name-求助 PHP如何接受ajax的post 表单并分行保存TXT呢
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2022-04-29 11:00:51
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namephp表单post
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)
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