Re: display a hiddenfield

Dominique ANOKRE wrote:

I use a simple form like this : print("<form method=\"post\" action=\"image.php\">\n"); print("<input type=\"hidden\" name=\"hiddenField\">\n"); This will result in

<input type="hidden" name="hiddenField"> That is a hidden field without a value. This will get to the browser.

$_POST["hiddenField"] = $image_avant; This will assign a value to $_POST["hiddenField"], which means just that, nothing more. It does not change html output.

You need to use: print("<input type=\"hidden\" name=\"hiddenField\" value=\"". htmlspecialchars($image_avant) ."\">\n");

print("<input type=\"submit\" name=\"avant\" value=\"Avant\">\n"); print("</form>\n"); and when i click on the submit button, i want to display the value of my hiddenField (doing by the file image.php) : <?php echo $_POST["hiddenField"]; ?> but nothing is displaying !!!!!! Please help !