name-求助 PHP如何接受ajax的post 表單並分行儲存TXT呢

namephp表單post

JS代碼如下

 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("姓名必須填寫");        return;    }    if (Global.IsNullOrWhitesapce(Mobile)) {        Global.Tip.TipError("手機必須填寫");        return;    }    if (/^1[3|4|5|8][0-9]\d{4,8}$/.test(Mobile) == false) {        Global.Tip.TipError("手機格式不正確");        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('提交失敗!');            GlobalPlayer.DisableNext();        }    });}/** * 名稱：申請提交成功 */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('提交失敗!' + data.Message);        GlobalPlayer.DisableNext();    }}

哪個好人幫我寫個完整的PHP接受源碼吧 本人PHP小白（已哭暈）

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