为什么我的 MySQL `NOT IN` 查询使用 `COUNT()` 会导致'操作数应包含 1 列”错误？

NOT IN 具有聚合函数的子查询中的 MySQL“操作数应包含 1 列”错误

在 MySQL 中使用 NOT IN 子查询需要仔细注意列计数。 当子查询使用 COUNT() 等聚合函数返回多列时，会出现一个常见错误“操作数应包含 1 列”。 MySQL 的 NOT IN 运算符需要单列比较。

根本原因：

该错误源于主查询和子查询返回的列数不匹配。 包含聚合函数的 NOT IN 子查询会生成包含多个列的结果集，与主查询的 id 子句中的单列 WHERE 冲突。

示例：

考虑这个有问题的查询：

SELECT *
FROM campaigns
WHERE id NOT IN
(
    SELECT
        e.id_campaign,  
        d.name,        
        d.frequency,   
        d.country,     
        d.referral,    
        d.bid,         
        d.status,      
        COUNT(e.id) AS countcap  
    FROM campaigns d
    LEFT JOIN served e
        ON d.id = e.id_campaign
    WHERE
        d.status = 'Active'
    GROUP BY
        e.id_campaign
    HAVING
        countcap < p>The intention is to select campaigns *not* included in the subquery's results. The subquery, however, returns eight columns, causing the "Operand should contain 1 column" error because `NOT IN` expects a single-column comparison against the `id` column in the `campaigns` table.</p><p>**Resolution:**</p><p>The solution involves restructuring the subquery to return only the `id_campaign` column:</p><pre class="brush:php;toolbar:false"><code class="sql">SELECT *
FROM campaigns
WHERE id NOT IN
(
    SELECT e.id_campaign
    FROM campaigns d
    LEFT JOIN served e ON d.id = e.id_campaign
    WHERE d.status = 'Active'
    GROUP BY e.id_campaign
    HAVING COUNT(e.id) < 
</code>

Alternatively, for situations requiring multiple columns, use `EXISTS` or `NOT EXISTS` for a more efficient and accurate solution:

<code class="sql">SELECT *
FROM campaigns c
WHERE NOT EXISTS
(
    SELECT 1
    FROM campaigns d
    INNER JOIN served e ON d.id = e.id_campaign
    WHERE d.id = c.id
      AND d.status = 'Active'
      AND COUNT(e.id) < 
</code>

This revised approach avoids the column count mismatch and provides a cleaner solution for scenarios involving aggregate functions within subqueries used with `NOT IN`.

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