mysql如何将group_concat（）结果放入列表答案

【问题标题】：mysql How to get group_concat () result into listmysql如何将group_concat（）结果放入列表
【发布时间】：2017-11-14 17:22:06
【问题描述】：

我有一个名为abc 的表，其值为：

date        |   Code |  Class  |    subitem |   Item_No |
2017-04-02  |   FII  | INDIR   |    YAA     |   28      |
2017-04-09  |   FII  | INDIR   |    YAA     |   28      |
2017-04-16  |   FII  | INDIR   |    YAA     |   14      |
2017-04-23  |   FII  | INDIR   |    YAA     |   24      |
2017-04-30  |   FII  | INDIR   |    YAA     |   41      |
2017-04-02  |   RRT  | INDIR   |    BKK     |   8       |
2017-04-09  |   RRT  | INDIR   |    BKK     |   3       |
2017-04-16  |   RRT  | INDIR   |    BKK     |   16      |
2017-04-23  |   RRT  | INDIR   |    BKK     |   15      |
2017-04-30  |   RRT  | INDIR   |    BKK     |   32      |
2017-04-09  |   RRT  | INDIR   |    WESEL   |   14      |   
2017-04-16  |   RRT  | INDIR   |    WESEL   |   7       |   
2017-04-23  |   RRT  | INDIR   |    WESEL   |   2       |   
2017-04-02  |   RRT  | INDIR   |    YAA     |   67      |
2017-04-09  |   RRT  | INDIR   |    YAA     |   98      |
2017-04-16  |   RRT  | INDIR   |    YAA     |   135     |
2017-04-23  |   RRT  | INDIR   |    YAA     |   103     |
2017-04-30  |   RRT  | INDIR   |    YAA     |   119     |
2017-04-02  |   IDS  | INDIR   |    YAA     |   8       |

我想要的输出是这样的：

CODE    |   ITEM_NO |   GROUP   |   DATE
FII     |   103     |   A1      |   2017-04-02
RRT     |   103     |   A1      |   2017-04-02
IDS     |   103     |   A1      |   2017-04-02
FII     |   126     |   A2      |   2017-04-09
RRT     |   126     |   A2      |   2017-04-09
FII     |   149     |   A3      |   2017-04-16
RRT     |   149     |   A3      |   2017-04-16
FII     |   127     |   A4      |   2017-04-23
RRT     |   127     |   A4      |   2017-04-23
FII     |   160     |   A5      |   2017-04-30
RRT     |   160     |   A5      |   2017-04-30

这是我的脚本：

select distinct
Group_concat(Code SEPARATOR ', ') Code,
sum(Item_No) totally,
CONCAT("A",CAST(@jim:=@jim +1 AS CHAR)) Group,
Date Date
from tableT
where Date between '2017-04-01' and '2017-04-30'
and SubItem = 'YAA'
and dept IN ('FII','RRT','IDS')
group by  Date

但我只能得到这样的结果：

CODE            | ITEM_NO   |   GROUP   |   DATE
FII, RRT, IDS   |   103     |   A1      |   2017-04-02
FII, RRT        |   126     |   A2      |   2017-04-09
FII, RRT        |   149     |   A3      |   2017-04-16
FII, RRT        |   127     |   A4      |   2017-04-23
FII, RRT        |   160     |   A5      |   2017-04-30

还有其他方法可以得到 RESULT 吗？谢谢你的帮助

【问题讨论】：

我发现数据集和预期结果之间没有相关性
Deptpivot？部门？子项？团体？没有任何意义
见meta.stackoverflow.com/questions/333952/…
@Strawberry，对不起，我写的有些正确。真实的是代码而不是 Dept Pivot，是的，数据集与所需结果之间没有相关性。但是当我在本地运行时，结果是让我 19 行

标签： mysql casting group-concat

【解决方案1】：

答案在链接所在的 SO 中 Mysql string split

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 (col1, col2)
   SELECT col1, strSplit(col2, ',', i) FROM BadTable
  WHERE strSplit(col2, ',', i) IS NOT NULL;
SET i = i + 1;
UNTIL ROW_COUNT() = 0
  END REPEAT;
END $$
DELIMITER ;

【讨论】：

【解决方案2】：

为此，您可以在组选择和代码选择之间使用连接

  select 
  t1.deptPivot  Code,
  sum(t1.Item_No) totally,
  t2.Group,
  t1.DATE
  from tableT t1
  INNER JOIN (
        select 
        CONCAT("A",CAST(@jim:=@jim +1 AS CHAR)) Group,
        Date Date
        from tableT
        where Date between '2017-04-01' and '2017-04-30'
        and SubItem = 'YAA'
        and dept IN ('FII','RRT','IDS')
        group by  Date, 

  ) t2 on t1.Date = t2.Date
  where Date between '2017-04-01' and '2017-04-30'
  and SubItem = 'YAA'
  and dept IN ('FII','RRT','IDS')
  group by  code, group, Date

【讨论】：