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?


当前回答

完美的工作! !

带有JOIN的POSTGRE SQL - UPDATE

下面代码-检查列和id的定位如下:

如果你把它放在下面,那么只有它会工作!

---IF you want to update FIRST table
UPDATE table1
SET attribute1 = table2.attribute1
FROM table2
WHERE table2.product_ID = table1.product_ID;

OR

---IF you want to update SECOND table
UPDATE table2
SET attribute1 = table1.attribute1
FROM table1
WHERE table1.product_ID = table2.product_ID;

其他回答

第一个表名:tbl_table1 (tab1)。 第二表名:tbl_table2 (tab2)。

将tbl_table1的ac_status列设置为“INACTIVE”

update common.tbl_table1 as tab1
set ac_status= 'INACTIVE' --tbl_table1's "ac_status"
from common.tbl_table2 as tab2
where tab1.ref_id= '1111111' 
and tab2.rel_type= 'CUSTOMER';

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。

——目标:使用join (postgres)更新选定的列——

UPDATE table1 t1      
SET    column1 = 'data' 
FROM   table1    
       RIGHT JOIN table2   
               ON table2.id = table1.id   
WHERE  t1.id IN     
(SELECT table2.id   FROM   table2   WHERE  table2.column2 = 12345) 

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

1760行受影响

执行时间:29.18秒

UPDATE语法为:

[ WITH [ RECURSIVE ] with_query [, ...] ]
UPDATE [ ONLY ] table [ [ AS ] alias ]
    SET { column = { expression | DEFAULT } |
          ( column [, ...] ) = ( { expression | DEFAULT } [, ...] ) } [, ...]
    [ FROM from_list ]
    [ WHERE condition | WHERE CURRENT OF cursor_name ]
    [ RETURNING * | output_expression [ [ AS ] output_name ] [, ...] ]

在你的情况下,我认为你想要的是:

UPDATE vehicles_vehicle AS v 
SET price = s.price_per_vehicle
FROM shipments_shipment AS s
WHERE v.shipment_id = s.id 

或者如果你需要连接两个或多个表:

UPDATE table_1 t1
SET foo = 'new_value'
FROM table_2 t2
    JOIN table_3 t3 ON t3.id = t2.t3_id
WHERE
    t2.id = t1.t2_id
    AND t3.bar = True;