Find Jobs
Hire Freelancers

Assist us in resolving this javascripting issue

$10 USD

已取消
已发布超过 7 年前

$10 USD

货到付款
Help needed at [login to view URL] Javascript excerpt: / drag and drop functionality starts here / $('#' + [login to view URL] + ' tbody').sortable({ start: function(event, ui) { [login to view URL] = [login to view URL](); }, stop: function(event, ui) { [login to view URL]("Start position: " + [login to view URL]); [login to view URL]("New position: " + [login to view URL]()); if([login to view URL] != [login to view URL]()) { var table = $('#' + [login to view URL]).DataTable(); var start_row = [login to view URL][0].data(); var end_row = [login to view URL][[login to view URL]()].data(); [login to view URL](start_row); [login to view URL](end_row); /* [login to view URL]("target row id: " + [login to view URL]([login to view URL]()).data('unique_id')); [login to view URL]("target row id: " + [login to view URL]([login to view URL]()).data('unique_id')); $.ajax({ url: postDT.ajax_url, type: 'POST', dataType: 'json', data: { action: 'DTMoveRow', startrow: start_row['id'], endrow: end_row['id'] }, success: function (return_data) { [login to view URL]("success"); [login to view URL](return_data); [login to view URL](return_data['success']); [login to view URL](return_data['error']); wpDataTables[tableid].fnDraw(false); }, error: function(e) { [login to view URL](e); } }); SwapRowValues([login to view URL], [login to view URL],[login to view URL], dttableid, [login to view URL], [login to view URL]()); */ } }, appendTo: "parent", helper: "clone" }).disableSelection(); // drag and drop ui ends here ================================================== The original code, which I was debugging but am now replacing, is the call to SwapRowValues([login to view URL], [login to view URL],[login to view URL], dttableid, [login to view URL], [login to view URL]()); The function I'm eliminating got some data from the database and then invoked php function to do an update, which was failing. My new design will have a much simpler php function that is called from this code snippet. Unfortunately, I'm having trouble getting proper values to pass to my php function. As you can see I'm using [login to view URL] to help me know what is going on. The code above generates an error on the line var start_row = [login to view URL][0].data(); I think the table object must not be getting fetched properly, but I copied its code from the function that used to be called here, substituting the parameters from the function call for the relevant arguments from the function... I'm hoping I did something dumb that a simple code read-through will reveal. But I'd also love to have an experienced javascript debug person help me out with this on the test site.
项目 ID: 12575353

关于此项目

远程项目
活跃7 年前

想赚点钱吗?

在Freelancer上竞价的好处

设定您的预算和时间范围
为您的工作获得报酬
简要概述您的提案
免费注册和竞标工作

关于客户

UNITED STATES的国旗
Waco, United States
5.0
32
付款方式已验证
会员自2月 13, 2009起

客户认证

谢谢!我们已通过电子邮件向您发送了索取免费积分的链接。
发送电子邮件时出现问题。请再试一次。
已注册用户 发布工作总数
Freelancer ® is a registered Trademark of Freelancer Technology Pty Limited (ACN 142 189 759)
Copyright © 2024 Freelancer Technology Pty Limited (ACN 142 189 759)
加载预览
授予地理位置权限。
您的登录会话已过期而且您已经登出,请再次登录。