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id (int 11) NewsIdn (vArchar 10) CommentsIdn (VARchar 10) // ForCommentsIdn (varchar 10) //have CommentsIdn comments(answers) reply to which users Text (varchar 100) DateCreate (datetime)
-- comment 1
-----answer on comment 1
--------answer on answer on comment 1 (№1)
--------answer on answer on comment 1 (№2)
--------answer on answer on comment 1 (№3)
----------answer on answer №3 on answer on comment 1
-------------
---------------N
***
***
***
-- comment 2
-----answer on comment 2
--------answer on answer on comment 2 (№1)
--------answer on answer on comment 2 (№2)
--------answer on answer on comment 2 (№3)
----------answer on answer №3 on answer on comment 2
-------------
---------------N
***
***
***
-- comment N
-----answer on comment N
--------answer on answer on comment N (№1)
--------answer on answer on comment N (№2)
--------answer on answer on comment N (№3)
----------answer on answer №3 on answer on comment N
AND IE.
SELECT * From COMMENTS WHERE NewsIdn='1122121' // value NewsIdn as example
请告诉我怎么做？
@L_512_13@表
id(PK) name
新闻表
id (int 11) Text DateCreate (datetime) user_id(FK wITh user table)
评论表
id (int 11) Text DateCreate (datetime) news_id(int 11)(FK with news table) user_id
id (int 11) comments_id(FK with comments table) Text user_id(int)(FK with user table) DateCreate (datetime)
对于循环,只需在注释表中使用一个循环.然后你得到每个评论的所有答案.
$query=query("select id from news");
while($q=MysqL_fetch_assoc($query)){
//Here is each news
$query1=query("SELECT comments_id,comments FROM comments c WHERE c.news_id=$q['id']");
//Here is each comment
while($q1=MysqL_fetch_assoc($query1)){
$query2=query("
SELECT sub.text,u.user_name FROM sub_comments AS sub
LEFT JOIN comments AS c
ON c.id=sub.COMments_id
INNER JOIN user AS u
ON u.id=sub.user_id
WHERE c.id=$q1['comments_id']
");
while($q2=MysqL_fetch_assoc($query2)){
//Here are sub comments for each comments
PRint $q2['text'];
}
}
}
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