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

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;

其他回答

第一种方式比第二种方式慢。

第一:

DO $$ 
DECLARE 
  page int := 10000;
  min_id bigint; max_id bigint;
BEGIN
  SELECT max(id),min(id) INTO max_id,min_id FROM opportunities;
  FOR j IN min_id..max_id BY page LOOP 
    UPDATE opportunities SET sec_type = 'Unsec'
    FROM opportunities AS opp
    INNER JOIN accounts AS acc
    ON opp.account_id = acc.id
    WHERE acc.borrower = true
    AND opp.sec_type IS NULL
    AND opp.id >= j AND opp.id < j+page;
    COMMIT;            
  END LOOP;
END; $$;

第二:

DO $$ 
DECLARE 
  page int := 10000;
  min_id bigint; max_id bigint;
BEGIN
  SELECT max(id),min(id) INTO max_id,min_id FROM opportunities;
  FOR j IN min_id..max_id BY page LOOP
    UPDATE opportunities AS opp 
    SET sec_type = 'Unsec'
    FROM accounts AS acc
    WHERE opp.account_id = acc.id
    AND opp.sec_type IS NULL
    AND acc.borrower = true 
    AND opp.id >= j AND opp.id < j+page;
    COMMIT;            
  END LOOP;
END; $$;

下面的链接提供了一个示例,可以帮助您更好地理解如何使用update和join postgres。

UPDATE product
SET net_price = price - price * discount
FROM
product_segment
WHERE
product.segment_id = product_segment.id;

参见:http://www.postgresqltutorial.com/postgresql-update-join/

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;

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

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;