On Jul 13, 2009, at 4:40 PM, Daniel Brown wrote:

On Mon, Jul 13, 2009 at 17:24, Govinda<govinda.webdnat...@gmail.com> wrote:

"No database selected"

   You're getting warmer....

   How would you select a MySQL database in PHP?  (Hint: there's a
built-in function.)

To try and follow your lead and focus just on what you are saying, to be rid of that first error that said "No database selected",
 I changed the code back to this:

        if (!$db_billing) { die('Could not connect: ' . mysql_error()); }

        //$sql = "SHOW TABLES FROM billing LIKE 'mytable'";
        $sql = "SHOW TABLES"; //line 237
        //$result = mysql_query($sql);
        $result = mysql_query($sql,$db_billing) or die(mysql_error());
//$result = $db_billing->query("$sql") or die(mysql_error()); // line 241
        foreach(mysql_fetch_assoc($result) as $k => $v) { //line 243
        //{DESCRIBE | DESC} tbl_name [col_name | wild]
        $ssql = "DESCRIBE ".mysql_real_escape_string($v);
        //$ssql = "DESCRIBE mytable ".mysql_real_escape_string($v); //line 246
        $rresult = mysql_query($ssql);
        echo "<b>".$k."</b>:<br />\n";
        echo "<pre>\n";
        print_r(mysql_fetch_assoc($rresult)); //line 250
        echo "</pre>\n";
        echo "<br />\n";

The question is why does line 240 think that no db is selected?!
I read the docs for mysql_query and it says:
resource mysql_query  ( string $query  [, resource $link_identifier  ] )
so that 2nd param is optional; "If the link identifier is not specified, the last link opened by mysql_connect() is assumed". But anyway I stuck it in there too, to test, and I get the same result, i.e. "No database selected".
I tried line 241 too.  No luck.
So I am really stumped (lame as that feels/sounds). Line 237 is just a string assignment. Line 240 *IS* being handed the $link to use, which AFAIK means a reference to the database as well as the table, so why the error?

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