Passing a form variable twice

judson

I need to pass two form variables twice. What I am doing is allowing users to design signs online. They have the option for a border color or no border. This is an example:

if $POST['bordercolor'] == "none"
then bordercolor="$POST['bg_color']"
else bordercolor="$_POST['bordercolor']"

What this is doing is making the border color the same as the background if they choose no border. The syntax in the example is not correct, I know that. What I am wondering is can I declare the variable as a seperate variable and pass that value in the second instance, like so:

##at the topof the page##

$POST['bordercolor'] == $bordercolor
$POST['bg_color'] == $background

##in table##

if $_POST['bordercolor'] == "none"
then bordercolor="$background"
else bordercolor="$bordercolor"

Again, I know the syntax is incorrect, I know the correct synax (for this case anyways). I just need to know how to pass the variable twice. I am not using a database.

Pig

Sorry, I don't understand what you are asking, or how passing it twice comes into play. Can you explain it a different way? Are you just asking how to pass form variables through 2 pages?

judson

On design1.php, the user specifies the background size, border color, background color, letter size, font and letter color. This passes the form variables to design2.php. This is what I am trying to do: If the user chooses no border, the border color should default to the backgound color. I keep getting the border showing up with a '' value. I didn't know if it was because I was trying to pass the background color variable twice or what. When I substitute the php code in the border color value, thereby removing the second reference, it works. So my if/then statement syntax is correct. Might be a coding error, I don't know! Thanks for the help.

Pig

If you post up the actual code you are using, it would help alot.

yelvington

You can refer to a variable as many times as you want.

if ($POST['bordercolor'] == 'none')
$bordercolor = $POST['bg_color'];
else $bordercolor = $_POST['bordercolor'];

is perfectly fine.

If you are having problems, it is likely that your issue stems from the HTML in your form.