php JQuery UI 保存可排序列表

JQuery UI Saving Sortable List

提问by Laurence

I know this question has been asked before but the solutions did not work for me. I am trying to save the new ordering of items to the database.

我知道以前有人问过这个问题，但解决方案对我不起作用。我正在尝试将项目的新排序保存到数据库中。

I have simplified it very considerably but this is the basic idea of it. I have a form with a sortable list embedded in it.

我已经大大简化了它，但这是它的基本思想。我有一个表单，其中嵌入了一个可排序的列表。

<form id="itemlist">
    <ul id="itemsort">
       <li id="Item_1">Item<input type="hidden" name="itemid[]" value="itemsRowID01"/></li>
       <li id="Item_2">Item<input type="hidden" name="itemid[]" value="itemsRowID02"/></li>
       <li id="Item_3">Item<input type="hidden" name="itemid[]" value="itemsRowID03"/></li>
       <li id="Item_4">Item<input type="hidden" name="itemid[]" value="itemsRowID04"/></li>
    </ul>
</form>

I have JQuery and JQuery UI Loaded and the The Following code enables the sortable list function and posts the item ids and New sort order to a php script. the "editor" variable is a public variable that is set on load it works fine. The sorting works fine but the neworder value that posts doesn't seem to change when I re-order the list.

我加载了 JQuery 和 JQuery UI，以下代码启用了可排序列表功能，并将项目 ID 和新排序顺序发布到 php 脚本。“编辑器”变量是在加载时设置的公共变量，它工作正常。排序工作正常，但当我重新排序列表时，发布的 neworder 值似乎没有改变。

//sorting feature  
    $("#itemsort").live('hover', function() {
        $("#itemsort").sortable({ 
            opacity:.5,
            update : function () {          

                var neworder =  $('#itemsort').sortable('serialize');
                var inputs = serializePost('#itemlist');

                $.post("core/actions.php",{
                   'order': editor,
                   'inputs': inputs,
                   'neworder': neworder},function(){

                       alert("Order saved.", 1);

                });
            } 
        });
    });

On actions.php...

在actions.php...

    if(isset($_POST['order'])){

            //set a variable for each post
            $batchid = $_POST['inputs']['itemid'];

            parse_str($_POST['neworder'], $neworder);

            //count the number of entries to be ordered
            $count = count($batchid);        

            //use the count to create an incremental loop for each item to be updated.
            $i=0;
            while ($i <= $count) {

   $query ="UPDATE {$_POST['order']} SET order=$neworder[item][$i] WHERE id=$batchid[$i]";
                ++$i;
            }
        }

I'm not sure why the order I get for each item will not change.

我不知道为什么我为每个项目得到的订单不会改变。

Any Ideas?

有任何想法吗？

-L

-L

回答by Laurence

$("#list").live('hover', function() {
        $("#list").sortable({

            update : function () {

                var neworder = new Array();

                $('#list li').each(function() {    

                    //get the id
                    var id  = $(this).attr("id");
                    //create an object
                    var obj = {};
                    //insert the id into the object
                    obj[] = id;
                    //push the object into the array
                    neworder.push(obj);

                });

                $.post("pagewhereyouuselist.php",{'neworder': neworder},function(data){});

            }
        });
    });

Then in your PHP file, or in this example "pagewhereyouuselist.php"

然后在你的 PHP 文件中，或者在这个例子中“pagewhereyouuselist.php”

$neworderarray = $_POST['neworder'];
//loop through the list of ids and update your db
foreach($neworderarray as $order=>$id){    
    //you prob jave a connection already i just added this as an example
    $con = mysql_connect("host","username","password");

    if (!$con){
         die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("my_db", $con);

    mysql_query("UPDATE table SET order = {$order} WHERE id = {$id}");
    mysql_close($con);

}

that should do it i didn't test it as it is an example connection. the actual script I am actually using is more specific to my program this is a simplified version to show the concept

应该这样做我没有测试它，因为它是一个示例连接。我实际使用的实际脚本更特定于我的程序，这是一个简化版本以展示概念

回答by Artem Oliynyk

Try this:

尝试这个：

Your HTML fields

您的 HTML 字段

<form id="itemlist" method="POST">
    <ul id="itemsort">
        <li id="Item_1">Item 1<input type="hidden" name="itemid[]" value="itemsRowID01"/></li>
        <li id="Item_2">Item 2<input type="hidden" name="itemid[]" value="itemsRowID02"/></li>
        <li id="Item_3">Item 3<input type="hidden" name="itemid[]" value="itemsRowID03"/></li>
        <li id="Item_4">Item 4<input type="hidden" name="itemid[]" value="itemsRowID04"/></li>
    </ul>
</form>

JS to send order:

JS发送订单：

$("#itemsort").live( 'hover', function() {
    $("#itemsort").sortable({
        update: function () {          
            var inputs = $('#itemlist').serialize();
            $.post("./jq-ui-test.php", inputs, alert("Order saved.") );
        } 
    });
});

Saving order:

保存顺序：

if( isset( $_POST['itemid'] ) && is_array( $_POST['itemid'] ) ) {
    foreach( $_POST['itemid'] as $order => $item ) {
        $order = intval( $order );
        $item_esc = mysql_real_escape_string( $item );
        $sql_query = "UPDATE {$_POST['order']} SET order={$order} WHERE id = '{$item_esc}'";
    }
}

Also, if you want that start order start from 1 ( not from 0 ) change $order = intval( $order );to $order = intval( $order ) + 1;

另外，如果您希望开始顺序从 1 （而不是从 0 ）更改$order = intval( $order );为$order = intval( $order ) + 1;