获取MySQL错误似乎与我正在做的事情无关

我有这个用PHP编写的MySQL查询:

$insert_suggested_solution_comment = "insert into suggested_solution_comments 
(problem_id , commenter_id, comment, solution_part , date , guid) values ( ".$problem_id." ,  ".$member_id." , '".$comment."' , ".$recent_topic_id.",
NOW() , '".$md5c."' )" ;

但我得到的错误就是这个:

Field 'suggested_solution_id' doesn't have a default value

但那怎么可能呢？:)我从未在查询中引用suggest_solution_id.

我三重检查,是的,这个错误确实来自这个查询.有谁知道为什么会这样？

这就是表格的样子:

mysql> describe suggested_solution_comments;
+-----------------------+----------------+------+-----+---------+----------------+
| Field                 | Type           | Null | Key | Default | Extra          |
+-----------------------+----------------+------+-----+---------+----------------+
| comment_id            | int(10)        | NO   | PRI | NULL    | auto_increment |
| problem_id            | int(10)        | NO   |     | NULL    |                |
| suggested_solution_id | int(10)        | NO   |     | NULL    |                |
| commenter_id          | int(10)        | NO   |     | NULL    |                |
| comment               | varchar(10000) | YES  |     | NULL    |                |
| solution_part         | int(3)         | NO   |     | NULL    |                |
| date                  | date           | NO   |     | NULL    |                |
| guid                  | varchar(50)    | YES  | UNI | NULL    |                |
+-----------------------+----------------+------+-----+---------+----------------+

谢谢!