2009-11-30 76 views
0

嗨我已經創建了下面的代碼來檢查用戶是否有效登錄到數據庫,我使用SSL來保護連接,但我不知道這是否仍然是我已經完成的一個好方法。任何人都可以提供建議謝謝。再次這是用ajax實現登錄的安全/好方法嗎?

感謝號碼:

(對不起它升技長)

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<link rel="stylesheet" type="text/css" href="login_styles.css" /> 
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script> 
<script src="../jquery/ui/effects.core.js"></script> 
<script src="../jquery/ui/effects.shake.js"></script> 

<title>Login</title> 

</head> 

<body> 

<script type="text/javascript" > 
var clicked = 0; 

$(document).ready(function() 
{ 
    document.form.username.focus(); 
}) 
    function make_request() 
{ 
    try 
    { 
     // Firefox, Opera 8.0+, Safari 
     httpxml=new XMLHttpRequest(); 
    } 
    catch (e) 
    { 
     // Internet Explorer 
     try 
    { 
     httpxml=new ActiveXObject("Msxml2.XMLHTTP"); 
    } 
    catch (e) 
    { 
    try 
    { 
     httpxml=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    catch (e) 
    { 
     alert("Your browser does not support AJAX!"); 
     return false; } 
    } 
    } 
} 

function checkCanSubmit() 
{ 
    if (fnameok && lnameok && emailok && projectnameok && descriptionok) 
    { 
    document.getElementById("button").disabled= false; 

    } 
    else 
    { 
    document.getElementById("button").disabled= true; 
    } 

} 

function counting() 
{ 
clicked++; 
} 

function check_login(username, password) 
{ 


make_request() 
$("#login_loading").show(); 
var parameters = 'username=' + document.getElementById("username").value + '&password=' + document.getElementById("password").value; 
httpxml.onreadystatechange = stateck;  
httpxml.open('POST', 'login.php', true); 
httpxml.setRequestHeader('Content-Type', "application/x-www-form-urlencoded"); 
httpxml.setRequestHeader('Content-Length', parameters.length); 
httpxml.send(parameters); 

function stateck() 
{ 

if(httpxml.readyState==4) 

    { 
    if (httpxml.responseText.indexOf("Login Successful") >= 0) 
    { 
    $("#login_loading").hide("fast"); 
    $("#login_error").hide("slow"); 
    $("#forgot").hide("slow"); 
    $("#button").fadeOut("slow");  
    $("#loading_big").fadeIn("slow"); 
    setTimeout("window.location.href = '../members'", 2000); 


    } 
    else 
    { 
     focus(); 
     if (clicked > 1) 
     { 
     $("#login_error").effect("shake", { times:2 }, 100); 
     $("#login_loading").fadeOut(1000); 
     $("#forgot").show("slow"); 

     } 
     else 
     {   
     $("#login_loading").fadeOut(1000); 
     $("#login_successful").hide("normal"); 
     $("#login_error").show("slow"); 
     } 

    }   

    } 
} 
} 

function focus() 
{  
    document.form.username.focus(); 
} 

</script> 


    <div class="login_loading" id="login_loading"><img src="images/login_loader.gif"  alt="loading1" id="loading1" /></div> 
    <div class="loading_big" id="loading_big"> Login successful please wait . . .<br /><img src="images/loading_big.gif" alt="loading" /></div> 
    <div class="form"> 
    <table class="form" id="form" cellpadding="10" > 
    <form id="form" name="form" method="post" action="login.php" /> 
    <tr> 

    <td>Username</td> 
    <td><input type="text" name="username" id="username" size="26" /></td> 

    </tr> 

    <tr> 

    <td>Password</td> 
     <td><input type="password" name="password" id="password" size="26"/> 

     <div id="login_error" class="login_error" ><img src="images/error.png" alt="Error" id="error" />Username or password incorrect.</div> 
     <div id="forgot" class="login_error">Forgot Password?<a href="forgotpass.php"></a></div> 
     <div id="login_successful" class="login_successful" >Login successful please wait . . . </div></td> 

</tr> 

<tr> 
<td> 
<input type="submit" name="button" id="button" value="Submit" onclick="check_login(username, password); counting(); return false;"/> 
</td> 
</tr> 
</form> 
</table> 
</div> 
</body> 
</html> 

代碼格式的推移這裏亂七八糟的升技。

回答

0

除了一些明顯的清理點,以及由於您無論如何都要導入jQuery,您不妨使用其Ajax方法而不是手工編寫自己的代碼,因此我在這裏看不到任何可怕的錯誤。有一個巨大的警告 - "members"最好驗證用戶有一個合法的登錄會話。如果用戶訪問../members而沒有首先進入此登錄頁面或登錄失敗後會發生什麼?他們是否重定向到登錄頁面?

+0

嗨,是的我有完整的PHP登錄系統工作我只是想添加一個Ajax前端。檢查成員中的每個頁面是否有效會話等 – Elliott 2009-11-30 21:07:11

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