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。

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。

下面是一个简单的SQL，使用Name中的Middle_Name字段更新Name3表中的Mid_Name:

update name3
set mid_name = name.middle_name
from name
where name3.person_id = name.person_id;

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。

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

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。

——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

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

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;