在JOIN的ON子句中使用别名

Question

堆栈溢出的新功能（通常是编码）。

我做了一些研究，但是找不到以下问题的答案：

如何在应用于维的函数结果上而不是维本身上联接两个表？

即我想在函数lower（）的小写结果上联接以下两个表，而不是在大小写不明确的维度上联接。

SELECT
lower(first_name) as firstname
,lower(last_name) as lastname
,lower(email) as email1
,total_donated
From BensData.Donations As a

JOIN EACH

(Select
lower(first_name) as first
,lower(last_name) as last
,lower(email) as email2
,sum(amount) as total_donated
From BensData.Donations 
GROUP BY email2, first, last) As b

ON a.email1=b.email2 AND a.firstname=b.first AND a.lastname=b.last

它不允许我加入在第一个表（a）中创建的别名，但是，如果我加入表a中的原始维度（first_name和last_name），则结果基于大小写不明确的维度，并给出不良结果。

我希望这很清楚。

谢谢你的帮助！

Answer 1

尝试使用两个子查询，如下所示：

SELECT
a.firstname
,a.lastname
,a.email1
,a.total_donated
FROM

(SELECT
lower(first_name) as firstname
,lower(last_name) as lastname
,lower(email) as email1
,total_donated
From BensData.Donations) As a

JOIN EACH

(Select
lower(first_name) as first
,lower(last_name) as last
,lower(email) as email2
,sum(amount) as total_donated
From BensData.Donations 
GROUP BY email2, first, last) As b

ON a.email1=b.email2 AND a.firstname=b.first AND a.lastname=b.last

在原始查询中，a只是BensData.Donations的别名，因此您只能联接该表中存在的字段。

Answer 2

我从未听说过join each ，也没有将其记录为MySQL连接的语法（请参见此处）。

试试这个from子句：

From BensData.Donations a JOIN
     (Select lower(first_name) as first, lower(last_name) as last,
             sum(amount) as total_donated
      From BensData.Donations 
      GROUP BY first, last
     ) b
     ON a.firstname = b.first AND a.lastname = b.last

您的版本在列之间的子查询中也缺少逗号。

为了清楚起见，您可能应该将from子句写为：

     ON lower(a.firstname) = b.first AND lower(a.lastname) = b.last

或从子查询中删除lower() 。 否则，您将依赖于服务器，数据库和表的默认归类来确定on子句的实际作用。

Answer 3

感谢大家的帮助！

特别扣人心弦的人向我指出了正确的方向！ 他的代码和我的代码的主要区别在于，我的代码没有在第一个SELECT子句的每个维的前面附加表别名（例如** a。** fistname，** a。** lastname，- ->名字，姓氏）

由于某种原因，BigQuery一直由于表别名而给我一个错误。

这是起作用的代码。

SELECT
firstname
,lastname
,email1
,total_donated
FROM

(SELECT
lower(first_name) as firstname
,lower(last_name) as lastname
,lower(email) as email1
From BensData.Donations) As a

JOIN EACH

(Select
lower(first_name) as first
,lower(last_name) as last
,lower(email) as email2
,sum(float(amount)) as total_donated
From BensData.Donations 
GROUP BY email2, first, last) As b

ON a.email1=b.email2 AND a.firstname=b.first AND a.lastname=b.last

感谢你的帮助！