm   | thing 3 | Overflow  |
| 11773 | lvl3      | 89756 | 11773 | lvl3_nm   | thing 4 | Overflow  |
| 11773 | lvl4      | 89833 | 11773 | lvl4_nm   | thing 5 | Overflow  |
| 11773 | lvl5      | 11773 | 11773 | lvl5_nm   |         | Overflow  |
| 11773 | lvl6      |       | 11773 | lvl6_nm   |         | Overflow  |
| 11773 | lvl7      |       | 11773 | lvl7_nm   |         | Overflow  |
| 11773 | lvl8      |       | 11773 | lvl8_nm   |         | Overflow  |
| 11773 | lvl9      |       | 11773 | lvl9_nm   |         | Overflow  |
| 11773 | lvl10     |       | 11773 | lvl10_nm  |         | Overflow  |
| 11773 | lvl11     |       | 11773 | lvl11_nm  |         | Overflow  |
| 11773 | lvl12     |       | 11773 | lvl12_nm  |         | Overflow  |
| 11773 | lvl13     |       | 11773 | lvl13_nm  |         | Overflow  |
| 11773 | lvl14     |       | 11773 | lvl14_nm  |         | Overflow  |
| 11773 | lvl15     |       | 11773 | lvl15_nm  |         | Overflow  |
| 11773 | lvl16     |       | 11773 | lvl16_nm  |         | Overflow  |
| 11773 | lvl17     |       | 11773 | lvl17_nm  |         | Overflow  |
| 11773 | lvl18     |       | 11773 | lvl18_nm  |         | Overflow  |
+-------+-----------+-------+-------+-----------+---------+-----------+

映射表

+--------+-------+-----------+
| Number | Name  | Hierarchy |
+--------+-------+-----------+
|  82282 | Super | User      |
+--------+-------+-----------+

预期结果：

+-------+-----------+-------+-------+-----------+----------+
| unit  | variable1 | Value | unit2 | variable2 |  value2  |
+-------+-----------+-------+-------+-----------+----------+
| 96702 | lvl0      | 80000 | 96702 | lvl0_nm   | thing 6  |
| 96702 | lvl1      | 80393 | 96702 | lvl1_nm   | thing 7  |
| 96702 | lvl2      | 85176 | 96702 | lvl2_nm   | thing 8  |
| 96702 | lvl3      | 82282 | 96702 | lvl3_nm   | thing 9  |
| 96702 | lvl4      | 87927 | 96702 | lvl4_nm   | thing 10 |
| 96702 | lvl5      | 88172 | 96702 | lvl5_nm   | thing 11 |
| 96702 | lvl6      | 81025 | 96702 | lvl6_nm   | thing 12 |
| 96702 | lvl7      | 96702 | 96702 | lvl7_nm   |          |
| 96702 | lvl8      |       | 96702 | lvl8_nm   |          |
| 96702 | lvl9      |       | 96702 | lvl9_nm   |          |
| 96702 | lvl10     |       | 96702 | lvl10_nm  |          |
| 96702 | lvl11     |       | 96702 | lvl11_nm  |          |
| 96702 | lvl12     |       | 96702 | lvl12_nm  |          |
| 96702 | lvl13     |       | 96702 | lvl13_nm  |          |
| 96702 | lvl14     |       | 96702 | lvl14_nm  |          |
| 96702 | lvl15     |       | 96702 | lvl15_nm  |          |
| 96702 | lvl16     |       | 96702 | lvl16_nm  |          |
| 96702 | lvl17     |       | 96702 | lvl17_nm  |          |
| 96702 | lvl18     |       | 96702 | lvl18_nm  |          |
+-------+-----------+-------+-------+-----------+----------+

Answer 1

您的代码，

SELECT 
    DISTINCT A.[unit],
    B.[Hierarchy] 
FROM [Mapping$] as A inner join [Segmentation$] as B ON (A.[unit2] = B.[Number] OR A.[Value] = B.[Number])
WHERE B.[Hierarchy] <> ''    
Order By A.[unit] asc;

由于您使用的是Inner Join，因此不需要WHERE B.[Hierarchy] <> ''。仅当匹配时，联接才会产生结果。

要获得预期的结果，可以使用LEFT JOIN或子查询。

SELECT 
    DISTINCT A.[unit],
    B.[Hierarchy] 
FROM [Mapping$] as A **LEFT JOIN** [Segmentation$] as B ON (A.[unit2] = B.[Number] OR A.[Value] = B.[Number])
Order By A.[unit] asc;

这将生成唯一的A.Unit和B.Hierarchy的列表。

您还可以使用子查询。（如果segmentation.Number不是唯一的，请使用前1个。否则，您的子查询将返回多行并抱怨该行）

SELECT 
    A.*,
    (SELECT B.[Hierarchy] 
     FROM [Segmentation$] as B
     WHERE (A.[unit2] = B.[Number] OR A.[Value] = B.[Number])) AS Hierarchy
FROM [Mapping$] as A
Order By A.[unit] asc;

Excel Jet Access查询，模拟vlookup

1 个答案: