var xmlHttp; ????if (window.ActiveXObject) { ????xmlHttp = new ActiveXObject("Microsoft.XMLHTTP"); ????} else if (window.XMLHttpRequest) { ????xmlHttp=new XMLHttpRequest(); ????}function getAppCaptcha(cdata){ ???????????if (isCaptcha) {return} ???????????????isCaptcha = true; ????????????xmlHttp.open("POST",aaa.php‘); ????????xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); ???????xmlHttp.setRequestHeader("X-Requested-With","XMLHttpRequest"); ???????xmlHttp.send(disposeData(cdata)); ????????xmlHttp.onreadystatechange = function(data) { ???????????if ((xmlHttp.readyState == 4) && (xmlHttp.status == 200)) { ????????????var sdata = JSON.parse(data.currentTarget.response); ????????????????if (sdata.error_code == 0) { ???????????????????//验证成功 ???????????????????//干点什么 ???????????????????isClick = false; ???????????????}else if(sdata.error_code == 1048){//超过四次 ???????????????????isClick = true; ???????????????????document.querySelector(‘.pop_click_box‘).style.display = ‘block‘ ???????????????}else{ ???????????????????//验证失败 ???????????????????tipfun(sdata.error_msg); ???????????????} ???????????} else { ???????????????isClick = false; ???????????} ???????} ????}上面代码由于PHP做了ajax校验需要
xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); ???????xmlHttp.setRequestHeader("X-Requested-With","XMLHttpRequest");
---------------------------------------------------------------------------------------PHP判断ajax请求的原理:
在发送ajax请求的时候，我们可以通过XMLHttpRequest这个对象，创建自定义的header头信息， 在jquery框架中，对于通过它的$.ajax, $.get, or $.post方法请求网页内容时，它会向服务器传递一个HTTP_X_REQUESTED_WITH的参数，php中就是在header一层判断是否是ajax请求，对应的根据$_SERVER[‘HTTP_X_REQUESTED_WITH‘]判断。
注意:
- 一般情况下$_SERVER[‘HTTP_X_REQUESTED_WITH‘]默认是XMLHttpRequest
- $_SERVER[‘HTTP_X_REQUESTED_WITH‘]也可以自定义创建的，使用XMLHttpRequest.setRequestHeader(name,value);
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提交的数据需要disposeData方法来处理成对应字符串才可以
//传参封装 ???function disposeData(data){ ???????var strData = ‘‘,num = 0; ???????for(var key in data){ ???????????if(num == 0){ ???????????????strData = key + ‘=‘ + data[key]; ???????????}else{ ???????????????strData += ‘&‘ + key + ‘=‘ + ?data[key]; ???????????} ???????????num++; ???????} ???????return strData; ???}类似这样的字符串给到后台name=地方&bankCard=2321323123123&phone=13800138000
原生ajax提交php后台接收不到问题
原文地址：https://www.cnblogs.com/lichuntian/p/8724116.html