inserting data in mysql table using php

[roshanjameer]

hi,

iam making a website for school.on one web page,there are 2 radio buttons.1 is for science and one for computers.like this
subjects
1.science
2.computers

the user can click any of radiobutton and press submit button it will moves to the second web page.on that again i have 2 radiobuttons.1 for sing and one for music like this.
hobbies
1.sing
2.music.

here he clicks one radiobutton and press the submit button,then his choosen subject and hobbies should be stored in mysql table.i did all this.but here what i actually want to do is,if he selects only subject and ignore the hobbies then the subject should be stored in db.if ignore the subject and chooses the hobbies only then it should be stored in db.i used sessions to move data from one page to another.
i dont know how to do this.please any of your help is appreciated.

thanks
mrjameer

[rincewind456]

This may help you in the right direction

<?php
session_start();
error_reporting(E_ALL);

if (!isset($_POST['subject']) && !isset($_POST['hobbies'])) {	
// If neither submit has been pressed print this 
	print $subject = <<<_SUBJECT_
<form action="" method="post">
Science<input type="radio" name="science">
<br>
Computers<input type="radio" name="computers">
<br>
<input type="submit" name="subject" value="Submit">
</form>
_SUBJECT_;


} elseif (isset($_POST['subject']) && !isset($_POST['hobbies'])) {

// Add POST values to SESSION
if (isset($_POST['science'])) {$_SESSION['science'] = true;}
if (isset($_POST['computers'])) {$_SESSION['computers'] = true;}

$_SESSION['subject'] = true;
// If the first page has been submitted, but not the second Print this
	print $hobbies = <<<_HOBBIES_
<form action="" method="post">
Singing<input type="radio" name="singing">
<br>
Music<input type="radio" name="music">
<br>
<input type="submit" name="hobbies" value="Submit">
</form>
_HOBBIES_;

} elseif (isset($_SESSION['subject']) && isset($_POST['hobbies'])) {
	//If both submits have been pressed do the sql insert

$array = array(); //initialse the array variable

if (isset($_POST['singing'])) {$_SESSION['singing'] = true;}
if (isset($_POST['music'])) {$_SESSION['music'] = true;}

//go through the returned values and build query array
$message = 'You chose the following:<br> ';
if ($_SESSION['science'] == true) {
	$message .='Science<br>';
	$array['science'] = 1;
} else { $array['science'] = 0;}

if ($_SESSION['computers'] == true){
	$message .='Computers<br>';
	$array['computers'] = 1;
} else{	$array['computers'] = 0;}

if ($_SESSION['singing'] == true) {
	$message.='Singing<br>';
	$array[ 'singing'] = 1;
} else {$array[ 'singing'] = 0;}

if ($_SESSION['music'] == true) {
	$message .='Music<br>';
	$array['music'] = 1;
} else {$array['music'] = 0;
}
print $message; // Print message saying what has been selected

//loop through $array and build INSERT query
$sql = 'INSERT INTO table_name (';
foreach ($array as $key => $value){
	$sql .= '\''.$key.'\', ';
}
$sql .= ') VALUES(';
foreach ($array as $key => $value){
	$sql .= $value.', ';
}
$sql .=')';
//print $sql; 	//uncomment for debugging

// do your $sql insert here

session_destroy();
}

I have put it all on one page, no real need for three pages. Bear in mind that this is just a basic example, it does not for instance have any checking to prevent sql injection, and it of course uses a database table where all the possible values are either set to true or false.