Re: Mysql Variable Displayed in php

From: Date: Tue, 13 May 2003 01:58:04 +0000
Subject: Re: Mysql Variable Displayed in php
References: 1  Groups: php.db 
Request: Send a blank email to php-db+get-27686@lists.php.net to get a copy of this message
Thanks Jennifer i'll try this one. Regards, Delz ----- Original Message ----- From: "Jennifer Goodie" <goodie@apollointeractive.com> To: "Delz" <phpuser@modeglobal.com>; <php-db@lists.php.net> Sent: Tuesday, May 13, 2003 1:54 AM Subject: RE: [PHP-DB] Mysql Variable Displayed in php > while ($row = mysql_fetch_row($result)) { // Begin while > echo($row[0]); //this is one way > echo($row['overall']); //this is another way > } // end while > > > > -----Original Message----- > > From: Delz [mailto:phpuser@modeglobal.com] > > Sent: Friday, May 09, 2003 8:38 PM > > To: php-db@lists.php.net > > Subject: [PHP-DB] Mysql Variable Displayed in php > > > > > > RE: [PHP-DB] mysql variable displayed in phpHi All, > > > > I have this following script in php connecting to a mysql > > database and declared a variable named "overall" in the mysql statement: > > How do I display the result in php generated by the mysql > > variable "overall" so I can display it in my browser. I'm new to > > php & can't > > seem to output the result. Any help will be appreciated. > > > > ############################################################ > > <? > > > > $connection = mysql_connect("myhost","myuser","mypasswd"); > > if (!$connection) { > > echo "Could not make a connection!"; > > exit; > > } > > > > $db = mysql_select_db("mydb",$connection); > > > > if (!$db) { > > echo "Could not select database!"; > > exit; > > } > > > > > > $query = "select @overalltotal :=SUM(total) as overall FROM mytable"; > > $result = mysql_db_query("mydb", $query); > > > > if ($result) > > { > > > > > > while ($row = mysql_fetch_row($result)) { // Begin while > > > > > > } // end while > > > > > > } else { echo "problems...."; } > > mysql_close(); > > > > ?> > > >

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