Maintain CHECKED state in form?

patrickackerman

New to PHP and not all that versed in programming.
I've created a form that contains a number of checkbox input fields. The form posts to $PHP_SELF and will mail the form if complete or reload the form if certain fields are incomplete. In the case when the page reloads how do you identify that a check box input field in a form is "CHECKED" and have the input maintain a checked state when the form reloads?

Thanks.-Patrick

Anon

Use an if isset() check:

if ( isset ($checkbox_variable_name) ) {
// output checked stuff, etc.
}

The variable will actually be set to 'on' but if the checkbox was not set the variable won't even exist.

patrickackerman

Thanks Ken. I'm still unlcear though how to use the isset function. This is my code:

<body>
<?
$form_block = "
<form method=\"post\" action=\"$PHP_SELF\">
Name<input type=\"text\" name=\"name\" VALUE=\"$name\"><br>
<input type=\"checkbox\" name=\"audition\" value=\"Auditioning\">Audition Information<br>
<input type=\"checkbox\" name=\"volunteer\" value=\"Volunteer Opportunities\">Volunteer Opportunities
<input type=\"submit\" name=\"Submit\" value=\"Send This Message\">
</form>
";

if ($name == ""){
$name_err = "Please enter your name.";
$send = "no";
}

if ($send != "no") {

	$msg = "Sender's Name:$name\n";
	$msg .= "Send Info About:$audition, $volunteer\n";

	mail("patrick@patrickackerman.com", "", $msg, "");
	echo "Thank you";

} else if ($send == "no") {

	echo "$name_err";
	echo "$form_block";
	}

?>
</table>
</body>

Anon

If the checkbox has been checked then the name of the checkbox wil exist as a variable.

eg: you have a checkbox where name="audition"

If it was checked it will have the value 'on' (or perhaps 'Auditioning' in your code). Otherwise the variable does not exist. The proper way to check for existence is to use the isset() function (check the manual).

So in your reloaded form you would have code that might look like:

...
<input type="checkbox" name="audition" <? if ( isset( $audition ) ) PRINT "CHECKED"; ?>>
...

Clean it up and format it whatever way you like. That's the guts of how you do it.

patrickackerman

Thanks. I tried putting the isset function within the input field, but it wasn't processing the function, just printing it into the HTML. It would recognize the $audition value being checked, but not print. So instead I moved the function to before the form area and I just created a new variable to pass the "CHECKED" value to the input field.

if(isset($audition)) {$concertchecked="CHECKED";}

I liked your way better, but just couldn't get the function to be processed. This is what I was trying

Quotes were escaped. Don't know what else to try. If you have any other suggestions, I'd appreciate it. Thanks again for all of your help.
-Patrick