Sql 合并所有子查询中的结果

我有一个postgre表，其表示形式与下面的数据类似

餐饮服务表：

-----------------------
| Name    | Option     |
-----------------------
| jane    | social      |
| jane    | vegan       |
| jane    | gmo-free    |
| jane    | italian     |
| jack    | social      |
| jack    | corporate   |
| jack    | gmo-free    |
| jack    | greek       |
| rodz    | social      |
| rodz    | wedding     |
| rodz    | gmo-free    |
| rodz    | vegan       |
| rodz    | french      |

这是我试图运行的“伪”查询

SELECT * FROM caters 
WHERE option is either ['italian', 'french'] 
AND WHERE option is both ['wedding', 'social']

此伪查询应返回

rodz

。因为它既有意大利语也有法语，既有婚礼也有社交活动

这是我试图编写的查询，以完成我的sudo查询

SELECT c.name FROM caters c
WHERE c.option in ('italian', 'french')
GROUP BY c.name
HAVING array_agg(c.option) @> array['wedding', 'social']

无论如何，这都不会返回任何结果。单独运行查询

SELECT c.name FROM caters c
WHERE c.option in ('italian', 'french')
GROUP BY c.name

结果:

-----------
| Name    |
-----------
| jane    | // has italian
| rodz    | // has french

-----------
| Name    |
-----------
| rodz    | // has wedding and social

另一个问题

SELECT c.name FROM caters c
GROUP BY c.name
HAVING array_agg(c.option) @> array['wedding', 'social']

结果:

-----------
| Name    |
-----------
| jane    | // has italian
| rodz    | // has french

-----------
| Name    |
-----------
| rodz    | // has wedding and social

因此，我可以单独看到查询是正确的。这让我思考得很好，如果我有两个查询给了我正确的结果，我只需要过滤掉两个查询中的结果，为什么我不加入它们呢

所以我试过了

SELECT c.name FROM caters c
JOIN caters c1 
ON c1.name = c.name and c1.option = c.option
WHERE c1.option in ('italian', 'french')
GROUP BY c.name
HAVING array_agg(c.option) @> array['wedding', 'social']

但这也没有产生任何结果。你知道我该怎么做吗

注意：每次运行查询时，查询都是动态的。使用的值可能不同，有时可能是5种语言，有时是2种语言，如本例中的

（“意大利语”、“法语”）

。举一个我所说的动态查询的例子，另一个查询可以是

SELECT * FROM caters 
WHERE option is either ['italian'] 
AND WHERE option is both ['corporate', 'social']
// returns none
----------------------------------------------------------
SELECT * FROM caters 
WHERE option is either ['french', 'greek'] 
AND WHERE option is either ['gmo-free', 'vegan'] 
AND WHERE option is both ['corporate', 'social']
// returns jack
----------------------------------------------------------
SELECT * FROM caters WHERE option is ['social']
// returns jack, and rodz

---

您可以尝试使用相关子查询

select distinct name from tablename a 
where option in ('italian', 'french') and exists
(
  select 1 from tablename b where a.name=b.name and option in ('wedding', 'social')
  group by b.name having count(distinct option)=2
)

输出：

name
rodz

这里有一种方法：

SELECT c.name
FROM caters c
WHERE c.option in ('italian', 'french', 'wedding', 'social')
GROUP BY c.name
HAVING COUNT(*) FILTER (WHERE c.option IN ('italian', 'french')) >= 1 AND
       COUNT(*) FILTER (WHERE c.option IN ('wedding', 'social')) = 2;

---

我可以调整此查询以使用动态值吗？我更新了我的问题，以便更好地解释我所说的动态values@a_horse_with_no_name哈哈，对不起，英语不是我的第一语言，我没有意识到我有一个

p

和一个

e

伪。更新了拼写错误。我的错误我没有提供查询的所有需要-主要的是动态的。我更新了我的问题以反映这一点。@user10457989。您只需要根据“动态”标准调整列表。