Question

The code below works great. It prints out items in a MySQL database as an HTML table. However, the database has entries where the first column (the variable “site”) is blank. How can I have this table exclude rows / entries where “site” is blank?

$result=mysql_query("SHOW TABLES FROM database LIKE '$find'")
or die(mysql_error());


if(mysql_num_rows($result)>0){
while($table=mysql_fetch_row($result)){
print "<p class=\"topic\">$table[0]</p>\n";
$r=mysql_query("SELECT * , itemsa - itemsb AS itemstotal FROM `$table[0]` ORDER BY itemstotal DESC");



print "<table class=\"navbar\">\n";
while($row=mysql_fetch_array($r)){


$itemstotal = $row['itemsa'] - $row['itemsb']; 

print "<tr>";


print "<td class='sitename'>".'<a type="amzn" category="books" class="links2">'.$row['site'].'</a>'."</td>";

print "<td class='class1'>".'<span class="class1_count" id="class1_count'.$row['id'].'">'.number_format($itemstotal).'</span>'."</td>";
print "<td class='selector'>".'<span class="button" id="button'.$row['id'].'">'.'<a href="javascript:;" class="cell1" id="'.$row['id'].'">'.Select.'</a>'.'</span>'."</td>";
}
print "</tr>\n";
}
print "</table>\n";



}
else{
print "";
}

Answer 1

Add a where clause to the SQL query.

WHERE `site` <> ''
or
WHERE `site` != ''

should work.

However, if you want the rows where site is blank for other PHP options, it would be better to filter out the blank sites in the PHP and not the MySQL query.