Re: trying to dump out table info

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

On Mon, Jul 13, 2009 at 17:24, Govinda<govinda.webdnatalk@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: $db_billing=mysql_connect(localhost,metheuser,mypass,billing); if (!$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?