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

对于那些想要做一个JOIN,更新你的连接返回行使用:

UPDATE a
SET price = b_alias.unit_price
FROM      a AS a_alias
LEFT JOIN b AS b_alias ON a_alias.b_fk = b_alias.id
WHERE a_alias.unit_name LIKE 'some_value' 
AND a.id = a_alias.id
--the below line is critical for updating ONLY joined rows
AND a.pk_id = a_alias.pk_id;

这是上面提到的,但只是通过一个评论..因为它是至关重要的,以获得正确的结果张贴新的答案,工作

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

在上面所有的答案中添加一些非常重要的东西,当你想要更新连接表时,你可能会遇到两个问题:

您不能使用您想要更新的表来JOIN另一个表 Postgres希望在JOIN之后有一个ON子句,因此不能只使用where子句。

这意味着基本上,以下查询是无效的:

UPDATE join_a_b
SET count = 10
FROM a
JOIN b on b.id = join_a_b.b_id -- Not valid since join_a_b is used here
WHERE a.id = join_a_b.a_id
AND a.name = 'A'
AND b.name = 'B'
UPDATE join_a_b
SET count = 10
FROM a
JOIN b -- Not valid since there is no ON clause
WHERE a.id = join_a_b.a_id 
AND b.id = join_a_b.b_id
a.name = 'A'
AND b.name = 'B'

相反,你必须像这样使用FROM子句中的所有表:

UPDATE join_a_b
SET count = 10
FROM a, b
WHERE a.id = join_a_b.a_id 
AND b.id = join_a_b.b_id 
AND a.name = 'A'
AND b.name = 'B'

这对一些人来说可能很简单,但我被这个问题困住了,想知道发生了什么,所以希望它能帮助到其他人。

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/

开始吧:

update vehicles_vehicle v
set price=s.price_per_vehicle
from shipments_shipment s
where v.shipment_id=s.id;

我能做到的最简单。

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';