Re: [PHP-DB] trying to dump out table info

[Govinda]

On Jul 13, 2009, at 2:58 PM, Daniel Brown wrote:
On Mon, Jul 13, 2009 at 16:53, Govinda<govinda.webdnat...@gmail.com>  
wrote:
Hi all

I am not sure why this is failing/giving error.  I am quite new at
PHP/Mysql, and apologize for what will undoubtedly end up being a  
trivial
error on my part.

   Hey, I recognize that code.... ;-P

:-)

Here is the gifted code:

$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
if (!$db_billing) { die('Could not connect: ' . mysql_error()); }

       $sql = "SHOW TABLES";

       $result = mysql_query($sql) or die(mysql_error());

   Note the addition to the $result definition.  This will cause the
script to exit with the MySQL error message (if any) if the issue is
in the query.

when I make that change, it returns this:

"No database selected"

so I try to modify like 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'";
        
        $result = mysql_query($sql) or die(mysql_error());
        
        foreach(mysql_fetch_assoc($result) as $k => $v) { //line 64
        //{DESCRIBE | DESC} tbl_name [col_name | wild]
        //$ssql = "DESCRIBE ".mysql_real_escape_string($v);
        $ssql = "DESCRIBE myTable ".mysql_real_escape_string($v);
        $rresult = mysql_query($ssql);
        echo "<b>".$k."</b>:<br />\n";
        echo "<pre>\n";
        print_r(mysql_fetch_assoc($rresult)); //line 71
        echo "</pre>\n";
        echo "<br />\n";
        }

and now the script returns this:

Tables_in_billing (myTable):

Warning:  mysql_fetch_assoc(): supplied argument is not a valid MySQL  
result resource in /home/meee/public_html/somedir/test.php on line 72


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