php 上一篇与下一篇文章代码
来源:程序员人生 发布时间:2013-11-16 05:39:07 阅读次数:2385次
基本知识:就是对id对进行order by id desc 或 order by id asc进行排序,然后再判断比当前id > or小于当前文章id的,实例的sql语句如下:
select * from news where id<$id order by id desc limit 0,1
select * from news where id>$id order by id desc limit 0,1
表的结构 `string_find`
- CREATE TABLE IF NOT EXISTS `string_find` (
- `id` int(4) NOT NULL auto_increment,
- `charList` varchar(100) default NULL,
- PRIMARY KEY (`id`)
- ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=7 ;
导出表中的数据 `string_find`
- INSERT INTO `string_find` (`id`, `charList`) VALUES
- (1, '邓先驱者'),
- (2, '邓先驱者'),
- (5, 'fdafdsaf'),
- (6, 'www.phpfensi.com');
好了万事俱备了,下面来看一下操作方法:
- */
- mysql教程_pconnect('localhost','root','root') or die(mysql_error());
- mysql_select_db('cc');
- mysql_query("set names 'gbk'");
- $cid =5;
- $sql ="select * from string_find where id>$cid order by id desc limit 0,1";
- $sql1 ="select * from string_find where id<$cid order by id asc limit 0,1";
-
- $result = mysql_query( $sql );
- if( mysql_num_rows( $result ) )
- {
- $rs = mysql_fetch_array( $result );
- echo "上一篇".$rs[0];
- }
- else
- {
- echo "没有了";
- }
-
- $result1 = mysql_query( $sql1 );
- if( mysql_num_rows( $result1 ) )
- {
- $rs1 = mysql_fetch_array( $result1 );
- echo "下一篇".$rs1[0];
- }
- else
- {
- echo "没有了";
- }
生活不易,码农辛苦
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