「no no column」from select

Question

我有兩個表，分別是firstname,lastname和amount。

我可以做一個左連接：

select * from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname); 

這給了良好的效果：

|# |firstname  |lastname |amount|dp.firstname |dp.lastname |dp.amount 
|----+--------------+------------+------+-------------+-------------+--------- 
|1 |saumeh synah |s*****  |50.0 |    |    | 
|2 |Neda   |M*****  |1000.0|    |    | 
|3 |Mansoor  |B********** |100.0 |    |    | 
|4 |Hanna   |W****  |50.0 |Hanna  |W****  |50.0 
|5 |Kristen  |A****  |40.0 |Kristen  |A****  |40.0 
|6 |David   |B******  |10.0 |David  |B******  |10.0 

現在我想選擇只是從dp缺少的行。

所以我寫：

select * from ( 
select * from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname)) 
where dp.amount = null; 

，但我得到的錯誤

no such column: dp.amount 

爲什麼不呢？

Answer 1

錯誤歸因於派生表。剛剛嘗試一個簡單的查詢：

select spi.* 
from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname) 
where dp.amount IS null ; 

Answer 2

dp.amount不存在如您已在select * from (select * from...)

內心的組括號中創建了派生表中創建一個派生表，以便大家參考與DP沒有按」在那些父母的外面存在。把你的where dp.amount = null放在最後一個圓括號內，並將你的分號推到外面。

select * from ( 
select * from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname) 
-- this will never find results as NULL will never equal anything 
-- where dp.amount = null 
where dp.amount IS NULL) AS D; 

Answer 3

實際上，您不需要使用sub query，其中dp不存在。

select * from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname) 
WHERE dp.amount IS NULL 

編輯：這裏有一些nice examples about JOIN

Answer 4

不知道如果我是100％的理解，但將這項工作？

select dp.firstname,dp.lastname,dp.amount 
from dp,spi 
where dp.firstname=spi.firstname 
and dp.lastname=spi.lastname 
and dp.amount is null; 

Answer 5

select * from (
    select * from .... 
    where dp.amount is null; 
) 

Answer 6

可能是更好的名字嵌套查詢：

select * from ( 
select spi.firstname, spi.lastname, spi.amount, dp.firstname f, dp.lastname l, dp.amount a from spi left join dp 
on spi.amount = dp.amount 
and lower(spi.firstname) = lower(dp.firstname) 
and lower(spi.lastname) = lower(dp.lastname)) as x 
where x.a = null;