Php jquery可拖动列表位置保存到数据库
我有以下代码,用于使用jQueryUI元素从数据库获取数据并对可拖动列表进行排序Php jquery可拖动列表位置保存到数据库,php,jquery,sql,database,Php,Jquery,Sql,Database,我有以下代码,用于使用jQueryUI元素从数据库获取数据并对可拖动列表进行排序 <script src="http://code.jquery.com/jquery-1.9.1.js"></script> <script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script> <style> #sortable { list-style-type: none; mar
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<style>
#sortable { list-style-type: none; margin: 0; padding: 0; width: 60%; }
#sortable li { margin: 0 3px 3px 3px; padding: 0.4em; padding-left: 1.5em; font-size: 1.4em; height: 18px; background-color:#CCC;}
#sortable li span { position: absolute; margin-left: -1.3em; }
</style>
<script>
$(function() {
$( "#sortable" ).sortable();
$( "#sortable" ).disableSelection();
});
</script>
<?php
$con=mysqli_connect("localhost","root","","db_name");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$user_id = $_SESSION['user_id'];
$result = mysqli_query($con,"SELECT * FROM users WHERE user_id = '$user_id'");
echo "<ul id='sortable'>";
while($row = mysqli_fetch_array($result))
{
echo "<li class='ui-state-default'>" . $row['Name'] . ' ' . $row['UserName'] . $row['sort'] ."</li>";
}
echo "</ul>";
mysqli_close($con);
?>
示例结果
user_id Name UserName Password sort
1 AAA aa *** 1
2 BBB bb *** 2
3 CCC cc *** 3
4 DDD dd *** 4
我想问的是,我可以使用jquery
draggable
属性对列表项进行重新排序,但是如果对列表项进行了重新排序,我如何将sort
number保存到数据库中呢?jquery UI sortable有一个停止回调,当您停止移动排序表时会调用它。它还有一个序列化函数,您可以结合使用该函数将排序表的顺序发送到更新数据库的进程。您可以这样使用它:
要使用序列化
,需要修改DOM中的可排序元素,使其包含对用户id的引用。例如
<li class="ui-state-default" id="id_<?php echo $row['user_id'] ?>">...</li>
然后需要一个进程来接收这些数据并更新数据库。我会把大部分的事情留给你,但这里有一个最小的例子
<?php
// Store user IDs in array. E.g. array(0 => "1", 1 => "3", 2 => "10"....)
$userIds = $_POST['id'];
// Connect to your database
$conn = mysqli_connect("localhost","root","","db_name");
foreach ($userIds as $key => $userId) {
$sequence = $key + 1;
$stmt = $conn->prepare("UPDATE users SET sort = $sequence WHERE user_id = ?");
$stmt->bind_param("i", (int) $userId);
$stmt->execute();
}
$stmt->close();
使用此事件处理程序DB密码通过ajax发送数据q8scool\u考试
这是一个家庭作业问题吗?我知道这个答案有点旧,但对任何新出现的人来说都是一个提示。不要使用mysqli\u connect
var $sortable = $( "#sortable" );
$sortable.sortable({
stop: function ( event, ui ) {
// parameters will be a string "id[]=1&id[]=3&id[]=10"...etc
var parameters = $sortable.sortable( "serialize" );
// send new sort data to process
$.post("/sort/my/data.php?"+parameters);
}
});
<?php
// Store user IDs in array. E.g. array(0 => "1", 1 => "3", 2 => "10"....)
$userIds = $_POST['id'];
// Connect to your database
$conn = mysqli_connect("localhost","root","","db_name");
foreach ($userIds as $key => $userId) {
$sequence = $key + 1;
$stmt = $conn->prepare("UPDATE users SET sort = $sequence WHERE user_id = ?");
$stmt->bind_param("i", (int) $userId);
$stmt->execute();
}
$stmt->close();