PHP+Mysql+AJAX登录验证
html部分
<html> <head> <script src = "login.js"> </script> </hdad> <body> <form> UserName: <input id = "UserName" name = "UserName" type = "text" /> <br> <br> PassWord: <input id = "PassWord" name = "PassWord" type = "text" /> <br> <br> <input type = "button" value = "submit" onclick = "validation();" /> </form> <div id = "info"> </div> </body> </html>
js部分
function validation(){
var UserName = document.getElementById("UserName").value;
var PassWord = document.getElementById("PassWord").value;
var postStr = "UserName=" + UserName + "&PassWord=" + PassWord;
//the notes are another way to achieve this function
//ajax("login.php",postStr);
ajax("login.php",postStr,function(result){
document.getElementById("info").innerHTML = result;
});
}
function ajax(url,postStr,onsuccess){
//function ajax(url,postStr){
var xmlhttp = window.XMLHttpRequest?new XMLHttpRequest():new ActiveXObject("Microsoft.XMLHTTP");
xmlhttp.open("POST",url,true);
xmlhttp.onreadystatechange = function(){
if(xmlhttp.readyState == 4){
if(xmlhttp.status == 200){
//document.getElementById("info").innerHTML = xmlhttp.responseText;
onsuccess(xmlhttp.responseText);
}
else{
alert("AJAX ERROR!");
}
}
}
xmlhttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
xmlhttp.send(postStr);
}
//http://www.cnblogs.com/liuswi/p/4067881.html
php部分
<?php
$con = mysql_connect("localhost","root");
mysql_select_db("my_db",$con);
//the original code which noted cannot be achieved!
//$result = mysql_query("SELECT * FROM users WHERE Username = "$_POST[UserName]"");
$result = mysql_query("SELECT * FROM users WHERE UserName ="".$_POST["UserName"].""");
if ($row = mysql_fetch_array($result)) {
if($row["PassWord"] == $_POST["PassWord"]){
echo "Petch Success!";
}
else{
echo "ERROR!";
}
}
else{
echo "UserName EMPTY!";
}
mysql_close($con);
?>声明：该文观点仅代表作者本人，牛骨文系教育信息发布平台，牛骨文仅提供信息存储空间服务。
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