我有两组代码Loadmore Button和Jqueury Drag N Drop,但它们并不是一样的。
如何将它们合并为一个代码?
以下是问题:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"
type="text/javascript">
Loadmore按钮需要此行并杀死拖放元素的UI功能。
拖动N Drop:
<title>jQuery Dynamic Drag'n Drop</title>
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript" src="jquery-ui-1.7.1.custom.min.js">
</script><script type="text/javascript">
$(document).ready(function(){
$(function() {
$("#contentLeft ul").sortable({ opacity: 0.6, cursor: 'move', update: function() {
var order = $(this).sortable("serialize") + '&action=updateRecordsListings';
$.post("updateDB.php", order, function(theResponse){
$("#contentRight").html(theResponse);
});
}
});
});
});
</script>
DND updateDB.php
<?php
require("db.php");
$action = mysql_real_escape_string($_POST['action']);
$updateRecordsArray = $_POST['recordsArray'];
if ($action == "updateRecordsListings"){
$listingCounter = 1;
foreach ($updateRecordsArray as $recordIDValue) {
$query = "UPDATE records SET recordListingID = " . $listingCounter . " WHERE recordID = " . $recordIDValue;
mysql_query($query) or die('Error, insert query failed');
$listingCounter = $listingCounter + 1;
}
echo '<pre>';
print_r($updateRecordsArray);
echo '</pre>';
echo 'If you refresh the page, you will see that records will stay just as you modified.';
}?>
DND主要PHP:
<ul>
<?php
$query = "SELECT * FROM records ORDER BY recordListingID ASC";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
?>
<li id="recordsArray_<?php echo $row['recordID']; ?>"><?php echo $row['recordID'] . ". " . $row['recordText']; ?></li>
<?php } ?>
</ul>
Loadmore:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"
type="text/javascript"></script>
<script>
$(function() {
var page = 1;
$("#LoadMore").click(function(){
$.ajax({
type:"GET",
url:"page4.php",
data:{page:page},
success: function(response) {
$("#data_grid").append(response);
page++;
}
});
});
});</script>
Loadmore page4.php
<?php
//set argument as your mysql server
$connect = mysql_connect("mysql_server","mysql_user","mysql_password");
mysql_select_db("database_name",$connect);
$page = isset($_GET["page"]) ? $_GET["page"] : 1;
$limit = 25;
$offset = ($page - 1) * $limit;
$sql = "SELECT * FROM table2 limit $offset, $limit";
$result = mysql_query($sql);
$numRows = mysql_num_rows($result);
if($numRows>0) {
while($row = mysql_fetch_array($result)) {
//get field data and set to the following row
echo "<tr><td>field 1</td><td>field 2</td><td>field 3</td></tr>";
//edit row as you table data
}
} else {
echo "<tr><td colspan='3'> No more data </td></tr>";
}
exit;
?>
我错过了什么?我能解决这个问题吗?我是否要废弃Jquery Drag N Drop?有没有我应该使用的替代方案?
背景:用户将按照他们认为合适的顺序对列出的项目进行评级。总可能的项目超过300,000,但用户可能只使用前几百个,其余的用作可搜索的数据库,以将奇数项目添加到列表中。我可以将有问题的2个代码分开工作,但是当我尝试组合时会产生冲突。
感谢您抽出时间协助这个项目。
答案 0 :(得分:0)
看这里
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"
type="text/javascript">
和
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
是相同的(jQuery文件)
所以只使用一个
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"
type="text/javascript">