Retaining checkbox state on redisplay

[Jankie]

Greetings everyone !
I seek some help to retain checkbox state on redisplay.I only have one optional checkbox to indicate whether it's the user's first visit to the site

Form1:

<INPUT type="checkbox" name="guest" value="yes">
 

[Please use CODE tags when posting source code. Thanks! --pbmods]

I tried it as below on Form2,but didnt work for me

Form2:

<INPUT type="checkbox" name="guest" value="yes" <?php if(guest == "yes"){echo " CHECKED";}?>>

Any ideas?

Thank you

[epots9]

it would be:
[PHP]
if($_POST["guest"] == "yes")
[/PHP]

or:
[PHP]
if($_GET["guest"] == "yes")
[/PHP]

if your first form has the method set to post then use the first example, if it is set to get then use the second example.

good luck

[Jankie]

Thank you very much Epots9 ! really appreciate it
And good luck to you, too

[pbmods]

Heya, Jankie.

If you're curious, when you do this:

<?php if(guest == "yes"){echo " CHECKED";}?>

What PHP is actually doing is looking for a constant named 'guest'. Since (presumably), you didn't define a constant (using the define keyword), PHP will instead use the string literal 'guest'. Since 'guest' will never == "yes", your script will never echo " CHECKED".

Here's an explanation of why this happens (used to explain array indexes, but the principle is the same for strings):
http://us2.php.net/manual/en/languag....array.foo-bar

P.S., for full XHTML compliance, you should echo " checked=\"checked\"" instead ~_^

[Jankie]

pbmods
Interesting explanation,
i actually meant it the correct way,but sometimes as it turns out in my first code,the concept alone does not kill it,php-aware is essential
Thank you all for sharing this insight !