CSS styles and snippets?

How do you get a snippet to respond to an class name assingned in an external css sheet included in the mother template.
i can not get the snippet to run if i put a class identifier in the script

Help please

Can you post a sample of your snippet code where you are putting the class identifier? make sure to enclose it in the tags for easier reading.

$global_dbh = mysql_connect(’JONSERV’, ’thegent’, ’gringo’);
mysql_select_db(’formula1’, $global_dbh);

function display_db_table($connection )
{
$query_string = "SELECT * FROM drivers";
$result_id = mysql_query($query_string);
$column_count = mysql_num_fields($result_id);

print("<table border=0>\n");
while ($row = mysql_fetch_row($result_id))
{
print("<tr align=left valign=top color=#ff0000>");
for ($column_num = 0;
$column_num < $column_count;
$column_num++)
print("<td
...class="Venue3L"...
>$row[$column_num]</td>\n");
print("</tr>\n");
}

that class is defined in an external css file which is included in the template containing snippet
}
display_db_table($global_dbh);

You have to escape the quotes, instead of class="Venue3L" use class=\"Venue3L\"

Thanks Doze
sorted