假设我的Postgres数据库中有两个表:
create table transactions
(
id bigint primary key,
doc_id bigint not null,
-- lots of other columns...
amount numeric not null
);
-- same columns
create temporary table updated_transactions
(
id bigint primary key,
doc_id bigint not null,
-- lots of other columns...
amount numeric not null
);
两个表只有一个主键,没有唯一索引。
我需要使用以下规则将updated_transactions
中的行追加到transactions
中:
transactions
和updated_transactions
中的列值不匹配doc_id
等其他列(amount
除外)应匹配amount
和id
列id
中的 updated_transactions
值来自序列。
业务对象只填充updated_transactions
然后合并
使用upsert查询将新行或更新的行放入transactions
。
所以我原来保持不变的交易保持id
完整,并保持更新
被分配了新的id
s。
在MSSQL和Oracle中,它将是merge
语句,类似于:
merge into transactions t
using updated_transactions ut on t.doc_id = ut.doc_id, ...
when matched then
update set t.id = ut.id, t.amount = ut.amount
when not matched then
insert (t.id, t.doc_id, ..., t.amount)
values (ut.id, ut.doc_id, ..., ut.amount);
在PostgreSQL中,我想它应该是这样的:
insert into transactions(id, doc_id, ..., amount)
select coalesce(t.id, ut.id), ut.doc_id, ... ut.amount
from updated_transactions ut
left join transactions t on t.doc_id = ut.doc_id, ....
on conflict
on constraint transactions_pkey
do update
set amount = excluded.amount, id = excluded.id
问题在于do update
子句:excluded.id
是一个旧值
来自transactions
表格,而我需要updated_transactions
的新值。
ut.id
子句, do update
值是不可访问的,而且我唯一能做到的
use是excluded
行。但是excluded
行只有coalesce(t.id, ut.id)
表达式,返回现有行的旧id
值。
是否可以使用upsert查询更新id
和amount
列?
答案 0 :(得分:2)
在您用作键的那些列上创建唯一索引,并在upsert表达式中传递其名称,以便它使用它而不是pkey。
然后,如果没有找到匹配项,它将使用updated_transactions
中的ID插入行。如果找到匹配项,则可以使用excluded.id从updated_transactions
获取ID。
我认为left join transactions
是多余的。
所以看起来有点像这样:
insert into transactions(id, doc_id, ..., amount)
select ut.id, ut.doc_id, ... ut.amount
from updated_transactions ut
on conflict
on constraint transactions_multi_column_unique_index
do update
set amount = excluded.amount, id = excluded.id
答案 1 :(得分:1)
看起来可以使用writable CTEs而不是普通的upsert来完成任务。
首先,我会发布更简单的查询版本,以回答原始问题。此解决方案假定doc_id, unit_id
列代表候选键,但不要求在这些列上使用唯一索引。
测试数据:
create temp table transactions
(
id bigint primary key,
doc_id bigint,
unit_id bigint,
amount numeric
);
create temp table updated_transactions
(
id bigint primary key,
doc_id bigint,
unit_id bigint,
amount numeric
);
insert into transactions(id, doc_id, unit_id, amount)
values (1, 1, 1, 10), (2, 1, 2, 15), (3, 1, 3, 10);
insert into updated_transactions(id, doc_id, unit_id, amount)
values (6, 1, 1, 11), (7, 1, 2, 15), (8, 1, 4, 20);
将updated_transactions
合并到transactions
的查询:
with new_values as
(
select ut.id new_id, t.id old_id, ut.doc_id, ut.unit_id, ut.amount
from updated_transactions ut
left join transactions t
on t.doc_id = ut.doc_id and t.unit_id = ut.unit_id
),
updated as
(
update transactions tr
set id = nv.new_id, amount = nv.amount
from new_values nv
where id = nv.old_id
returning tr.*
)
insert into transactions(id, doc_id, unit_id, amount)
select ut.new_id, ut.doc_id, ut.unit_id, ut.amount
from new_values ut
where ut.new_id not in (select id from updated);
结果:
select * from transactions
-- id | doc_id | unit_id | amount
------+--------+---------+-------
-- 3 | 1 | 3 | 10 -- not changed
-- 6 | 1 | 1 | 11 -- updated
-- 7 | 1 | 2 | 15 -- updated
-- 8 | 1 | 4 | 20 -- inserted
在我的真实应用中doc_id, unit_id
并不总是唯一的,因此它们不代表候选键。为了匹配行,我考虑了行号,为按id
s排序的行计算。所以这是我的第二个解决方案。
测试数据:
-- the tables are the same as above
insert into transactions(id, doc_id, unit_id, amount)
values (1, 1, 1, 10), (2, 1, 1, 15), (3, 1, 3, 10);
insert into updated_transactions(id, doc_id, unit_id, amount)
values (6, 1, 1, 11), (7, 1, 1, 15), (8, 1, 4, 20);
合并查询:
with trans as
(
select id, doc_id, unit_id, amount,
row_number() over(partition by doc_id, unit_id order by id) row_num
from transactions
),
updated_trans as
(
select id, doc_id, unit_id, amount,
row_number() over(partition by doc_id, unit_id order by id) row_num
from updated_transactions
),
new_values as
(
select ut.id new_id, t.id old_id, ut.doc_id, ut.unit_id, ut.amount
from updated_trans ut
left join trans t
on t.doc_id = ut.doc_id and t.unit_id = ut.unit_id and t.row_num = ut.row_num
),
updated as
(
update transactions tr
set id = nv.new_id, amount = nv.amount
from new_values nv
where id = nv.old_id
returning tr.*
)
insert into transactions(id, doc_id, unit_id, amount)
select ut.new_id, ut.doc_id, ut.unit_id, ut.amount
from new_values ut
where ut.new_id not in (select id from updated);
结果:
select * from transactions;
-- id | doc_id | unit_id | amount
------+--------+---------+-------
-- 3 | 1 | 3 | 10 -- not changed
-- 6 | 1 | 1 | 11 -- updated
-- 7 | 1 | 1 | 15 -- updated
-- 8 | 1 | 4 | 20 -- inserted
参考文献: