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