mysql - SQL 选择以消除在下一列中具有 2 个其他值的重复值

Question

我已经构建了一个像这样的连接表。

表名： myTable

p_id | c_id
-----------
1     1
1     2
1     3
2     2    
2     3
3     2
3     3
3     4

我想选择不具有 c_id 3 和 4 的 p_id。在这种情况下，只有 p_id 3 同时具有 c_id 3 和 4，因此在 select 语句之后，查询应该返回 p_id 1 和 2。问题是我尝试不同一种方法，但它仍然行不通。我真的需要帮助。

我的查询

1.)SELECT DISTINCT p_id FROM myTable WHERE c_id != 3 AND course_id != 4;

问题：它仍然返回 3 作为结果之一，因为 3 的 c_id 为 2

Answer 1

像这样的东西：

SELECT DISTINCT p_id 
FROM   mytable 
WHERE  p_id NOT IN (SELECT p_id 
                    FROM   mytable 
                    WHERE  c_id IN ( 3, 4 ) 
                    GROUP  BY p_id 
                    HAVING Count(DISTINCT c_id) = 2)

SQLFiddle demo

Answer 2

尝试这个：

SELECT DISTINCT p_id 
FROM myTable 
WHERE c_id IN (3,4) 
GROUP BY p_id HAVING COUNT(DISTINCT c_id)<2

Answer 3

直接的解决方案是使用存在：

  select 
distinct p_Id
    from myTable t
   where not (exists (select 1
                        from myTable
                       where (c_id = 3) and
                             (p_id = t.p_id)) and 
              exists (select 1
                        from myTable
                       where (c_id = 4) and
                             (p_id = t.p_id)))

Answer 4

尝试这个：

SELECT mytable.p_id 
FROM   mytable 
       LEFT OUTER JOIN (SELECT v1.p_id 
                        FROM   (SELECT p_id 
                                FROM   mytable 
                                WHERE  c_id = 3) v1 
                                INNER JOIN (SELECT p_id 
                                            FROM   mytable 
                                           WHERE  c_id = 4) v2 
                                       ON v1.p_id = v2.p_id) v 
                    ON mytable.p_id = v.p_id 
WHERE  v.p_id IS NULL 
GROUP  BY mytable.p_id 

Answer 5

尝试这个：

select distinct mytable.p_id from mytable where c_id not in (3,4) and p_id <>3

这将给出没有 3 和 4 的结果