name-求助 PHP如何接受ajax的post 表单并分行保存TXT呢

namephp表单post

JS代码如下

<code> function SubmitApply() {    var form = Global.GetElement("#submitForm");    var UserName = Global.GetElement(form, 'input[name="Name"]').value;    var Mobile = Global.GetElement(form, 'input[name="Tel"]').value;    var Position = Global.GetElement(form, 'input[name="Position"]').value;    var Company = Global.GetElement(form, 'input[name="Company"]').value;    var ActId = Global.GetElement(form, 'input[name="Actid"]').value;    if (Global.IsNullOrWhitesapce(UserName)) {        Global.Tip.TipError("姓名必须填写");        return;    }    if (Global.IsNullOrWhitesapce(Mobile)) {        Global.Tip.TipError("手机必须填写");        return;    }    if (/^1[3|4|5|8][0-9]\d{4,8}$/.test(Mobile) == false) {        Global.Tip.TipError("手机格式不正确");        return;    }    var ajaxRequest = new HttpAjaxRequest();    ajaxRequest.ajax({        dataType: 'json',        url: '',        data: { name: UserName, mobile: Mobile, companyname: Company, position: Position, actid: ActId },        successed: OnSubmitAppplySuccesse,        error: function () {            Global.Tip.TipError('提交失败!');            GlobalPlayer.DisableNext();        }    });}/** * 名称：申请提交成功 */function OnSubmitAppplySuccesse(data) {    if (data.Code == 0) {        Global.Tip.TipMessage('提交成功!');        var form = Global.GetElement("#submitForm");        GlobalPlayer.EnableNext();        GlobalPlayer.PlayNext();        form.reset();    } else {        Global.Tip.TipError('提交失败!' + data.Message);        GlobalPlayer.DisableNext();    }}</code>

哪个好人帮我写个完整的PHP接受源码吧 本人PHP小白（已哭晕）