sql将值拆分为多行

qnyhuwrf  于 2021-06-18  发布在  Mysql
关注(0)|答案(9)|浏览(509)

我有一张table:

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
8yparm6h

8yparm6h1#

我的变量:以表名、字段名和分隔符为参数的存储过程。灵感来自posthttp://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(规范化):

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;
1rhkuytd

1rhkuytd2#

下面是我的尝试:第一个select将csv字段显示给split。使用递归cte,我们可以创建一个数字列表,该列表限制为csv字段中的字数。术语的数量只是csv字段的长度和它本身的长度之差,所有分隔符都被删除了。然后与这些数字结合,子串索引提取出这个项。

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
yduiuuwa

yduiuuwa3#

这是我的解决办法

-- 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
pcrecxhr

pcrecxhr4#

最初的问题是针对mysql和sql的。下面的例子是针对mysql的新版本。不幸的是,不可能在任何sql server上使用通用查询。有些服务器不支持cte,有些服务器没有子字符串索引,还有一些服务器具有用于将字符串拆分为多行的内置函数。
---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。它在版本5.7-上不起作用。旧版本不支持公共表表达式(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;
nnvyjq4y

nnvyjq4y5#

如果 name 列是一个json数组(如 '["a","b","c"]' ),然后可以使用json_table()对其进行解压缩(mysql 8.0.4以后提供):

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    |

db fiddle视图
如果以简单的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    |

db fiddle视图

kqlmhetl

kqlmhetl6#

我已经从这里引用了更改过的列名。

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 ;
bksxznpy

bksxznpy7#

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
tgabmvqs

tgabmvqs8#

如果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,则可以使用以下解决方案:

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

请看这里的小提琴。
如果无法创建表,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

这里是小提琴的一个例子。

j9per5c4

j9per5c49#

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   |

第二,把str分开:

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   |

相关问题