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PHP实现 上一篇、下一篇的代码

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Release: 2016-06-13 09:43:57
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代码:

  1. //----显示上一篇、下一篇文章代码 START----   
  2.   
  3. $sql_former = "select * from article where id; //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   
  4. $sql_later = "select * from article where id>$id "//下一篇文章sql语句   
  5. $queryset_former = mysql_query($sql_former); //执行sql语句   
  6. 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   
  7. $result = mysql_fetch_array($queryset_former);   
  8. echo "上一篇  "$result[title].
    "
    ;   
  9. else {echo "上一篇 没有了
    "
    ;}   
  10.   // www.jbxue.com
  11. $queryset_later = mysql_query($sql_later);   
  12. if(mysql_num_rows($queryset_later)){   
  13. $result = mysql_fetch_array($queryset_later);   
  14. echo "下一篇  "$result['title']."
    "
    ;   
  15. else {echo "下一篇 没有了
    "
    ;}   
  16. ?>   
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