i cant get it to output

[gotornot]

Im trying to run a simple if command and this doesnt seem to work im trying to get it to hiighlight the option already selected in the db any ideas why?

function selected()
{
global $selectedno;
//for loop to loop through 10
$a = 11;
for ($i = 0; $i <10; $i++)
{
$a = --$a;
$num = $a;
if ($selectedno == $num)
{
$bash = ' selected="selected"';
}
echo '<option value="' . $a . '"' . $bash . '>' . $a . ' ' . $selectedno . ' </option>';
}
}

[rincewind456]

If you change your code to this :

function selected($selectedno)
{
	//for loop to loop through 10
	$a = 11;
	for ($i = 0; $i <10; $i++)
	{
		$a = --$a;
		$num = $a;
		if ($selectedno === $num)
		{
			$bash = ' selected="selected"';
		}
		echo '<option value='.$a.''.$bash.'>'.$a.'</option>'."\n";
		$bash ='';
	}
}

$number = 5;
selected($number);

It negates the need for a global variable.

BTW you have to reset the variable $bash to empty after each iteration to ensure that only one item is selected, and you dont need the double quotes around the item number.

[gotornot]

Im not sure but the other page declairs the number already in the database so how does this interact?

im calling the function to solely counth through quanitites and show up the one tht has already been selected

[rincewind456]

If you're getting the variable $selectedno from somewhere else then you would use it as so:

selected($selectedno);