How to do an update + join in PostgreSQL?
基本上,我想这样做:
1 2 3 | UPDATE vehicles_vehicle v JOIN shipments_shipment s ON v.shipment_id=s.id SET v.price=s.price_per_vehicle; |
我很确定它可以在MySQL(我的背景)中工作,但它似乎不适用于postgres。 我得到的错误是:
1 2 3 | ERROR: syntax error at OR near"join" LINE 1: UPDATE vehicles_vehicle v JOIN shipments_shipment s ON v.shi... ^ |
当然有一种简单的方法可以做到这一点,但我找不到合适的语法。 那么,我如何在PostgreSQL中编写这个?
UPDATE语法是:
1 2 3 4 5 6 7 | [ 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,我认为你想要这个:
1 2 3 4 | UPDATE vehicles_vehicle AS v SET price = s.price_per_vehicle FROM shipments_shipment AS s WHERE v.shipment_id = s.id |
让我通过我的例子再解释一下。
任务:正确的信息,其中abiturients(即将离开中学的学生)早些时候向大学提交申请,而不是他们获得学校证书(是的,他们获得的证书早于他们颁发的证书(通过指定的证书日期)。所以,我们将增加申请提交日期以适应证书签发日期。
从而。下一个类似MySQL的声明:
1 2 3 4 5 6 7 8 9 10 | 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; |
以这种方式成为类似PostgreSQL的
1 2 3 4 5 6 | 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 |
正如您所看到的,原始子查询
Mark Byers的答案在这种情况下是最佳的。
虽然在更复杂的情况下,您可以使用返回rowid和计算值的select查询,并将其附加到更新查询,如下所示:
1 2 3 4 5 6 7 8 9 10 | 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 t1 SET VALUE = t.calculatedvalue FROM t WHERE id = t.rowid |
此方法允许您开发和测试您的选择查询,并分两步将其转换为更新查询。
因此,在您的情况下,结果查询将是:
1 2 3 4 5 6 7 8 9 | 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,列别名是必需的,否则PostgreSQL会抱怨列名的含糊不清。
对于那些真正想要做
1 2 3 4 5 6 | 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,您可以在等号右侧的
等号左侧的字段不需要表引用,因为它们被认为来自原始的"a"表。
对于那些想要进行JOIN的人来说,只更新你的联接返回的行:
1 2 3 4 5 6 7 8 | 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,但只是通过评论..因为获得正确的结果发布新的答案,这是至关重要的
开始了:
1 2 3 4 | UPDATE vehicles_vehicle v SET price=s.price_per_vehicle FROM shipments_shipment s WHERE v.shipment_id=s.id; |
简单,因为我可以做到。多谢你们!
也可以这样做: del>
1 2 3 4 5 | -- Doesn't work apparently UPDATE vehicles_vehicle SET price=s.price_per_vehicle FROM vehicles_vehicle v JOIN shipments_shipment s ON v.shipment_id=s.id; |
但是那时你已经有两次车辆表,并且你只允许别名一次,你不能在"set"部分使用别名。 del>
下面的链接有一个示例可以解决并帮助更好地理解如何使用postgres
1 2 3 4 5 6 | UPDATE product SET net_price = price - price * discount FROM product_segment WHERE product.segment_id = product_segment.id; |
请参阅:http://www.postgresqltutorial.com/postgresql-update-join/
这是一个简单的SQL,它使用Name中的Middle_Name字段更新Name3表上的Mid_Name:
1 2 3 4 | UPDATE name3 SET mid_name = name.middle_name FROM name WHERE name3.person_id = name.person_id; |