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PHP怎么接受ajax的post表单并分行保存TXT呢

时间:2021-07-01 10:21:17 帮助过:23人阅读

求助 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)
------解决思路----------------------
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);

------解决思路----------------------
你没有指定工作方式,默认应该的 GET
用 $_POST 当然收不到数据

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