Re: Form variable to PHP script problems

From: Date: Mon, 05 Aug 2002 05:04:00 +0000
Subject: Re: Form variable to PHP script problems
References: 1  Groups: php.general 
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Search the archives for "register globals"... this has been discussed in around 2000+ emails to the group. In short, include this line at the top of the form processing script: <? $nome = $_POST['nome']; $senha = $_POST['senha']; ?> ... or refer to the POSTed (form) variables as $_POST['variable_name'] rather than $variable_name. Justin French on 05/08/02 9:29 AM, Renato B. Miranda (renato_bm@yahoo.com.br) wrote: > Hello ! > I´m developing a dinamic site and I started to use PHP, MySQL and > Apache. > The things are going well, I´m already using database, inserting, > everything .... > But when I try to pass some variables from a Form to a PHP isn´t working > ! > Somebody could help me ? > Here is my code ..... > > HTML > <html> > <p align="center"><font face="Arial Black" > size="5">Banco de Dados > Cesop</font></p> > <p align="center">&nbsp;</p> > <form action="validaUsuario.php" method="post" > > <p align="center">Usuário:&nbsp; <input type="text" > NAME="nome" > size="20"></p> > <p align="center">Senha:&nbsp;&nbsp;&nbsp;&nbsp; <input > type="password" NAME="senha" size="20"></p> > <p align="center"><input type="submit" value="Submeter" > name="B1"></p> > > </form> > </body> > </html> > > PHP > <? > $connection = mysql_connect("localhost", "renato", "ren9ato8") > or die ("Unable to connect to server."); > $db = mysql_select_db("cesop", $connection) > or die ("Unable to select database."); > > echo $senha; > > $sql = "SELECT nome > FROM users > WHERE nome='$nome' and password='$senha'"; > $result = mysql_query($sql, $connection) > or die ("Unable to get results."); > $num = mysql_numrows($result) > or die ("You're not authorized to be here. If you feel you have > recieved this > message in error, please contact the <a > > href=\"mailto:tyler.longren@midiowa.net\">webmaster</a>"); > > if ($num == 1) { > echo "<p>You can be here<br>"; > echo "Your username is $nome</p>"; > } > ?> > > Thanks ! >

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