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我已經看過多個主題，試圖解決我的難題，但我無法理解它。對於初學者;我已經嘗試過嬰兒牀並修改了http://phpseason.wordpress.com/2013/02/17/live-username-availability-checking-using-ajax-and-jquery/，我將js1.7.1保存在../js/jquery-1.7.1.min.js中，我的ajax_check_login存儲在../customers/中，就像account_creation一樣。Jquery Ajax用戶名驗證不起作用
我試圖讓它做的是驗證，如果用戶名可在網頁上直播。注意：當我點擊我的提交按鈕時，它會將我帶到../customers/ajax_check_login.php，並正確顯示1或0。
我也在我的調試嘗試中註釋了各種各樣的位。
我對我的主帳戶創建頁面有以下代碼;
<?php
session_start();
?>
<!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" />
<title>Account Creation</title>
<script type="text/javascript" src="../js/jquery-1.7.1.min.js"></script>
<!-- <script type="text/javascript" src="js/jquery-ui-1.8.17.custom.min.js"></script> -->
<script type="text/javascript" src="../js/jquery-1.7.1.min.js">
$(document).ready(function(){
$("#wantedusername").change(function(){
/* $("#message").html("<img src='ajax-loader.gif' /> checking...");*/
$("#message").html("checking...");
var wantedusername=$("#wantedusername").val();
$.ajax({
type:"POST",
url:"../customers/ajax_check_login.php",
data:"wantedusername="+wantedusername,
success:function(data){
if(data==0){
/* $("#message").html("<img src='tick.png' /> Username available");*/
$("#message").html(" Username available");
}
else{
/* $("#message").html("<img src='cross.png' /> Username already taken"); */
$("#message").html("Username already taken");
}
}
});
});
});
</script>
<?php include "../header.php" ; ?></head>
<body>
<header class="headerbg">
<section class="headertop">
<?php include '../nav.php'?>
</section>
</header><!--header and menu wrapper end-->
<align ="center">
Thank you for showing interest in registering for a Crosstrend Analysis ltd account. Please note that these require administrator authorisation before the account will become active, this can take up to 72 hours.
VERSION CONTROL 1.3
<div align="center">
<table width="300" border="0" align="Center" cellpadding="0" cellspacing="1" bgcolor="="#cccccc">
<tr>
<form id="acctcreation" name="acctcreation" method="post" action="../customers/ajax_check_login.php">
<td>
<table width = "100%" border ="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3"><p><Strong> New Account Request </strong></p></td>
</tr>
<tr>
<td width = "78">Username</td>
<td widith = "6">:</td>
<td width = "294"><input name="wantedusername" type="text" id="wantedusername"></td>
<td id="message"></td>
</tr>
<tr>
<td>Password</td>
<td>:</td>
<td><input name="wantedpassword" type="password" id="wantedpassword"></td>
</tr>
<tr>
<td>Password Verification</td>
<td> : </td>
<td><input name ="verificationpassword" type="password" id ="verificationpassword"></td>
</tr>
<tr>
<td>Email address</td>
<td>:</td>
<td><input name="emailaddress" type="text" id="emailaddress"></td>
</tr>
<tr>
<td>Company</td>
<td>:</td>
<td><input name="company" type="text" id="company"></td>
</tr>
<tr>
<td> </td>
<td> </td>
<td><input type="submit" name="Submit" value="Submit for authentication"> </td>
</tr>
</table>
</td>
</form>
</tr>
</table>
</div>
而且我ajax_check_login.php是
<?php
include "../connect.php";
/* print "<script type=\"text/javascript\">";
print "alert('I have been called')";
print "</script>"; */
if(isset($_POST['wantedusername'])){
$wantedusername = mysqli_real_escape_string($mysqli, $_POST['wantedusername']);
$usernamecheckstring = "SELECT 1 FROM table WHERE username='".$wantedusername."'";
$usernamecheckquery = mysqli_query($mysqli, $usernamecheckstring);
$rows =mysqli_num_rows($usernamecheckquery);
echo $rows;
}
?>
任何幫助，將不勝感激一如既往。
問候 Maudise
您需要刪除'SRC =」 ../ JS/jQuery的1.7。 1.min.js「'來自包含JavaScript代碼的腳本標記 - 」src'屬性僅在引用外部文件時有效。 –
添加爲答案，解決它，我已經添加了該行，並刪除了好幾次，它似乎沒有幫助。我花了15分鐘寫在這裏，它的工作原理！謝謝 – Maudise