I am having probs with the ifstatement below. It works or it displays information after it and no errors arise. But the field is empty. And it displays a 1 for using the <li> and it should just skip the line because I have checked my database and the field is empty. Can someone help me with. Thanks in advance
Sincerely John
<?php
$themixture = urldecode($thedrinkname);
$linkID = @mysql_connect("mysql.addr.com", "blabla", "******");
mysql_select_db("blabla", $linkID);
$resulta = mysql_query("SELECT drinkname, liqour, one, two, three, four, five, six, seven, eight, nine, ten, directions FROM mixedrinks WHERE drinkname = '$thedrinkname' and liqour = '$major'", $linkID );
print "<table>";
for ($x= 0; $x < mysql_num_rows($resulta); $x++)
{
$row = mysql_fetch_assoc($resulta);
$teni = mysql_result($resulta, 0, "ten");
print "<tr>";
print "<td><font size=2 face=tahoma,arial><b><br>DRINK NAME: " . $row[drinkname] . "<br>";
if (($teni == NULL) or ($teni == ''))
{
continue;
}
else
{
print "<li>";
print mysql_result($resulta, 0, "ten");
}
print "</td>";
print "</tr>";
}
print "</table>";
?>