我们使用分页123按钮,在这里如果我们点击下一个按钮也意味着它应该转到下一页。但是这里没有进入下一页,因为在这里我将sql查询中的_POST变量传递给where条件,所以如果点击下一页,它不会显示相关页面。 我们认为当我们点击其他页面后,POST变量值不会更新。这是我的代码。提前致谢。使用分页和MySQL查询
if(isset($_POST['submit']))
{
$bool = true;
$gender=$_POST['gender'];
}
if(($bool == true)
{
$bool1=true;
$connection = mysql_connect("localhost","root","Sakthi");
if (!$connection)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("nursingcarein",$connection);
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * 3;
$gender1=$gender;
$profession1=$profession;
$state1=$state;
/**** Here gender value doesnt update when click the next page i think we have a error in this query ****/
$sql = "SELECT id,description,name FROM nursereg WHERE gender='$gender1' LIMIT $start_from, 3";
$rs_result = mysql_query ($sql,$connection);
echo "<table>
<tr><td>Image</td><td>Description</td></tr>";
while($row=mysql_fetch_array($rs_result)) {
echo" <tr>";
echo"<td>"; echo "<img src='image1.php?id=".$row['id']."'>"; echo"</td>";
echo"<td>"; echo $row['name']; echo"</td>";
echo"<td>"; echo $row['description']; echo"</td>";
echo"</tr>";
};
echo"</table>";
$sql = "SELECT COUNT(Name) FROM nursereg";
$rs_result = mysql_query($sql,$connection);
$row = mysql_fetch_row($rs_result);
$total_records = $row[0];
$total_pages = ceil($total_records/5);
for ($i=1; $i<=$total_pages; $i++) {
$pageno = $i;
echo "<a href='index2.php?page=".$i."'>".$i."</a> ";
};
};
?>
如果你害怕$ _POST不会改变它的值,你总是可以使用'var_dump'或'echo'函数来检查它。 – Voitcus 2013-03-15 15:56:58