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;

我举个例子再解释一下。

任务:正确的信息，在哪里abiturients(即将离开中学的学生)提交申请大学早于他们获得学校证书(是的，他们获得证书早于他们颁发的证书(指定证书日期)。因此，我们将增加申请提交日期，以适应证书颁发日期。

因此。下一个mysql类语句:

UPDATE applications a
JOIN (
    SELECT ap.id, ab.certificate_issued_at
    FROM abiturients ab
    JOIN applications ap 
    ON ab.id = ap.abiturient_id 
    WHERE ap.documents_taken_at::date < ab.certificate_issued_at
) b
ON a.id = b.id
SET a.documents_taken_at = b.certificate_issued_at;

以这样的方式变得像postgresql

UPDATE applications a
SET documents_taken_at = b.certificate_issued_at         -- we can reference joined table here
FROM abiturients b                                       -- joined table
WHERE 
    a.abiturient_id = b.id AND                           -- JOIN ON clause
    a.documents_taken_at::date < b.certificate_issued_at -- Subquery WHERE

可以看到，原来的子查询JOIN的ON子句已经变成了WHERE条件之一，由AND与其他子查询连接，这些子查询已经从子查询中移动，没有任何变化。并且不再需要将表与自身连接(就像在子查询中那样)。

对于那些真正想要做JOIN的人，你也可以使用:

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;

如果需要，可以在等号右侧的SET部分中使用a_alias。 等号左边的字段不需要表引用，因为它们被认为来自原始的“a”表。

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。

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。