求助 PHP如何接受ajax的post 表单并分行保存TXT呢

Question

JS代码如下

 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();
    }
}

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）

Answer 1

http://blog.csdn.net/chennan0769/article/details/12093773