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新手分页有关问题,PHP分页代查询功能,没有查询结果就出错

WBOY
WBOY원래의
2016-06-13 10:14:231214검색

新手分页问题,PHP分页代查询功能,没有查询结果就出错?

PHP code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> include("conn.php");function pager($rows,$page_size){ global $page,$select_from,$select_limit,$pagenav; $page_count = ceil($rows/$page_size); if($page = $page_count) $page = $page_count; $select_limit = $page_size; $select_from = ($page - 1) * $page_size.','; $pre_page = ($page == 1)? 1 : $page - 1; $next_page= ($page == $page_count)? $page_count : $page + 1 ; $pagenav .= "第 $page/$page_count 页 共 $rows 条记录 "; $pagenav .= "<a href="?page=1">首页</a> "; $pagenav .= "<a href="?page=%24pre_page">前一页</a> "; $pagenav .= "<a href="?page=%24next_page">后一页</a> "; $pagenav .= "<a href="?page=%24page_count">末页</a>"; $pagenav.=" 跳到<select name="topage" size="1" onchange='window.location=\"?page=\"+this.value'>\n"; for($i=1;$i$i\n"; else $pagenav.="<option value="$i">$i</option>\n"; } } ///////////////////////////// 利用pager函数计算出 $select_from 从哪条记录开始检索、$pagenav 输出分页导航 $countsql="select * from manager where 0=0";  if ($_POST[chinaname]!=""){       $countsql=$countsql." and chinaname like '%$_POST[chinaname]%'";  }  if ($_POST[Tel]!=""){       $countsql=$countsql." and Tel like '%$_POST[Tel]%'";  }$rows = mysql_num_rows(mysql_query($countsql)); $page = $_GET['page'];pager($rows,2); $sql = "select * from manager where 0=0";   if ($_POST[chinaname]!=""){   $sql=$sql." and chinaname like '%$_POST[chinaname]%'";  }  if ($_POST[Tel]!=""){   $sql=$sql." and Tel like '%$_POST[Tel]%'";  }  $sql=$sql." limit $select_from $select_limit";    echo $sql;$rst = mysql_query($sql); while($row=mysql_fetch_array($rst)){</select>


现在的问题是当我模糊查询没有结果的时候,limit 这里显示的语句就是 limit -2,2
想现在这样的问题,应该在呢么解决呢

------解决方案--------------------
$page_count = ceil($rows/$page_size); 
没有结果的时候$rows为0,得到的$page_count 也等于0,所以下面的
$select_from = (0 - 1) * 2 $select_from就得到-2了
你可以在后面加个判断 $page_count为0的时候 不加limit子句
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