【发布时间】:2014-11-12 14:25:13
【问题描述】:
我有一个登录页面,要求用户输入代码。我想计算用户输入错误代码的次数,如果超过 3 次,则显示一个页面告诉他们重新启动进程,但我的计数器只输出 1,即使我多次输入错误代码。
<div class="container">
<div class="alert alert-success" role="alert" align="center">
<p>A code has been sent to your email</p>
<p>Please consult your email to proceed with the login</p>
</div>
<form class="form-signin" method="post">
<h2 class="form-signin-heading">Enter code</h2>
<input type="hidden" name="partnerEmail" value="<?php echo $_POST['partnerEmail']; ?>">
<input class="form-control" type="text" name="partnerCode" placeholder="Code">
<button class="btn btn-lg btn-primary btn-block" type="submit">Accept</button>
</form>
</div> <!-- /container -->
<?phpsession_start();echo $_SESSION['partnerEmail'];?>
<div class="container">
<?php
include_once "conn.php";
$partnerCode = $_POST['partnerCode'];
$partnerEmail = $_SESSION['partnerEmail'];
$sql = "SELECT * FROM partners WHERE partner_email='$partnerEmail' AND
partner_login_code='$partnerCode' AND partner_active ='yes'";
$counter = 0;
echo $counter;
$result = $conn->query($sql);
if (mysqli_num_rows($result)>0){
echo '<div class="alert alert-success" role="alert" align="center">
<p>Congratz - You logged in successfully!</p>
</div>';
}
else
{
echo '<div class="alert alert-danger" role="alert" align="center">
<p>ERROR - Please consult your email for the correct code!</p>
</div>';
$counter ++;
echo $counter;
?> <script> $(".alert").effect("shake");</script> <?php
}
echo $counter;
?>
【问题讨论】:
-
旁注:如果这是您的实际代码,您需要将这个
<?phpsession_start()分开。<?php session_start() -
我没有尝试更新查询。我正在尝试计算 else 语句通过了多少次。
-
<?phpsession_start()是你的实际代码吗?正如我已经说过的那样。
标签: php jquery mysql database counter