用户登录验证
发送ajax请求时,error函数的第一个参数的status和readystate分别是200和4,但是第二个参数返回parserror,请问是哪里出了问题
ajax请求的代码:
$.ajax({
type: "POST",
url: "php/login.php?action=login",
dataType: "json",
data: {"usernum":$usernum,"password":$password},
success: function(json){
if(json.success==1){
$(".loginBox2").css("display","none");
var div='
'
'
''+
'
'+json.username+',欢迎你!
'+'
上次登录:'+"2015-02-11"+'
'+''+
'
'
$(".left").prepend(div);
}else{
alert(json.msg);
return false;
}
} ,
error: function(XMLHttpRequest, textStatus, errorThrown) {
alert(XMLHttpRequest.status);
alert(XMLHttpRequest.readyState);
alert(textStatus);
}
});
php的代码:
<?php
session_start();
//包含数据库连接文件
include("conn.php");
$action = $_POST["action"];
if($action == "login"){
$usernum = htmlspecialchars($_POST["usernum"]);
$password = MD5($_POST["password"]);
$query = mysql_query("select * from userinfo where usernum='$usernum'");
$result = mysql_fetch_assoc($query);
if ($password == $result["password"] ){
$_SESSION["usernum"] = $result["usernum"];
$_SESSION["username"] = $result["username"];
$arr["success"] = 1;
$arr["username"] = $_SESSION["username"];
}
else {
$arr["success"] = 0;
$arr["msg"]="密码不正确!";
}
echo json_encode($arr);
}
?>