PHP Javascript：关闭弹出式窗口，在后台调用父窗口

Question

Let's say a parent window is www.mysite.com/index.php has several links. When user clicks on these links, a popup is opened.

The popup has some forms. When the user clicks submit button, another script is called within the popup like this:

if ((isset($_POST['Complete'])) && ($_POST['Complete']=='Complete')) {
    unset($_POST['Complete']);

    include 'result.php';

    return;
}

The result.php enters the data into database and pass the result to parent window with the following

<script language="JavaScript" type="text/javascript">
    window.opener.location.href = 'submission.php';
    window.close();
</script>

The parent window then show the final result, update the master database and emails the users.

Now here is the problem.

1. User clicks the link on parents.
2. Form opens, but user click close (browser cross button).
3. The user then closes parent window
4. The user then visit some other page of the site; say other.php
5. But the master database is getting updated and the user is getting the email (All null values though).

Since the form was never completed, the result.php should never be called. and if result.php is never called, submission.php should never be called.

What is happening here?

Answer 1

将关闭按钮的类型更改为button。