我想从数据库执行一些数据过滤。过滤后,会话变量丢失,我收到此错误:
警告:mysql_num_rows():提供的参数不是第34行的.. \ test.php中的有效MySQL结果资源。
有人可以帮忙吗?
FORM:
<form action= "test.php" method='get'>
<select name="Type">
<option value="winter" name="winter">winter</option>
</select>
<input type='submit' value = 'filter'>
</form>
CODE:
<?php
session_start();
$_SESSION['winter'] = $_GET['Type'];
$type = $_SESSION['winter'] ;
include "mysqlConnect.php";
//check for a page number. If not, set it to page 1
if (!(isset($_GET['pagenum']))){
$pagenum = 1;
}else{
$pagenum = $_GET['pagenum'];
}
//query for record count to setup pagination
$data = mysql_query("SELECT * FROM tblPhotos WHERE Type='$type' ");
$rows = mysql_num_rows($data);
//number of photos per page
$page_rows = 16;
//get the last page number
$last = ceil($rows/$page_rows);
//make sure the page number isn't below one, or more than last page num
if ($pagenum < 1){
$pagenum = 1;
}elseif ($pagenum > $last){
$pagenum = $last;
}
//Set the range to display in query
$max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows;
//get all of the photos
$dynamicList = "";
$sql = mysql_query("SELECT * FROM tblPhotos WHERE Type='$type' $max ");
//check for photos
$photoCount = mysql_num_rows($sql); //LINE 34
echo $photoCount;
if ($photoCount > 0){
while($row = mysql_fetch_array($sql)){
$photoID = $row["PhotoID"];
$photoName = $row["photoName"];
$category = $row["category"];
$dynamicList .= '
<div class="thumb">
<a href="photo.php?id=' . $photoID . '"><img class="clip" src="galleryPhotos/' . $photoID . '.jpg" alt="' . $photoName . '" width="175" border="0" /></a>
</div>
';
}
}else{
$dynamicList = "There are no photos at this time!";
}
mysql_close();
echo '<p style="text-align:center; font-weight:bold;">Page ' . $pagenum . ' of ' . $last . '</p>';
if ($pagenum == 1){
echo '<div class="pagination" align="center"><ul>';
}else{
echo '<div class="pagination" align="center"><ul><li><a href="' . $_SERVER['PHP_SELF'] . '?pagenum=1">« first</a></li>';
$previous = $pagenum-1;
}
//check if number of pages is higher than 1
if($last != 1){
//Loop from 1 to last page to create page number links
for($i = 1; $i <= $last; $i++){
echo '<li><a href="' . $_SERVER['PHP_SELF'] .'?pagenum=' . $i . '">' . $i . '</a></li>';
}
}
if ($pagenum == $last){
echo '</div>';
}else{
$next = $pagenum+1;
echo '<li><a href="' . $_SERVER['PHP_SELF'] . '?pagenum=' . $last . '">last »</a></li></ul></div>';
}
echo $dynamicList;
?>
答案 0 :(得分:0)
Your Code: mysql_query("SELECT * FROM tblPhotos WHERE Type='$type' $max ");
阅读错误消息,它告诉您查询有问题。查看了查询。我倾向于同意。