1 
options_table

options_id    | object_id  |   option_value
=========================================
1            |    1       |    drink
2            |    2       |    ice
3            |    1       |    bath
4            |    2       |    soda
5            |    2       |    drink
6            |    3       |    ice
7            |    4       |    bath
8            |    2       |    bath
9            |    1       |    storm


object_table

object_id    |   object_name
=============================
1            |    sun
2            |    moon
3            |    mars
4            |    jupiter

クエリ

SELECT object_table.object_name GROUP_CONCAT(options_table.option_value ) as object_options 
FROM options_table
LEFT JOIN object_table
ON object_table.object_id = options_table.object_id    
GROUP BY options_table.object_id

だから私は次のような結果を得る

object_name | object_options
=========================================
moon        | ice, soda, drink, bath
sun         | drink, bath, storm
mars        | ice
jupiter     | bath

ユーザーがオプション "drink" とオプション "bath" を持つすべてのオブジェクトを必要としているとします。だから私はこの結果しか得られません。

object_name | object_options
=========================================
moon        | ice, soda, drink, bath
sun         | drink, bath, storm

この結果を得るには、クエリをどのように編集する必要がありますか?

4

2 に答える 2

4

最も単純なもの: http://www.sqlfiddle.com/#!2/1b2e3/5

select obj.object_name, 
    group_concat(opt.option_value order by opt.options_id) as object_options
from options_table opt
join object_table obj using(object_id)
group by obj.object_id
having sum(opt.option_value in ('drink','bath')) = 2
order by obj.object_name;

出力：

| OBJECT_NAME |      OBJECT_OPTIONS |
-------------------------------------
|        moon | ice,soda,drink,bath |
|         sun |    drink,bath,storm |

クエリを実行するためにより ANSI-SQL の方法が必要な場合は、MySQLism (ブール値と整数の二重性) に依存せず、代わりに SUM で明示的な値を使用してください: http://www.sqlfiddle.com/#!1/14cf4 /1

select obj.object_name, 
    array_agg(opt.option_value order by opt.options_id) as object_options
from options_table opt
join object_table obj using(object_id)
group by obj.object_id, obj.object_name
having sum(case when opt.option_value in ('drink','bath') then 1 end) = 2
order by obj.object_name;
于 2012-08-07T12:51:53.757 に答える
0

別のアプローチ:

SELECT object_name, object_options FROM   
  (SELECT object_table.object_name, 
          GROUP_CONCAT(options_table.option_value ) as object_options
   FROM options_table
   LEFT JOIN object_table
     ON object_table.object_id = options_table.object_id    
   GROUP BY options_table.object_id) t   
WHERE FIND_IN_SET('drink', object_options)> 0 AND  FIND_IN_SET('bath', object_options)> 0
于 2012-08-07T12:57:02.763 に答える