我试图在登录后转到另一个页面,所以如何在使用Ajax和php成功登录后传递到另一个页面?当ajax向php发送请求然后得到响应时(如果响应是('true')或('false')这个响应取决于检查用户名和密码)所以我不知道如何处理这个响应在ajax中使用它作为传递到下一页的条件。请帮忙。
function doCallAjax(Mode) {
HttPRequest = false;
if (window.XMLHttpRequest) { // Mozilla, Safari,...
HttPRequest = new XMLHttpRequest();
if (HttPRequest.overrideMimeType) {
HttPRequest.overrideMimeType('text/html');
}
} else if (window.ActiveXObject) { // IE
try {
HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {}
}
}
if (!HttPRequest) {
alert('Cannot create XMLHTTP instance');
return false;
}
var url = 'PHP_Login.php';
var pmeters = "tEmail=" + encodeURI( document.getElementById("Email").value) +
"&tpassword=" + encodeURI( document.getElementById("password").value ) +
"&tMode=" + Mode;
HttPRequest.open('POST',url,true);
HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
HttPRequest.setRequestHeader("Content-length", pmeters.length);
HttPRequest.setRequestHeader("Connection", "close");
HttPRequest.send(pmeters);
HttPRequest.onreadystatechange = function()
{
if(HttPRequest.readyState == 3) // Loading Request
{
document.getElementById("mySpan").innerHTML = "Now is Loading...";
}
if(HttPRequest.readyState == 4) // Return Request
{
document.getElementById("mySpan").innerHTML = HttPRequest.responseText;
document.getElementById("Email").value = '';
document.getElementById("password").value = '';
}
}
}
答案 0 :(得分:1)
这个系统通常的工作方式是在后端进行身份验证,php通常会创建一个会话变量,无论是否登录,然后发回一个响应。获得响应后,您所要做的就是使用javascript的window.location()
或window.href
函数在javascript中重定向。