Table output of a database

[Anon]

Hello, I am trying to print the contents of a database into a 2 row table. I have it working, and it prints in both rows, but the problem that I am having is that it skips the first entry. Here is my code:

$j = 1;

while($data = mysql_fetch_array($result)) {
	$id = $data["id"];
	$company = $data["company"];
	if($j == 1) {
		echo "\t<tr>\n";
		echo "\t\t<td bgcolor=\"#dcdcdc\"><font face=\"Tahoma, Arial, sans-serif\" size=\"2\"><input type=\"checkbox\" name=\"$id\" value=\"$company\">&nbsp;&nbsp;$company</font></td>\n";
		$j = 2;
	}
	elseif($j == 2) {
		echo "\t\t<td bgcolor=\"#dcdcdc\"><font face=\"Tahoma, Arial, sans-serif\" size=\"2\"><input type=\"checkbox\" name=\"$id\" value=\"$company\">&nbsp;&nbsp;$company</font></td>\n";
		echo "\t</tr>\n";
		$j = 1;
	}
}

Just before that I have it connect to the database, yada yada. I have the output of the database, but not where $id = 1. When I just try to echo $id, it gives me 2 - 49 not 1 either. Any ideas?

Thanks,

Jacob Ham

[Anon]

Can we see the code from when you query to this loop? Because generally, your problem happens when you call mysql_fetch_array() before the loop but don't use it...

Diego

[Anon]

$sql_query = "SELECT * FROM loc_contacts";
$result = mysql_query($sql_query);
if(!$result) { echo sql_error(); }
$data = mysql_fetch_array($result);

That is query to the database.

[Anon]

There you go:
$data = mysql_fetch_array($result);

exactly what I predicted the problem was.

Diego