mysql_select_db using POST variable - php

Question

I am quite new to PHP, and still have long way to go. Probably, it is a logic failure or maybe I just dont know how exactly PHP works.
Anyway, The code I submited is from my admin control panel. I am trying to make a page to connect to db than show all db names and select one of them from radio buttons. And when pressed submit again to show tables.After showing table the next step will be to edit,add update them etc. I am not there yet. Because I could not make mysql_select_db to work. It gives error.
The error is;
Notice: Undefined variable: connect in /public_html/php/insert_delete_update_amend/mydata03.php on line 94
Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /public_html/php/insert_delete_update_amend/mydata03.php on line 94
Could not connect to db
Any help will be much appreciated.
P.S.: Also when the first form is submitted,first form disappears and second one appears and so on.
<?php
// form variables
$DisplayDBinfoForm = true;
$DisplayDBform = false;
if (isset($_POST['db_info_submit'])) { //when user presses db info submit button
//select which form to hide or appear
$DisplayDBinfoForm = false;
$DisplayDBform = true;
$hostname = $_POST['db_name'];
$db_user_name = $_POST['db_user_name'];
$db_user_password = $_POST['db_user_password'];
$connect = mysql_connect($hostname,$db_user_name,$db_user_password);
if(!$connect) die("Could not connect");
echo "<p><b>connected successfully</b></p>\n";
}
if($DisplayDBinfoForm) {
?>
<form name="dbinfo" method="POST" action="mydata03.php" onsubmit="return validateForm();">
Host Name: <input type="text" name="db_name" /><br />
<br />
DB User Name <input type="text" name="db_user_name" /><br />
<br />
DB User Password: <input type="text" name="db_user_password" /><br />
<br />
<input type="submit" name="db_info_submit"value="Login"><br />
</form>
<?php
}
if($DisplayDBform) {
?>
<form name="delete_table" id="delete_table" action="mydata03.php" method="post">
<table width="30%" border="1">
<tbody>
<?php
$query = "SHOW DATABASES";
$resultSet = mysql_query($query);
while($database = mysql_fetch_array($resultSet)) { // go through each row that was returned in $result
$dbname = $database[0];
echo "<tr><th>Database Name</th>
<th>Select</th>
</tr>
<tr>
<td>$dbname</td>
<td><input name=\"radDB\" id=\"radDB\" type=\"radio\" value=\"$dbname\"</td>
</tr>\n";
}
?>
</tbody>
</table>
<p>
<input name="btnSelectDB" type="submit" value="Select" />
</form>
</p>
<?php
}
if(isset($_POST['btnSelectDB'])) {
$DisplayDBinfoForm = false; // hide form
$DisplayDBform = false; // hide form
$db_name = $_REQUEST["radDB"]; // the db na,e
echo "The " . $db_name . " is selected\n";
$select_db = mysql_select_db($db_name,$connect);
if(!$select_db) die("Could not connect to db". mysql_error());
echo "<b>connected successfully to db</b>";
}
?>

Answer 1

First off, don't use mysql_* functions. That time has passed. Instead, use mysqli_* or PDO.
With that said, it appears that the initial connection to the database server complete fine, as your code doesn't crap out on you at:
$connect = mysql_connect($hostname,$db_user_name,$db_user_password);
if(!$connect) die("Could not connect");
Although at the later stage, when trying to connect to an actual database: $select_db = mysql_select_db($db_name,$connect); it does.
This leaves me to believe that the variable in which you set the database name $db_name = $_REQUEST["radDB"]; is not pulling the data in correctly.

Answer 2

try
$connect = mysql_connect($hostname,$db_user_name,$db_user_password) or die("Could not connect");
if( $connect)
echo "<p><b>connected successfully</b></p>\n";
}
also check that the
if( isset($_POST['db_name'] ) && isset($_POST['db_user_name']) && isset($_POST['db_user_password']){
$hostname = $_POST['db_name'];
$db_user_name = $_POST['db_user_name'];
$db_user_password = $_POST['db_user_password'];
}
Use of this extension(MYSQL_*) is discouraged. Instead, the MySQLi or PDO_MySQL extension should be used
Good Read
PDO Tutorial for MySQL Developers

Answer 3

The problem you have is that the database is only opened ($connect=) if you have done a post with 'db_info_submit', but you can still run line 94 if you have posted 'btnSelectDB'.
Simplest solution is to open the database outside the if ($_POST) statement at the top, in case it's needed below. Or wrap in either of the conditions required to open it (code shown below)
<?php
// form variables
$DisplayDBinfoForm = true;
$DisplayDBform = false;
if (isset($_POST['db_info_submit']) || isset($_POST['btnSelectDB'])) {
$connect = mysql_connect($hostname,$db_user_name,$db_user_password);
if(!$connect) die("Could not connect");
echo "<p><b>connected successfully</b></p>\n";
}
}
if (isset($_POST['db_info_submit'])) { //when user presses db info submit button
//select which form to hide or appear
$DisplayDBinfoForm = false;
$DisplayDBform = true;
$hostname = $_POST['db_name'];
$db_user_name = $_POST['db_user_name'];
$db_user_password = $_POST['db_user_password'];
}
if($DisplayDBinfoForm) {
?>
....
<?php
}
if(isset($_POST['btnSelectDB'])) {
$DisplayDBinfoForm = false; // hide form
$DisplayDBform = false; // hide form
$db_name = $_REQUEST["radDB"]; // the db na,e
echo "The " . $db_name . " is selected\n";
$select_db = mysql_select_db($db_name,$connect);
if(!$select_db) die("Could not connect to db". mysql_error());
echo "<b>connected successfully to db</b>";
}
?>
This isn't a direct answer, but it's the answer that's goign to help you the most.
If you have a function in your script that begins "myslq_" (except for 1 - mysql_real_esacape_string) then you're following an old example and are using code that is going to be depreciated. As you're just learning PHP and MySQL, get into the habits of using the more modern functions, either mysqli_ or PDO