Question

我遇到了一个奇怪的问题，我会尽力解释。我正在使用jQuery对话框进行表单提交，除了我第一次打开表单并单击提交时，它提交表单并正确输入数据库中的所有工作。但是，如果我再打开另一个表单并提交它会创建2个相同数据的帖子。它会在每次提交时将帖子递增1。如果我刷新浏览器，它会发布1个正确的条目。为什么我必须不断刷新浏览器才能正确发布？我已经发布了我正在使用的代码，如果有人能告诉我我的错误，我将不胜感激。非常感谢

PS。假设与db的所有连接都是正确的。

jQuery dialog code
==================================================

function feedbacknew() 
{    
    $("#form").dialog(
    {
        autoOpen: false,
        resizable: false,
        modal: true,
        title: 'Submit a feedback request',
        width: 460,
        height: 490,
        beforeclose: function(event, ui) { $("#message").html(""); }

    });

    $('#submit').click(function () 
    {
        var name = $('.uname').val();
        var company = $('#company').val();
        var email = $('.email').val();
        var position = $('.position').val();
        var feedback = $('.feedbacknew').val();
        var data = 'uname=' + encodeURIComponent(name) + "&company=" + encodeURIComponent(company) + "&email=" + encodeURIComponent(email) + "&position=" + encodeURIComponent(position) + "&feedback=" + encodeURIComponent(feedback);
        $.ajax(
        {
        type: "POST",
        url: "feedback.php",
        data: data,
        success: function (data) {
            $("#feedback").get(0).reset();
            $('#message').html(data);
            //$("#form").dialog('close');
            $("#flex1").flexReload();
            } 
        });
        return false;

    });

    $("#form").dialog('open');

}


![html form code
====================================================

// Feedback form
<div id="form" style="display:none;">
  <form method="post" id="feedback" class="webform" name="feedback">


        <label for="company">Company</label>
        <select name="company" id="company">
          <option SELECTED VALUE="">Select an option</option>
          <option value="Technical">Technical</option>
          <option value="Database">Database</option>
          <option value="Error">Error</option>
          <option value="Other">Other</option>
        </select>
        <label for="name">Full Name:</label>
        <input id="uname" name="uname"  type="text" class="text ui-widget-content ui-corner-all inputbox uname" value="<?php echo $_SESSION['kt_name_usr']; ?>" />
        <label for="email">Email address:</label>
        <input id="email" name="email" type="text" class="text ui-widget-content ui-corner-all inputbox email" value="<?php echo $_SESSION['kt_email_usr']; ?>"/>
        <label for="position">Position:</label>
        <input id="position" name="position" type="text" class="text ui-widget-content ui-corner-all inputbox position" />
        <label for="feedbacknew">Feedback:</label>
        <textarea id="feedbacknew" name="feedbacknew" cols="25" rows="3" type="text" class="text ui-widget-content ui-corner-all inputbox feedbacknew">Please make sure that any error messages or numbers are listed here </textarea><br />

        <button id="submit" class="submit">Submit</button>
        <div id="message"></div>
  </form>
</div>

deedback.php code
=============================================

<?php
$company = $_POST['company'];
$name = $_POST['uname'];
$email = $_POST['email'];
$position = $_POST['position'];
$feedback = $_POST['feedback'];

?>
<div style="padding:2px; color:#ff0000;font-size:12px;font-weight:normal"><?php echo "<br />" . "Thank you. You can now close this window." . "<br />"; ?></div>

<?php

mysql_query("INSERT INTO feedback (department, name, email, position, feedback, date) VALUES ('$company', '$name', '$email', '$position', '$feedback', NOW())");
?>][1]

重复的帖子项目

0 个答案: