回报是，如果只有一条记录被发现

Question

我有了这种结构的表格：

table 
id  | site_id 
------------------- 
240  | 1 
240  | 2 
240  | 3 
320  | 1 
320  | 2 
421  | 1 
520  | 2 
------------------- 

300K记录

我现在想编写一个查询返回一个是或否的每个记录（id）。

例如，如果与ID记录240 只有有SITE_ID 1然后返回“是”，如果它有2个，3等回报“否”

我不知道如何处理它但这里是一个结果示例：

result_table 
.-----------------------. 
| id  | result | 
|-----------------------| 
| 240  | No  | -- has a site_id 1, 2 and 3 
| 320  | No  | -- has a site_id 1 and 2 
| 421  | Yes  | -- has a site_id 1 only 
| 520  | No  | -- has a site_id 2 
'-----------------------' 

下面是该查询我到目前为止，但它似乎是不正确的

SELECT CASE WHEN count(id) > 1 THEN 'N' ELSE 'Y' END as Result 
FROM table sm 
WHERE sm.site_id IN (1) 
AND sm.site_id NOT IN (2,3,4,5,6,7,8,9) 
AND id = 240 

UPDATE

所以这里是我完整的查询，我从@gordon

SELECT 
    m.merchant_name, 
    m.merchant_id, 
    ut.realname, 
    a.affiliate_name, 
    (select (case when count(site_id) = 1 then 'Yes' else 'No' end) as result 
    from site_merchant sm 
    WHERE sm.merchant_id = m.merchant_id 
    group by merchant_id) as isTjoos, -- y or no 
    (select (case when count(site_id) = 2 then 'Yes' else 'No' end) as result 
    from site_merchant sm 
    WHERE sm.merchant_id = m.merchant_id 
    group by merchant_id) as isUCOnly 
-- isdlkonly -- y or no 
FROM merchant m 
LEFT JOIN merchant_editor_assignment mea ON mea.merchant_id = m.merchant_id 
LEFT JOIN user_table ut ON ut.user_id = mea.user_id 
LEFT JOIN affiliate a ON a.affiliate_id = m.affiliate_id_default 

Answer 1

您的查询似乎过于复杂的补充答案。刚开始，为什么IN(1)和NOT IN(2,3...9)？当你的“结果样本”显然不需要这个时，为什么只限于一个ID（AND id = 240）？这没有任何意义。这个怎么样？

SELECT CASE WHEN count(merchant_id) > 1 THEN 'N' ELSE 'Y' END as isTjoos 
FROM site_merchant 
GROUP BY site_id; 

Answer 2

我会使用Having Count语句。类似的东西：

SELECT site_id 
FROM site_merchant 
HAVING (count(merchant_id) > 1) 
GROUP BY site_id; 

Answer 3

SELECT 
    it, 
    CASE WHEN COUNT(CASE WHEN site_id THEN 1 END)=1 
      AND COUNT(CASE WHEN site_id!=1 THEN 1 END)=0 THEN 'Yes' 
     ELSE 'No' 
    END 
FROM sm 
GROUP BY it 

请参阅小提琴here。

Answer 4

SELECT ID, CASE WHEN EXTRA > 1 THEN 'No' ELSE 'Yes' END AS Result 
FROM 
(SELECT ID, Sum(site_id) AS Extra 
from myTable 
GROUP BY ID 
) AS Test 

编辑：我想这应该在MySQL中工作。尽管如此，我还没有努力。
这个想法是SUM上了site_id。对于只有 site_id = 1的记录，总和将为1.

Answer 5

我对此问题进行了解释，因为您希望只有一个值为site_id的id。我把示例中的问题，是一个例子，与SITE_ID = 1。要做到这一点：

你想用count(distinct)：

select id, (case when count(distinct site_id) = 1 then 'Yes' else 'No' end) as result 
from site_merchant sm 
group by id 

稍微更有效的版本是使用min()和max()，假设SITE_ID是从未NULL：

select id, (case when min(site_id) = max(site_id) then 'Yes' else 'No' end) as result 
from site_merchant sm 
group by id 

这是因为min和max通常需要少一点的处理比count(distinct)。

如果您要检查site_id是否为“1”，而不是其他任何内容，请将条件and min(site_id) = 1添加到when子句中。

如果你想检查SITE_ID为1，恰好有一个行，那么你可以做：

select id, (case when count(site_id) = 1 and min(site_id) = 1 then 'Yes' else 'No' end) as result 
from site_merchant sm 
group by id 

而且，如果你想检查恰好有一个行：

select id, (case when count(site_id) = 1 then 'Yes' else 'No' end) as result 
from site_merchant sm 
group by id 

Answer 6

这是我找到的解决方案。我用@Gordons查询上手少了什么是SITE_ID，而并不需要由集团：

SELECT 
    m.merchant_name, 
    m.merchant_id, 
    ut.realname, 
    a.affiliate_name, 
    (select (case when count(*)>0 then 'No' else 'Yes' end) as result 
    from site_merchant sm 
    WHERE sm.merchant_id = m.merchant_id 
    AND site_id != 1) as isTjoos, 
    (select (case when count(*)> 0 then 'No' else 'Yes' end) as result 
    from site_merchant sm 
    WHERE sm.merchant_id = m.merchant_id 
    AND site_id != 2) as isUCOnly, 
    (select (case when count(*)> 0 then 'No' else 'Yes' end) as result 
    from site_merchant sm 
    WHERE sm.merchant_id = m.merchant_id 
    AND site_id != 3) as isDLKonly 
FROM merchant m 
LEFT JOIN merchant_editor_assignment mea ON mea.merchant_id = m.merchant_id 
LEFT JOIN user_table ut ON ut.user_id = mea.user_id 
LEFT JOIN affiliate a ON a.affiliate_id = m.affiliate_id_default 

谢谢你的帮助。