If else help

Ray · Oct 19, 2005

ok.. here is my problem I get the following error from the code. I cant fiqure out what is wrong with my code.

Parse error: parse error, unexpected T_ELSE in /home/hughraho/public_html/inc/order/1.php on line 37

PHP:

if ($plan=="");
{
$plan = $_POST['plan'];
}
if ($plan=="1");
{
$acctplan = "hughraho_starter";
$vplan = "Starter";
}
if ($plan=="2");
{
$acctplan = "hughraho_advanced";
$vplan = "Advanced";
}

if ($plan=="3");
{
$acctplan = "hughraho_professional";
$vplan = "Professional";
}else;{
$acctplan = "hughraho_error";
$vplan = "Error";
	}

luxinterior · Oct 19, 2005

That error is usually a red-herring and is caused by other code not being properly terminated. You should learn to use SELECT CASE statements too, much cleaner. http://www.php.net/manual/en/control-structures.switch.php

Lux

Ray · Oct 19, 2005

thanks mate... ill try that... but i need to fiqure out this else command, bcecuase im having the same error on other pages

luxinterior · Oct 19, 2005

Try taking the semi colon off the end of the 'else;' statement.

Lux

Ray · Oct 19, 2005

I know for a fact it doesnt matter if thats there.

For my problem... I found that there was a ; next to all the if statements.. I have no clue why i pud them there while i new they didnt belong there

Peter · Oct 19, 2005

after all the if lines you have a ; which should not be there (should only be after each line within the statment)

the same goes for the else line as luxinterior stated.

Having these there you are confusing the php parser as it thinks it has reached the end of a line of code when it has not.

Regarding php errors sometimes being a red herring. The errors are ussually quite accurate sometimes the line number is wrong, if it is it's ussually the line before the error is on.

SiKing · Oct 19, 2005

Try...

PHP:

if (!isset($_POST['plan']))
{
$plan = $_POST['plan'];
}
elseif ($plan=="1")
{
$acctplan = "hughraho_starter";
$vplan = "Starter";
}
elseif ($plan=="2")
{
$acctplan = "hughraho_advanced";
$vplan = "Advanced";
}

elseif ($plan=="3")
{
$acctplan = "hughraho_professional";
$vplan = "Professional";
}
else
{
$acctplan = "hughraho_error";
$vplan = "Error";
}

Peter · Oct 19, 2005

that code is still wrong miseria

anyway why not just do the following :-

PHP:

switch ($plan) {
case 1:
   $acctplan = "hughraho_starter";
   $vplan = "Starter"; 
   break;
case 2:
   $acctplan = "hughraho_advanced";
   $vplan = "Advanced"; 
   break;
case 3:
   $acctplan = "hughraho_professional";
   $vplan = "Professional"; 
   break;
default:
   $acctplan = "hughraho_error";
   $vplan = "Error";
}

If else help

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