循环通过 MySQL 结果

Question

I'm not sure exactly how this is called but I'll try to describe as good as I can what I want to acheive.

So, first of all, there is a variable, called $id , which is actually $_GET['id'] . Assuming the user is entering the following page by requesting: /page.php?id=6 . Now what I need to do is to provide the information about the next 3 pages from database. Here is the database:

  TABLE `pages`

  id   |   page_name
  ______________________
  1    |   AAAAA
  2    |   BBBBB
  3    |   CCCCC
  4    |   DDDDD
  5    |   EEEEE
  6    |   FFFFF
  7    |   GGGGG
  8    |   HHHHH
  9    |   IIIII

So, while requesting the page with id 6 , the following script returns the next 3 pages (7,8,9):

 $res = mysql_query("SELECT * FROM `pages` WHERE `id`>'".intval($id)."' ORDER BY `id` DESC LIMIT 3");
 while($arr = mysql_fetch_assoc($res))
 {
       print("Page ID: ".$arr['id']."; Page Name: ".$arr['page_name']."\n");
 }

And here is the output:

 Page ID: 7; Page Name: GGGGG
 Page ID: 8; Page Name: HHHHH
 Page ID: 9; Page Name: IIIII

And it works fine until the $id is greater then 6. When it is ( /page.php?id={7/8/9} ), the output doesn't show 3 pages any more, but 2 pages, 1 page and respectively no output when $id is 9 .

So my question is: Is there a way to go back and start from the beginning when there are not enough results (less than 3) to display?

• When accessing /page.php?id=8 , the output should contain pages with id 9 , 1 and 2 .
• When accessing /page.php?id=9 , the output should contain pages with id 1 , 2 , 3 .
• When accessing /page.php?id=3 , the output should contain pages with id 4 , 5 , 6 and so on.

Answer 1

(SELECT *, 0 AS custom_order FROM `pages` WHERE `id`>'".intval($id)."' ORDER BY `id` ASC LIMIT 3)
UNION ALL
(SELECT *, 1 AS custom_order FROM `pages` ORDER BY `id` ASC LIMIT 3)
ORDER BY custom_order, id ASC
LIMIT 3

This way you always get 3 pages. If not enough next pages, you will get up to 3 from the beginning.

Answer 2

You could modify the query to be something like:

 select * from 
     (select *, id-$inval($id) as order_by 
         from pages were id > $inval($id) order by id asc limit 3
      union
      select *, id as order_by 
         from pages order by id asc limit 3 ) as pages
 order by order_by asc

Answer 3

I would solve this way (one possible issue is that the resultset could contain at most 6 records instead of 3):

$res = mysql_query("(SELECT * FROM `pages` WHERE `id`>'".intval($id)."' ORDER BY `id` ASC LIMIT 3) UNION DISTINCT (SELECT * FROM `pages` WHERE id>0 ORDER BY id ASC LIMIT 3)");
$counter = 0;
 while($arr = mysql_fetch_assoc($res) && $counter<3)
 {
       $counter++;
       print("Page ID: ".$arr['id']."; Page Name: ".$arr['page_name']."\n");
 }