我有一张看起来像这样的表：
ID | name | details
---------------------------
1.3.1-3 | Jack | a
5.4.1-2 | John | b
1.4.5 | Alex | c
如何分割它：
ID | name | details
---------------------------
1.3.1 | Jack | a
1.3.2 | Jack | a
1.3.3 | Jack | a
5.4.1 | John | b
5.4.2 | John | b
1.4.5 | Alex | c
如何在postgresql中执行此操作？
答案 0 :(得分：2)
CREATE TABLE tosplit
( id text NOT NULL
, name text
, details text
);
INSERT INTO tosplit( id , name , details ) VALUES
( '1.3.1-3' , 'Jack' , 'a' )
,( '5.4.1-2' , 'John' , 'b' )
,( '1.4.5' , 'Alex' , 'c' )
WITH zzz AS (
SELECT id
, regexp_replace(id, '([0-9\.]+\.)([0-9]+)-([0-9]+)', e'\\1', e'g') AS one
, regexp_replace(id, '([0-9\.]+\.)([0-9]+)-([0-9]+)', e'\\2', e'g') AS two
, regexp_replace(id, '([0-9\.]+\.)([0-9]+)-([0-9]+)', e'\\3', e'g') AS three
, name
, details
FROM tosplit
)
SELECT z1.id
-- , z1.one
, z1.one || generate_series( z1.two::integer, z1.three::integer)::text AS four
, z1.name, z1.details
FROM zzz z1
WHERE z1.two <> z1.one
UNION ALL
SELECT z0.id
-- , z0.one
, z0.one AS four
, z0.name, z0.details
FROM zzz z0
WHERE z0.two = z0.one
;
结果：
CREATE TABLE
INSERT 0 3
id | four | name | details
---------+-------+------+---------
1.3.1-3 | 1.3.1 | Jack | a
1.3.1-3 | 1.3.2 | Jack | a
1.3.1-3 | 1.3.3 | Jack | a
5.4.1-2 | 5.4.1 | John | b
5.4.1-2 | 5.4.2 | John | b
1.4.5 | 1.4.5 | Alex | c
答案 1 :(得分：1)
您可以根据id和-拆分.并与生成的系列连接：
CREATE TABLE tab(
ID VARCHAR(18) NOT NULL PRIMARY KEY
,name VARCHAR(8) NOT NULL
,details VARCHAR(11) NOT NULL
);
INSERT INTO tab(ID,name,details) VALUES ('1.3.1-3','Jack','a');
INSERT INTO tab(ID,name,details) VALUES ('5.4.1-2','John','b');
INSERT INTO tab(ID,name,details) VALUES ('1.4.5','Alex','c');
INSERT INTO tab(ID,name,details) VALUES ('1.7.11-13','Joe','d');
INSERT INTO tab(ID,name,details) VALUES ('1.7-13','Smith','e');
主要查询：
;WITH cte AS
(
SELECT *,
split_part(id, '-', 1) AS prefix,
split_part(reverse(split_part(id, '-', 1)),'.',1)::int AS start,
CASE WHEN split_part(id, '-',2) <> ''
THEN split_part(id, '-', 2):: int
ELSE NULL
END AS stop
FROM tab
)
SELECT
LEFT(prefix, LENGTH(prefix) - strpos(reverse(prefix), '.')) || '.' || n::text AS id,
name,
details
FROM cte
CROSS JOIN LATERAL generate_series(start,COALESCE(stop, start)) AS sub(n);
输出：
╔═════════╦════════╦═════════╗
║ id ║ name ║ details ║
╠═════════╬════════╬═════════╣
║ 1.3.1 ║ Jack ║ a ║
║ 1.3.2 ║ Jack ║ a ║
║ 1.3.3 ║ Jack ║ a ║
║ 5.4.1 ║ John ║ b ║
║ 5.4.2 ║ John ║ b ║
║ 1.4.5 ║ Alex ║ c ║
║ 1.7.11 ║ Joe ║ d ║
║ 1.7.12 ║ Joe ║ d ║
║ 1.7.13 ║ Joe ║ d ║
║ 1.7 ║ Smith ║ e ║
║ 1.8 ║ Smith ║ e ║
║ 1.9 ║ Smith ║ e ║
║ 1.10 ║ Smith ║ e ║
║ 1.11 ║ Smith ║ e ║
║ 1.12 ║ Smith ║ e ║
║ 1.13 ║ Smith ║ e ║
╚═════════╩════════╩═════════╝
答案 2 :(得分：1)
with elements as (
select id,
regexp_split_to_array(id, '(\.)') as id_elements,
name,
details
from the_table
), bounds as (
select id,
case
when strpos(id, '-') = 0 then 1
else split_part(id_elements[cardinality(id_elements)], '-', 1)::int
end as start_value,
case
when strpos(id, '-') = 0 then 1
else split_part(id_elements[cardinality(id_elements)], '-', 2)::int
end as end_value,
case
when strpos(id, '-') = 0 then id
else array_to_string(id_elements[1:cardinality(id_elements)-1], '.')
end as base_id,
name,
details
from elements
)
select b.base_id||'.'||c.cnt as new_id,
b.name,
b.details,
count(*) over (partition by b.base_id) as num_rows
from bounds b
cross join lateral generate_series(b.start_value, b.end_value) as c (cnt)
order by num_rows desc, c.cnt;
第一个CTE只根据.拆分ID。然后，第二个CTE计算每个ID的起始值和结束值，并从实际ID值“剥离”范围定义，以获得可以与最终select语句中的实际行索引连接的基数。
使用此测试数据：
insert into the_table
values
('1.3.1-3', 'Jack', 'details 1'),
('5.4.1-2', 'John', 'details 2'),
('1.4.5', 'Alex', 'details 3'),
('10.11.12.1-5', 'Peter', 'details 4'),
('1.4.10-13', 'Arthur','details 5'),
('11.12.13.14.15.16.2-7','Zaphod','details 6');
返回以下结果：
new_id | name | details | num_rows
--------------------+--------+-----------+---------
11.12.13.14.15.16.2 | Zaphod | details 6 | 6
11.12.13.14.15.16.3 | Zaphod | details 6 | 6
11.12.13.14.15.16.4 | Zaphod | details 6 | 6
11.12.13.14.15.16.5 | Zaphod | details 6 | 6
11.12.13.14.15.16.6 | Zaphod | details 6 | 6
11.12.13.14.15.16.7 | Zaphod | details 6 | 6
10.11.12.1 | Peter | details 4 | 5
10.11.12.2 | Peter | details 4 | 5
10.11.12.3 | Peter | details 4 | 5
10.11.12.4 | Peter | details 4 | 5
10.11.12.5 | Peter | details 4 | 5
1.4.10 | Arthur | details 5 | 4
1.4.11 | Arthur | details 5 | 4
1.4.12 | Arthur | details 5 | 4
1.4.13 | Arthur | details 5 | 4
1.3.1 | Jack | details 1 | 3
1.3.2 | Jack | details 1 | 3
1.3.3 | Jack | details 1 | 3
5.4.1 | John | details 2 | 2
5.4.2 | John | details 2 | 2
1.4.5.1 | Alex | details 3 | 1
使用cardinality(id_elements)需要Postgres 9.4。对于早期版本，需要将其替换为array_length(id_elements, 1))
最后一点：
如果将起始值和结束值存储在单独的（整数）列中，而不是将它们附加到ID本身，那么这将更容易批次。此模型违反了基本数据库规范化（第一范式）。
如果存储包含例如的解决方案，则该解决方案（或给出的答案中的任何解决方案）将会严重失败可以通过正确规范化数据来防止10.12.13.A-Z（非数字值）。
答案 3 :(得分：0)
您可以使用以下查询：
SELECT CASE
WHEN num = 0 THEN "ID"
ELSE CONCAT(LEFT("ID",
LENGTH("ID") + 1 -
position('.' IN REVERSE("ID"))),
num)
END,
"name",
"details"
FROM (
SELECT split_part("ID", '-', 1) AS "ID",
"name", "details",
generate_series(
CASE
WHEN position('-' in "ID") = 0 THEN 0
ELSE 1
END,
CASE
WHEN position('-' in "ID") = 0 THEN 0
ELSE CAST(split_part("ID", '-', 2) AS INT)
END) AS num
FROM mytable) AS t