passing variable from form

timmybuck

Hi guys,

I have a customer table and a customer form that takes customer details. i think i need to pass the details as variables so that i can get them and put them in my other page

how do i pass forename surname etc to paypal.php

this is my form

</TD>
</TR>

<tr> <td>Forename*</td> <td><input type="text" name="forename" maxlength="15" size="15"><br /></td> </tr>

<tr> <td>Surname</td> <td> <input type="text" name="surname" maxlength="15" size="15"><br /></td></tr>
<tr> <td>Age</td> <td> <input type="text" name="age" maxlength="2" size="2"><br /></td></tr>
<tr> <td>Telephone Number</td> <td> <input type="text" name="telephonenumber" maxlength="30" size="15"><br></td></tr>

<tr> <td>Email</td> <td> <input name="email" type="text" size="15" maxlength="30" onBlur="MM_validateForm('textfield','','NisEmail');return document.MM_returnValue">

<br></td></tr>

<tr> <td>Seat Catergory</td> <td> <input type="text" name="seatcategory" maxlength="30" size="15"><br></td></tr>
<tr>
<td></td>
</tr>
<tr>
<td></td>
</tr>

sidney

we need to see inserttutor.php

in there you can put info in to hidden fields and mahke another form and post to paypal.php

or sent the info in $_GET by attaching them to url eg?name=john

or you could use $_session vars

timmybuck

Hi Let me clarify, i think im confused myself...lol and tell me if it makes sense.

I have a booking1.php. This is a form that takes customer details when submitted this takes you to a insertdetails page. this page put the data in the db.

Now i would like to submit the customer details in to the paypal.php. This page has hidden fields. I would like to put the values in these hidden fields.

this is my insertdetails.php

<?php
if (!empty($POST['forename']) && !empty($POST['surname']) && !empty($POST['dob']) && !empty($POST['telephonenumber']) && !empty($POST['email'])) {
$forename = addslashes($POST['forename']);
$surname = addslashes($POST['surname']);
$dob = addslashes($POST['dob']);
$telephonenumber = addslashes($POST['telephonenumber']);
$email = addslashes($POST['email']);

$db = @mysql_connect('localhost', 'phpuser', 'phpuser03') or die('Could not connect to the database: '.mysql_error()); 
@mysql_select_db('booking') or die('Could not select the database: '.mysql_error()); 

$result = mysql_query('INSERT INTO `customer` (`forename`, `surname`, `date_of_birth`, `telephonenumber`, `email`)'. 
    "VALUES ('$forename', '$surname', '$dob', '$telephonenumber', '$email')") or die('Unable to insert record. '.mysql_error()); 

echo 'Your record was successfully inserted.';

} else {
echo 'You have not entered all the required details.<br />'."\n";
echo 'Please go back and try again.';
}
?>

and this is my paypal.php

<?php
//get values from previous page

$forename=$POST[forename];
$surname=$POST[surname];
$age=$POST[age];
$telephonenumber=$POST[telephonenumber];
$email=$_POST[email];

if (count($POST)) { echo '<pre>'; print_r($POST); exit; }

$quantity = $HTTP_GET_VARS['quantity'];
$total = $HTTP_GET_VARS['total'];
echo "<input type=\"hidden\" name=\"quantity\" value=\"$quantity\"><br>\n";
echo "<input type=\"hidden\" name=\"amount\" value=\"$total\"><br>\n";

//include database password information etc.
$hostname = "localhost";
$username = "phpuser";
$password = "phpuser03";
if(!($link = mysql_connect($hostname, $username,$password)))
die("Could not connect to database.");
$databasename = "booking";
if(!(mysql_select_db($databasename,$link)))
die("Could not open table.");

//now get customer info from database
$customerid = $HTTP_GET_VARS['customerID'];
$strsql= "SELECT forename, surname, age, telephonenumber,email
FROM
customer
WHERE CustomerID = '$customerid'";
if(!($rs= mysql_query($strsql, $link)))
die("Could not open table.");
//only one row should be returned
$row = @ mysql_fetch_array($rs);

// now complete the form
echo "<input type=\"hidden\" name=\"forename\"value=\"$row[forename]\"><br>\n";
echo "<input type=\"hidden\" name=\"surname\"value=\"$row[surname]\"><br>\n";
echo "<input type=\"hidden\" name=\"age\"value=\"$row[age]\"><br>\n";
echo "<input type=\"hidden\" name=\"telephonenumber\"value=\"$row[telephonenumber]\"><br>\n";
echo "<input type=\"hidden\" name=\"email\"value=\"$row[email]\"><br>\n";

?>

</form>

so i want to be able to pass the customer details to the hidden form

is what im saying make sense

sidney

yes where you get succsess response in insertdetails.php

just add another form with all the info in hidden fields

you can either have the users press a submit button or do an onload submit form to pay pal

hope that helps

timmybuck

just add another form with all the info in hidden fields

you can either have the users press a submit button or do an onload submit form to pay pal

what page do i do this on paypal.php ?

and what do u mean add a form

could u give me an example

sidney

wher is says - echo 'Your record was successfully inserted.';

in insertdetail.php

construct a form as you did before for paypal and bookings
with hidden fields same names as you are using
make action of form paypal.php

</form.
and repeat for all fields you want to pass on

timmybuck

does this look ok

sidney

and put the below in body tag if yo want onload submit

timmybuck

<?php
include("/Program Files\Apache Group\Apache2\htdocs\m82\software_common.inc");

intrahead("E-tickets.co.uk"); //escapes writing the HTML header code
?>
<table width="63%" height="493" border="0" align="center" cellpadding="2" cellspacing="2" bgcolor="#FFFFFF">
<tr align="left" valign="top" bgcolor="#FFFFFF">
<td align="center" bgcolor="#FFFFFF">
<p align="justify"></p>
<p align="justify">
<?php
if (!empty($POST['forename']) && !empty($POST['surname']) && !empty($POST['dob']) && !empty($POST['telephonenumber']) && !empty($POST['email'])) {
$forename = addslashes($POST['forename']);
$surname = addslashes($POST['surname']);
$dob = addslashes($POST['dob']);
$telephonenumber = addslashes($POST['telephonenumber']);
$email = addslashes($POST['email']);

$db = @mysql_connect('localhost', 'phpuser', 'phpuser03') or die('Could not connect to the database: '.mysql_error()); 
@mysql_select_db('booking') or die('Could not select the database: '.mysql_error()); 

$result = mysql_query('INSERT INTO `customer` (`forename`, `surname`, `date_of_birth`, `telephonenumber`, `email`)'. 
    "VALUES ('$forename', '$surname', '$dob', '$telephonenumber', '$email')") or die('Unable to insert record. '.mysql_error()); 

echo 'Your record was successfully inserted.'; ?>

<form action="paypal.php" name="form1" method="post">
<input name="forename" type="hidden" value="<?php echo $_POST['surname']; ?>">
<input name="surname" type="hidden" value="<?php echo $_POST['surname'];  ?>">
<input name="date_of_birth" type="hidden"value="<?php echo $_POST['date_of_birth']; ?>">
<input name="email" type="hidden" value="<?php echo $_POST['date_of_birth']; ?>">
<input name="telephonenumber" type="hidden" value="<?php echo $_POST['telephonenumber']; ?>">
</form>

<?php
} else {
echo 'You have not entered all the required details.<br />'."\n";
echo 'Please go back and try again.';
}
<script type="text/javascript" language="JavaScript">
//submit form
document.order.submit();
</script>
?>

this is what i have now

i get an error on thje javascript line

sidney

that needs to bee in body tag

but yours is generated by function i think

timmybuck

these are the instructions that im trying to follow to connect to paypal. my understanding is i can submit customer details and ticket details

thats why im trying to submit customer details to pay pal

http://www.devarticles.com/c/a/PHP/Quick-ECommerce-with-PHP-and-PayPal/1/

i get this now on the paypal.php

Array
(
[forename] => yup
[surname] => yup
[date_of_birth] =>
[email] =>
[telephonenumber] => 324324
)