Question

php Warning: mysql_query() expects parameter 2 to be resource

$database = "test";
$tablename = $_REQUEST['tablename'];
echo "<h2>Data from $tablename</h2>";
include('conn.php');

$query = "show columns from $tablename";
$result = mysql_query($query,$tablename);
$column = 0;
if ($tablename && $query)
{
echo "Found these entries in the database:<br><p></p>";
echo "<table width=90% align=center border=1><tr>";
while ($r = mysql_fetch_array($result))
{
echo "<td align=center bgcolor=#00FFFF>$r[0]</td>";
$colname[$column] = $r[0];
$column = $column + 1;
}
echo "</tr>";

Answer 1

mysql_query($query)就够了，不需要传$tableName

Answer 2

mysql_query()
第二个参数应该是database的handle，而上面的代码是表单提交的tablename,是字符串，不匹配
mysql_connect('host','user','password','database') 返回这样一个handle