[RESOLVED] State Array

[prkarpi]

I came across a code for choosing a state:

    $states_arr = array('NO'=>"I am not from US",'AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'... and so on);
    function showOptionsDrop($array){

    $string = '';
    foreach($array as $k => $v){        
        $string .= '<option value="'.$k.'"'.$s.'>'.$v.'</option>'."\n";     
    }
    return $string;
}

I want to make an HTML option "selected" into var ($s) on top when I pass the SQL variable ($st) into this script. I have modified the script:

    $states_arr = array('NO'=>"I am not from US",'AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'... and so on);
    function showOptionsDrop($array){
global $st; (MySQL var)
        $string = '';
        foreach($array as $k => $v){
if ($st = $k) {
$s = "selected";
}        

            $string .= '<option value="'.$k.'"'.$s.'>'.$v.'</option>'."\n";     

        }
        return $string;
    }

I've tried different variations, but it seems that $st is not registering within the function itself.

Ideas anyone? Thank you.

[rincewind456]

I would change your function to this

function showOptionsDrop($array,$mysqlVar){
	$string = '<select name="state">';
	foreach($array as $k => $v){
		if ($mysqlVar == $k) {  /* = assigns a value == compares a value*/
			$s = "selected";
		}
		$string .= '<option value="'.$k.'" '.$s.'>'.$v.'</option>'."\n"; /*Leave a space before selected*/
		unset($s);
	}
          $string .='</select>';
	return $string;

}
/*Usage*/
print showOptionsDrop($states_arr,$mysqlVar);

You were assigning a value within the if rather than comparing a value.

You also don't need a global var within the function, simply pass the mysql var as an argument.

And you need to [man]unset()[/man] the selected var or all the options after the first selected one will also be selected.

[prkarpi]

Great. Thanks rincewind456. I see my mistake. I didn't pass the MySQL var to the print output and used a wrong operator in IF statement. Thank you!