Mysql 连接两个表而不返回不需要的行

我的表结构如下所示：

      tbl.users                       tbl.issues
+--------+-----------+   +---------+------------+-----------+
| userid | real_name |   | issueid | assignedid | creatorid |
+--------+-----------+   +---------+------------+-----------+
|   1    |   test_1  |   |    1    |     1      |     1     |
|   2    |   test_2  |   |    2    |     1      |     2     |
+--------+-----------+   +---------+------------+-----------+

                           (results table)
+---------+------------+---------------+-----------+--------------+
| issueid | assignedid | assigned_name | creatorid | creator_name |
+---------+------------+---------------+-----------+--------------+
|    1    |     1      |    test_1     |    1      |    test_1    |
|    2    |     1      |    test_1     |    2      |    test_2    |
+---------+------------+---------------+-----------+--------------+

                (results table)
+---------+------------+-----------+-----------+
| issueid | assignedid | creatorid | real_name |
+---------+------------+-----------+-----------+
|    1    |     1      |     1     |   test_1  |
|    2    |     1      |     2     |   test_1  |
|    2    |     1      |     2     |   test_2  |
+---------+------------+-----------+-----------+

基本上，我想编写一个查询，该查询将以如下所示的结果表结束：

      tbl.users                       tbl.issues
+--------+-----------+   +---------+------------+-----------+
| userid | real_name |   | issueid | assignedid | creatorid |
+--------+-----------+   +---------+------------+-----------+
|   1    |   test_1  |   |    1    |     1      |     1     |
|   2    |   test_2  |   |    2    |     1      |     2     |
+--------+-----------+   +---------+------------+-----------+

                           (results table)
+---------+------------+---------------+-----------+--------------+
| issueid | assignedid | assigned_name | creatorid | creator_name |
+---------+------------+---------------+-----------+--------------+
|    1    |     1      |    test_1     |    1      |    test_1    |
|    2    |     1      |    test_1     |    2      |    test_2    |
+---------+------------+---------------+-----------+--------------+

                (results table)
+---------+------------+-----------+-----------+
| issueid | assignedid | creatorid | real_name |
+---------+------------+-----------+-----------+
|    1    |     1      |     1     |   test_1  |
|    2    |     1      |     2     |   test_1  |
|    2    |     1      |     2     |   test_2  |
+---------+------------+-----------+-----------+

我的SQL目前看起来是这样的：

SELECT 
  `issues`.`issueid`,
  `issues`.`creatorid`,
  `issues`.`assignedid`,
  `users`.`real_name`
FROM `issues`
JOIN `users` 
  ON ( `users`.`userid` = `issues`.`creatorid` )
  OR (`users`.`userid` = `issues`.`assignedid`)
ORDER BY `issueid` ASC
LIMIT 0 , 30 

这将返回如下内容：

      tbl.users                       tbl.issues
+--------+-----------+   +---------+------------+-----------+
| userid | real_name |   | issueid | assignedid | creatorid |
+--------+-----------+   +---------+------------+-----------+
|   1    |   test_1  |   |    1    |     1      |     1     |
|   2    |   test_2  |   |    2    |     1      |     2     |
+--------+-----------+   +---------+------------+-----------+

                           (results table)
+---------+------------+---------------+-----------+--------------+
| issueid | assignedid | assigned_name | creatorid | creator_name |
+---------+------------+---------------+-----------+--------------+
|    1    |     1      |    test_1     |    1      |    test_1    |
|    2    |     1      |    test_1     |    2      |    test_2    |
+---------+------------+---------------+-----------+--------------+

                (results table)
+---------+------------+-----------+-----------+
| issueid | assignedid | creatorid | real_name |
+---------+------------+-----------+-----------+
|    1    |     1      |     1     |   test_1  |
|    2    |     1      |     2     |   test_1  |
|    2    |     1      |     2     |   test_2  |
+---------+------------+-----------+-----------+

有人能帮我找到想要的结果表吗

SELECT DISTINCT (i.issueid, i.creatorid, i.assignedid, u.real_name)
FROM issues i, users u
WHERE u.userid = i.creatorid OR u.userid = assignedid
ORDER BY i.issueid ASC
LIMIT 0 , 30

不确定是否需要括号。

使用此选项：

SELECT 
  IssueID, 
  AssignedID, 
  CreatorID, 
  AssignedUser.real_name AS AssignedName, 
  CreatorUser.real_name AS CreatorName
FROM Issues
  LEFT JOIN Users AS AssignedUser
         ON Issues.AssignedID = AssignedUser.UserID
  LEFT JOIN Users AS CreatorUser
         ON Issues.CreatorID = CreatorUser.UserID
ORDER BY `issueid` ASC
LIMIT 0, 30

SELECT 
`issues`.`issueid`,
`issues`.`creatorid`,
`creator`.`real_name`,
`issues`.`assignedid`,
`assigned`.`real_name`
FROM `issues` i
INNER JOIN `users` creator ON ( `creator`.`userid` = `issues`.`creatorid` )
INNER JOIN `users` assigned ON (`assigned`.`userid` = `issues`.`assignedid`)
ORDER BY `issueid` ASC
LIMIT 0 , 30

这行吗

挑选 i、 发布ID， i、 转让， u1.真实姓名作为分配的姓名， i、 创造者， u2.真实姓名作为创建者姓名 来自用户u1 u1.userid=i.assignedid上的内部联接问题i u2.userid=i.creatorid上的内部加入用户u2 按i.issueid订购

选择
i、 issueid，
i、 分配ID，
a、 真实姓名，
i、 创建者ID，
c、 真实姓名
从
问题i
内部加入用户c
在c.userid=i.creatorid上
内部加入用户a
在a.userid=i.assignedid上
订购人

---

i、 issueid ASC

在常识方面，我们著名的网站创始人写了一篇关于这个主题的非常好的博客文章，我发现自己一遍又一遍地提到这个主题

---

很抱歉，这仍然会为issue#2返回一个双行。我猜issues.creatorid、issues.assignedid to users.userid中有两个外键，并且每个表中都不能为null。给定这些前提，使用左连接和内连接之间有什么区别吗？一个比另一个更自然，或者更快？