Array problem

[drummer101]

When echoing my array, it evaluates to, "Array"  ???

 

$group_id = mysql_query("SELECT phpbb_user_group.group_id FROM phpbb_user_group WHERE phpbb_user_group.user_id = '".$i['user_id']."'");
$last_online = date('g:ia l, M jS Y');
$j = mysql_fetch_array($group_id);
$group = array($group_id);
	if($password == $i['user_password']){
		echo $group;
		if(array_key_exists("6",$group)){
			echo "Works";
		} else {
			echo "Doesn't work ";
		}
                }

 

the sql statement evaluates to 6,19,2

 

Trying to figure out how to get the sql statement into a php array to case check.

[Daniel0]

You cannot echo an array like that. If you wish to represent an array as a string then use print_r(), var_dump or var_export().

[drummer101]

Any idea why the case check is evaluating to false?

 

or should I be using array($j) instead of array($group_id)?

[Daniel0]

Yes, you should use $j instead.

[drummer101]

Hrmm I'm having a problem getting the SQL statement into the php array. Only the last column, 6 will echo when I check it.

 

How can I store all three results in an array?

[Daniel0]

Loop through it. There are some examples here: http://php.net/mysql