PostgreSQL:使用聚合函数更新

New*_*elp 9 sql postgresql aggregate-functions

我想使用聚合函数更新paneldata设置列的表ibase.

UPDATE paneldata p
SET ibase=SUM(1/i.dist)
FROM ibaselang i
WHERE p.gid=i.gid
AND i.instp<p.period
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这导致了 ERROR: aggregate functions are not allowed in UPDATE

表定义

CREATE TABLE public.ibaselang
(
  gid integer,
  dist double precision,
  buildid integer,
  instp smallint
)
WITH (
  OIDS=FALSE
);
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解决方法

不幸的是,我不知道如何WHERE在子查询中实现我的函数.

Dat*_*ter 12

这是一个通用的示例。

UPDATE public.customer_value cv
SET total_value = sub_q.sum_val 
FROM 
    (
    SELECT SUM(order_amount) AS sum_val, o.customer_id 
    FROM public.orders AS o
    GROUP BY o.customer_id
    ) AS sub_q
WHERE sub_q.customer_id = cv.customer_id;
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如果您想完全尝试该示例,则可以创建如下的虚拟数据:

CREATE TABLE public.customer_value
(
  customer_id int 
, total_value numeric (10,2)
);

CREATE TABLE public.orders 
(
  customer_id int
, order_amount numeric(10,2)
);

INSERT INTO public.customer_value
(customer_id)
VALUES 
  (1)
, (2);


INSERT INTO public.orders
(customer_id, order_amount)
VALUES 
 (1, 10)
,(1, 10)
,(2, 7.5)
,(2, 7.5);
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Ham*_*one 5

我不知道这有多有效,但应该可以:

with updates as (
  select
    p.gid, sum (i.dist) as dist
  from
    paneldata p
    join ibaselang i on
      p.id = i.gid and
      i.instp < p.period
  where
    i.dist != 0
  group by
    p.gid
)
update paneldata p
set
  ibase = 1 / u.dist
from
  updates u
where
  p.gid = u.pid and
  u.dist != 0
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有关添加的其他一些注意事项:

  1. 如果您i.dist有很多零,where理论上删除子句中的零值可以加快查询速度
  2. 您可能存在除以零错误的潜在漏洞-也在update语句中处理。也许您的数据没有这种情况,但是最好进行处理以确保


sag*_*agi 4

尝试使用这样的相关查询:

UPDATE paneldata p
SET p.ibase=(select SUM(1/i.dist)
             FROM ibaselang i
             WHERE p.gid=i.gid
             AND i.instp<p.period)
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  • 但这将更新所有行!,必须遵守哪些条件? (6认同)