MySql將別名拆分為多個列

Question

select pd.products_name, 
GROUP_CONCAT(pag.customers_group_id SEPARATOR ',') group_id, 
pa.`options_values_price` Retail, 
GROUP_CONCAT(pag.options_values_price SEPARATOR ',') volume_and_designer


from products_attributes pa 

left join products_description pd 
on pa.products_id = pd.products_id and pd.language_id = '1' 

left join products_attributes_groups pag 
on pa.`products_attributes_id`= pag.`products_attributes_id`

where pa.products_id='225'

GROUP BY `pa`.`products_attributes_id` 

ORDER BY `pa`.`products_attributes_id` ASC

上面的查詢向我返回了這樣的輸出

| products_name | group_id |   Retail |   volume_and_sdesign |
-------------------------------------------------------------
| GOLD          |    1,2   |  15      |       30,35          |
| SILVER        |    2,1   |  16      |       40,45          |
| BRONZE        |    1,2   |  17      |       50,55          |

我想要實現的是在上面的表中添加2個別名，以便根據group_id列將最后一列（volume_and_sdesign）分成兩列（即volume，SDesign）。 1對應於音量，2對應於SDesign。

例如

Gold has group_id (1,2)
so its volume_and_sdesign (30,35) will make new columns 
volume = 30
SDesign = 35

Silver has group_id (2,1)
so its volume_and_sdesign (40,45) will make new columns 
volume = 45
SDesign = 40

 Bronze has group_id (1,2)
so its volume_and_sdesign (50,55) will make new columns 
volume = 50
SDesign = 55

所以，上面的表格看起來像這樣

| products_name | group_id |   Retail |   volume_and_sdesign |  volume |  SDesign|
-------------------------------------------------------------
| GOLD          |    1,2   |  15      |       30,35          |30       | 35      |
| SILVER        |    2,1   |  16      |       40,45          |45       | 40      |
| BRONZE        |    1,2   |  17      |       50,55          |50       | 55      |

任何幫助都感激不盡

Answer 1

您可以使用條件聚合 - 在聚合函數（例如max()中就是case ：

select pd.products_name, 
       group_concat(pag.customers_group_id SEPARATOR ',') as group_id, 
       pa.`options_values_price` as  Retail, 
       group_concat(pag.options_values_price SEPARATOR ',') as volume_and_designer,
       max(case when group_id = 1 then pag.options_values_price end) as volume,
       max(case when group_id = 2 then pag.options_values_price end) as SDesign
from products_attributes pa left join
     products_description pd 
     on pa.products_id = pd.products_id and pd.language_id = '1' left join
     products_attributes_groups pag 
     on pa.`products_attributes_id` = pag.`products_attributes_id`
where pa.products_id='225'
group by `pa`.`products_attributes_id` 
order by `pa`.`products_attributes_id` ASC

Answer 2

這可以使用LEFT ， RIGHT和SUBSTRING_INDEX來實現。

嘗試這個：

select
    pd.products_name, 
    GROUP_CONCAT(pag.customers_group_id SEPARATOR ',') group_id, 
    pa.`options_values_price` Retail, 
    GROUP_CONCAT(pag.options_values_price SEPARATOR ',') volume_and_designer,
    IF(LEFT(group_id,1) = 1, SUBSTRING_INDEX(volume_and_designer, ',', 1), SUBSTRING_INDEX(volume_and_designer, ',', -1)) as volume,
    IF(RIGHT(group_id,1) = 1, SUBSTRING_INDEX(volume_and_designer, ',', 1), SUBSTRING_INDEX(volume_and_designer, ',', -1)) as SDesign
from products_attributes pa 
left join products_description pd 
on pa.products_id = pd.products_id and pd.language_id = '1' 
left join products_attributes_groups pag 
on pa.`products_attributes_id`= pag.`products_attributes_id`
where pa.products_id='225'
GROUP BY `pa`.`products_attributes_id` 
ORDER BY `pa`.`products_attributes_id` ASC