PHP如何接受ajax的post 表单并分行保存TXT呢

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）

回复讨论(解决方案)

在贴吧看到你了。。


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('提交失败!');
            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);

在贴吧看到你了。。


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('提交失败!');
            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);

大神啊 接收端不错 就是这个JS 发不出去数据 一直提交的是空数据是怎么回事呀

你没有指定工作方式，默认应该的 GET
用 $_POST 当然收不到数据