我有桌子：
id | name
1 | a,b,c
2 | b
我想要这样的输出：
id | name
1 | a
1 | b
1 | c
2 | b
问问题
229168 次
10 回答
157
如果您可以创建一个数字表，其中包含从 1 到要拆分的最大字段的数字，您可以使用如下解决方案：
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
numbers inner join tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
请在此处查看小提琴。
如果您无法创建表，则解决方案可以是：
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
(select 1 n union all
select 2 union all select 3 union all
select 4 union all select 5) numbers INNER JOIN tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
一个例子小提琴是here。
于 2013-07-30T09:05:32.817 回答
14
如果该name列是 JSON 数组（如'["a","b","c"]'），那么您可以使用JSON_TABLE()提取/解压缩它（自 MySQL 8.0.4 起可用）：
select t.id, j.name
from mytable t
join json_table(
t.name,
'$[*]' columns (name varchar(50) path '$')
) j;
结果：
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
如果您以简单的 CSV 格式存储值，则首先需要将其转换为 JSON：
select t.id, j.name
from mytable t
join json_table(
replace(json_array(t.name), ',', '","'),
'$[*]' columns (name varchar(50) path '$')
) j
结果：
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
于 2019-12-06T10:10:28.877 回答
7
DELIMITER $$
CREATE FUNCTION strSplit(x VARCHAR(65000), delim VARCHAR(12), pos INTEGER)
RETURNS VARCHAR(65000)
BEGIN
DECLARE output VARCHAR(65000);
SET output = REPLACE(SUBSTRING(SUBSTRING_INDEX(x, delim, pos)
, LENGTH(SUBSTRING_INDEX(x, delim, pos - 1)) + 1)
, delim
, '');
IF output = '' THEN SET output = null; END IF;
RETURN output;
END $$
CREATE PROCEDURE BadTableToGoodTable()
BEGIN
DECLARE i INTEGER;
SET i = 1;
REPEAT
INSERT INTO GoodTable (id, name)
SELECT id, strSplit(name, ',', i) FROM BadTable
WHERE strSplit(name, ',', i) IS NOT NULL;
SET i = i + 1;
UNTIL ROW_COUNT() = 0
END REPEAT;
END $$
DELIMITER ;
于 2013-07-30T09:05:02.587 回答
7
这是我的尝试：第一个选择将 csv 字段呈现给拆分。使用递归 CTE，我们可以创建一个数字列表，限制在 csv 字段中的术语数量。术语的数量只是 csv 字段的长度与其本身的长度之差，所有分隔符都已删除。然后加入这个数字， substring_index 提取该术语。
with recursive
T as ( select 'a,b,c,d,e,f' as items),
N as ( select 1 as n union select n + 1 from N, T
where n <= length(items) - length(replace(items, ',', '')))
select distinct substring_index(substring_index(items, ',', n), ',', -1)
group_name from N, T
于 2019-08-01T14:07:42.050 回答
5
我的变体：以表名、字段名和分隔符作为参数的存储过程。灵感来自http://www.marcogoncalves.com/2011/03/mysql-split-column-string-into-rows/
delimiter $$
DROP PROCEDURE IF EXISTS split_value_into_multiple_rows $$
CREATE PROCEDURE split_value_into_multiple_rows(tablename VARCHAR(20),
id_column VARCHAR(20), value_column VARCHAR(20), delim CHAR(1))
BEGIN
DECLARE id INT DEFAULT 0;
DECLARE value VARCHAR(255);
DECLARE occurrences INT DEFAULT 0;
DECLARE i INT DEFAULT 0;
DECLARE splitted_value VARCHAR(255);
DECLARE done INT DEFAULT 0;
DECLARE cur CURSOR FOR SELECT tmp_table1.id, tmp_table1.value FROM
tmp_table1 WHERE tmp_table1.value IS NOT NULL AND tmp_table1.value != '';
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
SET @expr = CONCAT('CREATE TEMPORARY TABLE tmp_table1 (id INT NOT NULL, value VARCHAR(255)) ENGINE=Memory SELECT ',
id_column,' id, ', value_column,' value FROM ',tablename);
PREPARE stmt FROM @expr;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
DROP TEMPORARY TABLE IF EXISTS tmp_table2;
CREATE TEMPORARY TABLE tmp_table2 (id INT NOT NULL, value VARCHAR(255) NOT NULL) ENGINE=Memory;
OPEN cur;
read_loop: LOOP
FETCH cur INTO id, value;
IF done THEN
LEAVE read_loop;
END IF;
SET occurrences = (SELECT CHAR_LENGTH(value) -
CHAR_LENGTH(REPLACE(value, delim, '')) + 1);
SET i=1;
WHILE i <= occurrences DO
SET splitted_value = (SELECT TRIM(SUBSTRING_INDEX(
SUBSTRING_INDEX(value, delim, i), delim, -1)));
INSERT INTO tmp_table2 VALUES (id, splitted_value);
SET i = i + 1;
END WHILE;
END LOOP;
SELECT * FROM tmp_table2;
CLOSE cur;
DROP TEMPORARY TABLE tmp_table1;
END; $$
delimiter ;
用法示例（标准化）：
CALL split_value_into_multiple_rows('my_contacts', 'contact_id', 'interests', ',');
CREATE TABLE interests (
interest_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
interest VARCHAR(30) NOT NULL
) SELECT DISTINCT value interest FROM tmp_table2;
CREATE TABLE contact_interest (
contact_id INT NOT NULL,
interest_id INT NOT NULL,
CONSTRAINT fk_contact_interest_my_contacts_contact_id FOREIGN KEY (contact_id) REFERENCES my_contacts (contact_id),
CONSTRAINT fk_contact_interest_interests_interest_id FOREIGN KEY (interest_id) REFERENCES interests (interest_id)
) SELECT my_contacts.contact_id, interests.interest_id
FROM my_contacts, tmp_table2, interests
WHERE my_contacts.contact_id = tmp_table2.id AND interests.interest = tmp_table2.value;
于 2017-02-05T12:57:44.177 回答
2
最初的问题一般是针对 MySQL 和 SQL 的。下面的示例适用于 MySQL 的新版本。不幸的是，不可能在任何 SQL 服务器上工作的通用查询。有些服务器不支持 CTE，有些不支持 substring_index，还有一些内置函数可以将字符串拆分为多行。
---答案如下---
当服务器不提供内置功能时，递归查询很方便。它们也可能成为瓶颈。
以下查询是在 MySQL 版本 8.0.16 上编写和测试的。它不适用于版本 5.7-。旧版本不支持公用表表达式 (CTE)，因此不支持递归查询。
with recursive
input as (
select 1 as id, 'a,b,c' as names
union
select 2, 'b'
),
recurs as (
select id, 1 as pos, names as remain, substring_index( names, ',', 1 ) as name
from input
union all
select id, pos + 1, substring( remain, char_length( name ) + 2 ),
substring_index( substring( remain, char_length( name ) + 2 ), ',', 1 )
from recurs
where char_length( remain ) > char_length( name )
)
select id, name
from recurs
order by id, pos;
于 2019-05-29T20:11:02.430 回答
1
CREATE PROCEDURE `getVal`()
BEGIN
declare r_len integer;
declare r_id integer;
declare r_val varchar(20);
declare i integer;
DECLARE found_row int(10);
DECLARE row CURSOR FOR select length(replace(val,"|","")),id,val from split;
create table x(id int,name varchar(20));
open row;
select FOUND_ROWS() into found_row ;
read_loop: LOOP
IF found_row = 0 THEN
LEAVE read_loop;
END IF;
set i = 1;
FETCH row INTO r_len,r_id,r_val;
label1: LOOP
IF i <= r_len THEN
insert into x values( r_id,SUBSTRING(replace(r_val,"|",""),i,1));
SET i = i + 1;
ITERATE label1;
END IF;
LEAVE label1;
END LOOP label1;
set found_row = found_row - 1;
END LOOP;
close row;
select * from x;
drop table x;
END
于 2016-10-15T08:26:30.673 回答
0
这是我的解决方案
-- Create the maximum number of words we want to pick (indexes in n)
with recursive n(i) as (
select
1 i
union all
select i+1 from n where i < 1000
)
select distinct
s.id,
s.oaddress,
-- n.i,
-- use the index to pick the nth word, the last words will always repeat. Remove the duplicates with distinct
if(instr(reverse(trim(substring_index(s.oaddress,' ',n.i))),' ') > 0,
reverse(substr(reverse(trim(substring_index(s.oaddress,' ',n.i))),1,
instr(reverse(trim(substring_index(s.oaddress,' ',n.i))),' '))),
trim(substring_index(s.oaddress,' ',n.i))) oth
from
app_schools s,
n
于 2018-11-14T20:25:54.050 回答
0
最佳实践。结果：
SELECT
SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
(
SELECT @xi:=@xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT @xi:=-1) xc0
) a
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
首先，创建一个数字表：
SELECT @xi:=@xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT @xi:=-1) xc0;
| help_id |
| --- |
| 0 |
| 1 |
| 2 |
| 3 |
| ... |
| 24 |
其次，只需拆分str：
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
numbers_table
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
| oid |
| --- |
| ab |
| bc |
| cd |
于 2020-08-18T07:00:48.127 回答
-3
SELECT id, unnest(string_to_array(name, ',')) AS names
FROM datatable
希望这会有所帮助：D
于 2021-08-19T18:46:02.870 回答