Mysql中如何通过匹配同一个表来获取数据

我在Mysql数据库中有如下表结构 表：Mytable

--------------------------------
| uid | locationid | projectid |
--------------------------------
|  2  |      4     |    2     | 
|  2  |      4     |    2     | 
|  2  |      3     |    5     | 
|  2  |      3     |    5     | 
|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    1     | 
|  3  |      1     |    5     | 
|  3  |      1     |    2     | 
|  3  |      1     |    1     | 
|  3  |      1     |    1     | 
|  7  |      3     |    1     | 
|  5  |      1     |    5     | 
|  6  |      4     |    2     | 
|  5  |      1     |    5     | 
|  3  |      4     |    1     | 
|  3  |      3     |    3     | 
|  7  |      4     |    2     | 
|  7  |      4     |    3     | 

假设我传递了uid=2，那么对于uid=2，locationid和projectid的唯一组合是

|  2  |      4     |    2     | 
|  2  |      3     |    5     | 
|  2  |      1     |    2     | 
|  2  |      1     |    1     | 

现在，我希望所有UID与上述locationid和projectid组合匹配。意味着结果应该是

|  3  |      1     |    2     | 
|  3  |      1     |    1     | 
|  3  |      1     |    1     | 
|  6  |      4     |    2     | 
|  7  |      4     |    2     | 

|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    1     | 
|  5  |      1     |    5     | 
|  5  |      1     |    5     | 

如果我通过uid=3，那么结果应该是

|  3  |      1     |    2     | 
|  3  |      1     |    1     | 
|  3  |      1     |    1     | 
|  6  |      4     |    2     | 
|  7  |      4     |    2     | 

|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    2     | 
|  2  |      1     |    1     | 
|  5  |      1     |    5     | 
|  5  |      1     |    5     | 

为此，我使用了以下查询，但它给出了错误的结果，因为locationid和projectid的组合不匹配

 SELECT a.*
 FROM Mytable a, Mytable b 
 WHERE a.locationid = b.locationid
 AND a.projectid = b.projectid
 AND a.locationid IN (SELECT DISTINCT locationid FROM Mytable WHERE uid=$pmid)
 AND a.projectid IN(SELECT DISTINCT projectid FROM Mytable WHERE uid=$pmid)
 AND a.uid !=$pmid

这里$pmid是我传递的uid值。我的问题出了什么问题？我的质疑是否正确？ 请帮帮我

提前感谢。

这应该可以：

SELECT a.*
FROM Mytable a
     INNER JOIN (SELECT DISTINCT locationid, projectid
                 FROM   Mytable
                 WHERE  uid = 2
                ) b
        ON a.locationid = b.locationid
           AND a.projectid = b.projectid
           AND a.uid <> 2;

示例：试试这个：

SELECT t.* 
FROM   Mytable t
JOIN
       ( SELECT distinct locationid , projectid
         FROM Mytable
         WHERE uid=<uid>)a
ON     t.locationid =a.locationid 
AND    t.projectid=a.projectid
WHERE  t.uid != <uid>

请尝试此查询-

SELECT t1.* FROM mytable t1
  JOIN (SELECT * FROM mytable WHERE uid = 2 GROUP BY locationid, projectid) t2
    ON t1.uid <> t2.uid AND
       t1.locationid = t2.locationid AND
       t1.projectid = t2.projectid;

---

在WHERE子句的第二行指定您的uid。

问题不清楚！看一看第三个表ou提供的，有两个记录与3-1-1这应该是唯一的吗？感谢回复和给出解决方案