PHP怎麼接受ajax的post 表單並分行儲存TXT呢

求助 PHP如何接受ajax的post 表單並分行儲存TXT呢
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）
------解決思路----------------------
在貼吧看到你了。。


ajaxRequest.ajax({
dataType: 'json',
url: './save_to_txt.php',
data: { name: UserName, mobile: Mobile, companyname: Company, position: Position, actid: ActId },
successed: OnSubmitAppplySuccesse,
error: function () {
Global.Tip.TipError('提交失敗!');
GlobalPlayer.DisableNext();
}
});


save_to_txt.php:

$name = $_POST['name'];
$mobile = $_POST['mobile'];
$companyname = $_POST['companyname'];
$actid = $_POST['actid'];
$position = $_POST['position'];
 $str =$name. "\n".$mobile."\n".$companyname."\n".$actid."\n".$position;
$h = fopen('./t.txt','w');
fwrite($h,$str);
fclose($h);

------解決思路----------------------
你沒有指定工作方式，預設應該的 GET
用 $_POST 當然收不到資料

