我已经做了一些搜索，并根据这里的例子和其他在线资源多次重新处理我的sql，但每次运行此sql时，我都会得到相同的“#1064-您的sql语法有错误”错误。
我正在尝试使用内部连接连接四个表，它们应该始终有匹配的数据来进行键关闭（即，对于每个有效的令牌/token\u id，应该始终有一整行）。我使用的是MySQL5.7.26版本，下面是我尝试运行的查询：

SELECT
    i.name AS invitee_name,
    c.first_name AS child_first,
    c.last_name AS child_last,
    s.invite_status,
    c.avatar
  FROM
    Invites AS i, Tokens AS t, Children AS c, Invite_Statuses AS s
  WHERE
    t.token = sdie02d
  INNER JOIN
    t ON t.token_id = i.token_id
  INNER JOIN
    c ON c.child_id = i.child_id
  INNER JOIN
    s ON s.status_id = i.status_id

我（每次）收到的全部错误是：

# 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN
    t ON t.token_id = i.token_id
  INNER JOIN
    c ON c.child_id ' at line 11

表格如下所示：

Invites
+---------+----------+----------+-----------+---------------+----------+
| user_id | child_id | token_id | status_id |     email     |   name   |
+---------+----------+----------+-----------+---------------+----------+
|       9 |        2 |        1 |         1 | a@example.com | John Doe |
|       9 |        3 |        2 |         1 | b@example.com | Jane Doe |
+---------+----------+----------+-----------+---------------+----------+

Tokens
+----------+---------+
| token_id |  token  |
+----------+---------+
|        1 | 93kd8i0 |
|        2 | sdie02d |
|        3 | fsj2d9c |
+----------+---------+

Children
+----------+------------+-----------+--------+
| child_id | first_name | last_name | avatar |
+----------+------------+-----------+--------+
|        1 | Timmy      | Johnson   |      4 |
|        2 | Jenny      | Smith     |     32 |
|        3 | Jake       | Jones     |     12 |
+----------+------------+-----------+--------+

Invite_Statuses
+-----------+---------------+
| status_id | invite_status |
+-----------+---------------+
|         1 | invited       |
|         2 | accepted      |
|         3 | rejected      |
+-----------+---------------+

谢谢你的帮助。