[英]Aggregating consecutive rows in SQL
鑒於 sql 表(我使用的是 SQLite3):
CREATE TABLE person(name text, number integer);
並填充值:
insert into person values
('Leandro', 2),
('Leandro', 4),
('Maria', 8),
('Maria', 16),
('Jose', 32),
('Leandro', 64);
我想要的是獲取number
列的總和,但僅限於連續行,以便我可以得到保持原始插入順序的結果:
Leandro|6
Maria|24
Jose|32
Leandro|64
到目前為止我得到的“最接近”是:
select name, sum(number) over(partition by name) from person order by rowid;
但它清楚地表明我對 SQL 的理解還很遠,因為缺少最重要的功能(連續行的分組和求和),但至少順序是:-):
Leandro|70
Leandro|70
Maria|24
Maria|24
Jose|32
Leandro|70
最好答案不應該要求創建臨時表,因為預計輸出的順序總是與數據插入的順序相同。
您可以使用窗口函數來做到這一點:
然后按組分組並聚合:
select name, sum(number) total
from (
select *, sum(flag) over (order by rowid) grp
from (
select *, rowid, name <> lag(name, 1, '') over (order by rowid) flag
from person
)
)
group by grp
請參閱演示。
結果:
> name | total
> :------ | ----:
> Leandro | 6
> Maria | 24
> Jose | 32
> Leandro | 64
這是一種間隙和島嶼問題。 為此,您可以使用行號的差異:
select name, sum(number)
from (select p.*,
row_number() over (order by number) as seqnum,
row_number() over (partition by name order by number) as seqnum_1
from person p
) p
group by name, (seqnum - seqnum_1)
order by. min(number);
為什么這行得通有點難以解釋。 但是,當您查看子查詢的結果時,它變得非常明顯。 當名稱不變時,相鄰行的行號差異是恆定的。
這是一個 db<>fiddle。
我會將 create table 語句更改為以下內容:
CREATE TABLE person(id integer, firstname nvarchar(255), number integer);
然后你可以插入你的數據:
insert into person values
(1, 'Leandro', 2),
(2, 'Leandro', 4),
(3, 'Maria', 8),
(4, 'Maria', 16),
(5, 'Jose', 32),
(6, 'Leandro', 64);
之后,您可以通過以下方式查詢數據:
SELECT firstname, value FROM (
SELECT p.id, p.firstname, p.number, LAG(p.firstname) over (ORDER BY p.id) as prevname,
CASE
WHEN firstname LIKE LEAD(p.firstname) over (ORDER BY p.id) THEN number + LEAD(p.number) over(ORDER BY p.id)
ELSE number
END as value
FROM Person p
) AS temp
WHERE temp.firstname <> temp.prevname OR
temp.prevname IS NULL
為了更好地理解查詢,您可以單獨運行子查詢:
SELECT p.id, p.firstname, p.number, LEAD(p.firstname) over (ORDER BY p.id) as nextname, LAG(p.firstname) over (ORDER BY p.id) as prevname,
CASE
WHEN firstname LIKE LEAD(p.firstname) over (ORDER BY p.id) THEN number + LEAD(p.number) over(ORDER BY p.id)
ELSE number
END as value
FROM Person p
基於 Gordon Linoff 的回答 ( https://stackoverflow.com/a/64727401/1721672 ),我將內部選擇提取為 CTE,以下查詢效果很好:
with p(name, number, seqnum, seqnum_1) as
(select name, number,
row_number() over (order by number) as seqnum,
row_number() over (partition by name order by number) as seqnum_1
from person)
select
name, sum(number)
from
p
group by
name, (seqnum - seqnum_1)
order by
min(number);
產生預期結果:
Leandro|6
Maria|24
Jose|32
Leandro|64
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