mysql - 如果只找到一条记录，则返回 Yes

Question

我有一个具有这种结构的表：

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

30 万条记录

现在我正在尝试编写一个查询来为每条记录（id）返回一个是或否。

例如，如果 id 为 240 的记录只有site_id 1，则返回“是”，如果有 2、3 等等，则返回“否”

我不知道如何处理它，但这是一个结果示例：

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
'-----------------------'

这是我到目前为止的查询，但似乎不正确

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  

更新

所以这是我的完整查询，我添加了@gordon 的答案

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

我将这个问题解释为您想要只有一个 site_id 值的 id。我以问题中的示例为例，site_id = 1。为此：

你想使用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

一个稍微高效的版本是使用min()and max()，假设 site_id 从不为 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 通常比count(distinct).

如果要检查 site_id 是否为“1”而不是其他任何内容，则将条件添加and min(site_id) = 1到when子句中。

如果您想检查 site_id 是否为 1 并且恰好只有一行，那么您可以执行以下操作：

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，如果您想检查是否正好有一行：

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

Answer 2

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

请在此处查看小提琴。

Answer 3

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。我还没有研究它。
这个想法是SUM增加site_id。对于只有site_id = 1 的记录，总和将为 1。

Answer 4

这是我找到的解决方案。我使用@Gordons 查询开始，缺少的是 site_id，并且不需要 group by：

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

感谢您的帮助。

Answer 5

您的查询似乎过于复杂。刚开始，为什么IN(1)and NOT IN(2,3...9)？AND id = 240当您的“结果样本”显然不想要时，为什么要限制为单个 ID ( )？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 6

我会使用有计数语句。像这样的东西：

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