Question

For the code below, I am trying to use the IF statement

if(mysql_num_rows($resA)>0)

to avoid running the foreach loop if $entry is not in the “site” column in any table in my database. However, if $entry does not exist in the “site” column in any table in my database, I get the error message “Warning: Invalid argument supplied for foreach()”. Any idea why it is doing that?

Thanks in advance,

John

$result = mysql_query("SHOW TABLES FROM feather") 
or die(mysql_error()); 

while(list($table)= mysql_fetch_row($result))
{
  $sqlA = "SELECT COUNT(*) FROM `$table` WHERE `site` LIKE '$entry'";

  $resA = mysql_query($sqlA) or die("$sqlA:".mysql_error());
  list($isThere) = mysql_fetch_row($resA);
  $isThere = intval($isThere);
  if ($isThere)
  {
     $table_list[] = $table;
  }

}

if(mysql_num_rows($resA)>0){
foreach ($table_list as $table) { 
    $sql = "SELECT votes_up FROM `$table` WHERE `site` LIKE '$entry'"; 
    $sql1 = mysql_query($sql) or die("$sql:".mysql_error());
   while ($row = mysql_fetch_assoc($sql1)) {
       $votes[$table] = $row['votes_up'];
       $sum += $row['votes_up'];
   } 

}
}
else{
print "";
}

Answer 1

It’s a good idea to initialize variables:

$table_list = array(); // <-- Initialize variable.
while (list($table) = mysql_fetch_row($result)) {
    /* ... */
    $table_list[] = $table;
    /* ... */
}

if (mysql_num_rows($resA) > 0) {
    foreach ($table_list as $table) { 
        /* ... */
    }
}

You’re getting the error because no entries are added to $table_list which means the variable won’t exist when you run the foreach loop. Initialize it first to avoid this confusion.

The reason no entries were added to $table_list is because the count of all tables fetched was zero.