2012-05-07 137 views
-1

我无法让我的代码正常工作。当用户键入用户名时,它会检查数据库以查看它是否存在,以及是否将<P id="checkusername">设置为“用户名已接收”。如果没有将其设置为“未采取”。我看不出为什么我的代码无法正常工作。Ajax检查用户名是否存在

1 register.html

<html> 
    <body> 
    <CENTER> 
     <form name="register" action="register.php" method="post"> 
     Username: <input type="text" name="username" onkeyup="checkUsername(this.value)" /> 
     <P id="checkusername">checker</P> 
     <input type="submit" value="Login" />  
     </form> 
    </CENTER> 

    <script type="text/javascript"> 
     function checkUsername(){ 
     var xmlhttp; 
     var username=document.forms["register"]["username"].value; 
     if(username.length==0){ 
      document.getElementById("checkusername").innerHTML="Empty"; 
      return; 
     } 
     if(window.XMLHttpRequest){ 
      xmlhttp=new XMLHttpRequest(); 
     }else{ 
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
     } 

     var url = "login.php"; 
     var params = "header=checkusername&username="+username+"&password=""; 
     xmlhttp.open("POST", url, true); 
     xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
     xmlhttp.setRequestHeader("Content-length", params.length); 
     xmlhttp.setRequestHeader("Connection", "close"); 

     xmlhttp.onreadystatechange = function() {//Call a function when the state changes. 
    if(xmlhttp.readyState == 4 && xmlhttp.status == 200){ 
      document.getElementById("checkusername").innerHTML=xmlhttp.responseText; 
    } 
     } 
     xmlhttp.send(params); 
    } 
    </script> 
    </body> 
</html> 

2 register.php

<?php 
    $header = $_POST["header"]; 
    if(header=="checkusername"){ 
    checkusername($_POST["username"]); 
    }else{ 
    echo "No match: " . $header; 
    } 

    function connection(){ 
    $con = mysql_connect(URL, username, password); 
    if(!$con){ 
     die('Could not connect: ' . mysql_error()); 
    } 
    return $con; 
    } 
    function checkusername($username){ 
    $con = connection(); 
    mysql_select_db(database, $con); 
    $result = mysql_query("SELECT * FROM users WHERE username = \"" . $username . "\";"); 
    while($row = mysql_fetch_array($result)){ 
     if(($row['username'] == $username)){ 
     echo "Username Taken.<br/>"; 
     return; 
     } 
    } 
    echo "Not Taken."; 
    } 
    mysql_close(); 
?> 
+1

限制你的问题,以更小的代码示例。只有相关的部分。 –

+0

在你的while循环php文件,你总是输出“没有采取”,它应该在IF条件下..加..为什么你需要循环查看用户名是否匹配..你的SQL已经做到了..只是看看如果计数是0它没有采取。采取其 – Kamal

+0

SQL注入发现 – pomeh

回答