为什么我的模态不能通过html表单工作?

时间:2015-06-29 12:32:08

标签: php html mysql modal-dialog

为什么我的模态不起作用?使用POST方法是错误的吗?当我加载页面并单击忘记密码的链接时,最终显示模态,然后当我输入我的电子邮件时没有任何反应。有人能帮助我吗?

这是我的PHP脚本。

<?php
require 'db.php';
$msg='';
if(isset($_POST['submit']))
{
$emailRet = mysqli_real_escape_string($connection, $_POST['emailRetrieve']);

$sql2 = mysqli_query($connection, "SELECT Email_Address, Password FROM pawnshop WHERE Email_Address='".$emailRet."'");

while ($row = mysqli_fetch_array($sql2, MYSQL_ASSOC)) {
    $emailRetrieve = $row['Email_Address'];
    $passwordRetrieve = $row['Password'];
}

  $number = mysqli_num_rows($sql2);
  echo $number;

  if(mysqli_num_rows($sql2) < 1){  
    require 'smtp/Send_Mail.php';

    $to = $emailRetrieve;
    $subject = "PBMS Password Retrieval";
    $body ='Hi, This is your account information<br><br> Username:'.$emailRetrieve.'<br> Password:'.$passwordRetrieve.'';

    Send_Mail($to,$subject,$body);
    $msg='Check your email to get your password';
    echo 'send';
  }
  else
  {
    $msg='Check your email and try again';
  }

}
?>

这是我对Modal的HTML代码

    <html>
    <head>
      <link type='text/css' href='css/modal.css' rel='stylesheet' media='screen' />
      <script type='text/javascript' src='js/jqueryModal.js'></script>
      <script type='text/javascript' src='js/jquery.simplemodal.js'></script>
      <script type='text/javascript' src='js/modal.js'></script> 
    </head>
    <body>
    <form method="post">
    <div id='basic-modal'>
              <center>
                 <a href='#' class='basic custom-font-reg'>Forgot password?</a><br>
                 <a href='Registration.php' class='custom-font-reg'>Register for an account?</a>
              </center>
              </div>

              <hr class="hr-custom2">

              <div id="basic-modal-content">
                 <div class="getPasswordWrapper">
                 <hr class="PasswordHeaderColor"></hr>
                 <p class="enterEmailPasswordText">Enter your username to get your password</p>
                 <center>
                      <input type="text" name="emailRetrieve" class="getPassword" placeholder="Your Username\Email" required/>
                      <button type="submit" class="getPasswordButton">Send</button>
                      <span><?php echo $msg; ?></span>
                 </center>
                 </div>
              </div>
    </form>
    </body>
    </html>

2 个答案:

答案 0 :(得分:2)

首先,您没有提交&#34;提交&#34;与$_POST['submit']一起使用的name属性以及它失败的原因,因此if(isset($_POST['submit'])){...}中的任何内容都不会被执行。

这样做:

<button type="submit" class="getPasswordButton" name="submit">Send</button>

或输入:

<input type="submit" class="getPasswordButton" name="submit" value="Send">

另外,MYSQL_ASSOC需要MYSQLI_ASSOC。你不能混合使用MySQL函数。

但是,查看可能需要使用if(mysqli_num_rows($sql2) < 1){运算符更改的>。做< 1告诉MySQL它是否不存在; 洞察

答案 1 :(得分:1)

您的表单必须具有<form action="your/php/script.php" method="post"> 属性。其值必须是处理表单的脚本文件的名称。

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

或者,如果php脚本与表单在同一页面上,您可以写:

namespace oP
{
      enum adjustment
      {
         AUTO_OFF,
         AUTO_ONCE,
         AUTO_CONTINUOUS,
         AUTO_SEMI,
         ABSOLUTE,        // The line that the errors point to.
         NUDGE
      };
}