基本上,我想这样做:
update vehicles_vehicle v
join shipments_shipment s on v.shipment_id=s.id
set v.price=s.price_per_vehicle;
我很确定这在MySQL(我的背景)中可以工作,但在postgres中似乎不起作用。我得到的错误是:
ERROR: syntax error at or near "join"
LINE 1: update vehicles_vehicle v join shipments_shipment s on v.shi...
^
当然有一个简单的方法来做到这一点,但我找不到合适的语法。那么,我该如何在PostgreSQL中写这个呢?
在这种情况下,Mark Byers的答案是最优的。
尽管在更复杂的情况下,你可以使用select查询返回rowids和计算值,并将其附加到更新查询,如下所示:
with t as (
-- Any generic query which returns rowid and corresponding calculated values
select t1.id as rowid, f(t2, t2) as calculatedvalue
from table1 as t1
join table2 as t2 on t2.referenceid = t1.id
)
update table1
set value = t.calculatedvalue
from t
where id = t.rowid
这种方法允许您开发和测试选择查询,并在两个步骤中将其转换为更新查询。
所以在你的例子中,结果查询将是:
with t as (
select v.id as rowid, s.price_per_vehicle as calculatedvalue
from vehicles_vehicle v
join shipments_shipment s on v.shipment_id = s.id
)
update vehicles_vehicle
set price = t.calculatedvalue
from t
where id = t.rowid
请注意,列别名是必须的,否则PostgreSQL将抱怨列名的模糊性。
在PostGRE SQL / AWS (SQL工作台)中使用另一个表更新一个表。
在PostGRE SQL中,你需要在UPDATE Query中使用join:
UPDATE TABLEA set COLUMN_FROM_TABLEA = COLUMN_FROM_TABLEB FROM TABLEA,TABLEB WHERE FILTER_FROM_TABLEA = FILTER_FROM_TABLEB;
Example:
Update Employees Set Date_Of_Exit = Exit_Date_Recorded , Exit_Flg = 1 From Employees, Employee_Exit_Clearance Where Emp_ID = Exit_Emp_ID
表A - Employees列表A - Date_Of_Exit,Emp_ID,Exit_Flg表B是- Employee_Exit_Clearance列表B - Exit_Date_Recorded,Exit_Emp_ID
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