doufei8691 2013-12-11 21:57 采纳率: 100%
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在表单页面或弹出窗口中的div中回显sendmail.php表单中的var

The idea is to preview (this part works fine) the form somewhere so, if corrections are needed we can go back to the unrefreshed form to make corrections.

I have this form:

<form action="../../../../sendmail.php" id="Formulaire" method="post" name="Formulaire" onsubmit="return checkform()" target="_self">

I have a submit button to preview:

<input id="captcha" name="_Preview" type="submit" value="Preview" />

And here is my php code (from sendmail.php)

if(isset($_POST['_Preview'])) { echo $text; ?>

As stated above the preview (content of array $text) works well although it is presented on a blank page.

I like it to be presented either in a div on the form page or in a popup Window so when I close the popup I am back on the non-refreshed form.

I tryed different ways, my problem is everytime, after submitting the preview button I endup having the sendmail.php form in the background (blank of course or with the preview data). I don't know what approach to take. Thank you for your help.

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