PHP / SQL创建指向数据库中下一行的链接 [英] PHP/SQL creating links to next row in a database

问题描述

假设我有一个数据库：

  + ---------- + 
 |数据库| 
 + ---------- + 
 | id | 
 |图像| 
 |类别| 
 + ---------- + 
  

显示来自数据库的图片，我要添加

<上一张图片

和

>>

按钮。我可以只是取$ id并添加+1，但如果下一个ID不存在于DB中怎么办？

任何帮助感谢，谢谢

所以我发现这个：

  SELECT * 
 FROM database AS c 
 WHERE（id =（SELECT MAX（id）FROM database WHERE id  OR id =（SELECT MIN（id）FROM database WHERE id& id AND language ='en'））
  

但是如何使用链接：
< a href =domain.com/$c ??>下一页< / a> ？

 
  SELECT 
（SELECT ID FROM< TABLE-NAME> WHERE ID> $ currentID ORDER BY ID ASC LIMIT 1）
 AS NEXT_VALUE，
（SELECT ID FROM< ; TABLE-NAME> WHERE ID <$ currentID ORDER BY ID ASC LIMIT 1）
 AS PREV_VALUE 
 FROM DUAL; 
  

Lets say i have a databse:

+----------+
| Database |
+----------+
| id       |
| image    |
| category |
+----------+

Now i have a page that shows an image from the database, i want to add a

<< Previous image

and

Next image >>

button. I could just take the $id and add +1 , but what if the next ID does not exist in the DB ?

Any help appreciated, thanks

So i found this:

SELECT * 
  FROM database AS c 
 WHERE (id = (SELECT MAX(id) FROM database WHERE id < c.id AND language = 'en')
    OR  id = (SELECT MIN(id) FROM database WHERE id > c.id AND language = 'en'))

But how do i make a link out of it, like: <a href="domain.com/$c ??">Next</a> ?

解决方案

You'll have to use another select:

SELECT
    ( SELECT ID FROM <TABLE-NAME> WHERE ID > $currentID ORDER BY ID ASC LIMIT 1 )
    AS NEXT_VALUE,
    ( SELECT ID FROM <TABLE-NAME> WHERE ID < $currentID ORDER BY ID ASC LIMIT 1 )
    AS PREV_VALUE
FROM DUAL;

这篇关于PHP / SQL创建指向数据库中下一行的链接的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！