PHP实现 上一篇和下一篇的代码
<?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。注意是倒序,因为返回结果集时只用了第一篇文章,而不是最后一篇文章 $sql_later = "select * from article where id>$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 <a href='index.php?id=$result[id]'> ". $result[title]." </a><br>"; } else {echo "上一篇 没有了<br>";} // www.jbxue.com $queryset_later = mysql_query($sql_later); if(mysql_num_rows($queryset_later)){ $result = mysql_fetch_array($queryset_later); echo "下一篇 <a href='index.php?id=$result[id]'> ". $result['title']."</a><br>"; } else {echo "下一篇 没有了<br>";} ?>
下面是我的上一篇和下一篇
新手勿喷
<? $id = isset($_GET['id']) > 0 ? intval($_GET['id']) : ""; $query = mysql_query("SELECT id,title FROM article WHERE id>'$id' and class='".$_GET["catid"]."' ORDER BY id ASC LIMIT 1"); if(mysql_num_rows($query)){ $next = mysql_fetch_array($query); $next="下一篇 <a href='?catid=131&id=$next[id]'> ". $next[title]." </a> "; }else{ $next= "下一篇 没有了"; } $query = mysql_query("SELECT id,title FROM article WHERE id <'$id' and class='".$_GET["catid"]."' ORDER BY id DESC LIMIT 1"); if(mysql_num_rows($query)){ $prev = mysql_fetch_array($query); $prev="上一篇 <a href='?catid=131&id=$prev[id]'> ". $prev[title]." </a> "; }else{ $prev= "上一篇 没有了"; } ?>
页面里面:<div class="nws_fx_r nws_fx_c"><?= $prev;?> / <?= $next;?> </div>
转载于:https://www.cnblogs.com/yuesha/articles/5438519.html