用PHP实现 上一篇、下一篇的代码_PHP教程

//----显示上一篇、下一篇文章代码 START----

$sql_former = "select * from article where id$sql_later = "select * from article where id>$id "; //下一篇文章sql语句
$queryset_former = mysql_query($sql_former); //执行sql语句
if(mysql_num_rows($queryset_former)){ //返回记录数，并判断是否为真，以此为依据显示结果
$result = mysql_fetch_array($queryset_former);
echo "上一篇 ". $result[title]."
";
} else {echo "上一篇 没有了
";}

$queryset_later = mysql_query($sql_later);
if(mysql_num_rows($queryset_later)){
$result = mysql_fetch_array($queryset_later);
echo "下一篇 ". $result['title']."
";
} else {echo "下一篇 没有了
";}
?>

http://www.bkjia.com/PHPjc/326111.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/326111.htmlTechArticle复制代码 代码如下: ?php //----显示上一篇、下一篇文章代码 START---- $sql_former = "select * from article where id$id order by id desc "; //上一篇文章sql语句...