Help! MySQL, DropDown, PHP, Form!

josevalviana

Well I have this mysql database with some passwords. Now I have a form where a dropdown list is populated with the name of the users (there are repeated names). Well what I'm trying to do is after selecting an user in the dropdown list, the post will be directed to a result form where I'll have again the name of the user and now his password (for that exact row).

Here is the code for my first page:

<?php
include ("connect.php");
echo "<html>\n<head>\n<title>User Passwords</title>\n</head>\n<body>\n";
echo "<form action=\"password.php\" method=\"POST\">\n<select name=\"nomes\">\n";
while ($r = mysql_fetch_array($query))
{
$nome = $r["nome"];
echo "<option value=\"$nome\">$nome</option>\n";
}
echo "</select>\n";
echo "<p><input type=\"submit\" value=\"Dados de Acesso\" name=\"select\"></p>\n</form>\n</body>\n</html>";
?>

How should be the password.php file????

😕

bpat1434

<?php
// Get the posted name
$nome = $_POST['nomes'];

// Connect to database, run query finding user with name = $nome
// Echo the result

?>

~Brett

josevalviana

I could do that but it would give me lots of results... I just want that row I selected in the other form... that's the problem... repeated "nomes".... thanxs

bpat1434

Okay.... so you're saying that the value of $nomes gives a list like:

Miguel
Richard
John
Mark
John
Richard

Where there's more than record for each name? If that's so, then you need to change this: echo "<option value=\"$nome\">$nome</option>\n"; to

echo "<option value=\"$r['id']\">$nome</option>\n";

Then, in the password.php page, just grab the posted ID and select the 1 record with that ID.

~Brett

josevalviana

here is how i did it work.... please let me know if there is an easier way...

file index.php

<?php
$id = 0;
include ("connect.php");
echo "<html>\n<head>\n<title>Medical Journals Passwords</title>\n</head>\n<body>\n";
echo "<big>Favor escolha um jornal na lista abaixo:</big>\n";
echo "<form action=\"password.php\" method=\"POST\">\n<select name=\"nomes\">\n";
while ($r = mysql_fetch_array($query))
{
$nome = $r["nome"];
echo "<option value=\"$id\">$nome</option>\n";
$id ++;
}
echo "</select>\n";
echo "<p><input type=\"submit\" value=\"Dados de Acesso\" name=\"select\"></p>\n</form>\n</body>\n</html>";
?>

file password.php

<?php
include ("connect.php");
$chooser = $_POST['nomes'];
mysql_data_seek($query,$chooser);
$row = mysql_fetch_assoc($query);
$nome = $row['nome'];
$link = $row['link'];
$usuario = $row['usuario'];
$senha = $row['senha'];
$usar_proxy = $row['proxy'];
$qual_proxy = $row['qualproxy'];
echo "<html>\n<head>\n<title>Dados de Acesso</title>\n</head>\n<body>\n";
if ( $link ) {
echo "<a href=\"$link\">$nome</a><br >\n"; } else { echo "<big><i>$nome</i></big><br >\n"; }
if ( $usar_proxy == '1' ) {
echo "Proxy: " .$qual_proxy ."<br >\n"; }
if ( $usuario ) {
echo "Usuário: " .$usuario ."<br >\n"; } else { echo "Não necessita usuário<br >\n";
}
if ( $senha ) {
echo "Senha: " .$senha ."<br >\n"; } else { echo "Não necessita senha<br >\n";
}
echo "</body>\n</html>";
?>

that's it...

bpat1434

Does it work? Because I'm not seeing a query, but if it works, then leave it. Not necessarily an easier way, but you could use [man]extract/man to get the mysql results into variables in one line, rather than 10.

~Brett

josevalviana

yeah it works... the query is in the connect.php file... it's just $query = "SELECT * FROM passwords ORDER BY nome";

thanxs for your help... problem solved...

bpat1434

What?.... You want to get the password based upon the user input right? What username the user inputs is $POST['nomes']. So your query should be:

SELECT password 
  FROM `passwords`
  WHERE nome='$_POST['nomes']'
  ORDER BY nome

~Brett