Dynamic Dropdown send values to 1 form

jroryl

As you will see, I’m new to coding so be gentle ….

I have a MySQL database which has products and I have populated a dropdown that displays the products and price (concatenate the 2 $item & $price), this (apparently is working). What I can’t figure out is how in the heck do I pass the chosen dropdown value to a form (PayPal), currently its always passing the last value in the dropdown.

Database is like this:

Item price

Dropdown has all the above values but when choose any value and click submit, Product5 is sent to the PayPal form, basically the last one every time.

Here’s the code:

<?php
include('../../../config.php');

if (!$db_con)
{
die('Could not connect: ' . mysql_error());
}

function options(){

global $OPTION;

$result = mysql_query("SELECT * FROM products");

if (!$result)echo "Error in query";
else {
	$count=mysql_num_rows($result);
	if ($count==0) echo "Nothing found";
else {
	for ($i=0; $i<$count; $i++){

		$OPTION=mysql_result($result, $i, 'title');
		$VALUE=mysql_result($result, $i, 'price');
		echo "<OPTION VALUE=$value>$OPTION $ $VALUE</option>"; // Builds the Dropdown
	}
     }
    }
}

echo "<select name='select'>";

options();

echo "</select>";
?>
<form method="post" action="https://www.paypal.com/cgi-bin/webscr" target="post">
	<input type="hidden" name="cmd" value="_cart" />
	<input type="hidden" name="business" value="address@email.com" />
	<input type="hidden" name="amount" value="1.00">
	<input type="hidden" name="item_name" value="<?php echo $_POST['$OPTION']; ?>">  // This displays the last record no matter what user chooses
	<input type="hidden" name="item_number" value="item_number" />
	<input type="hidden" name="return" value="http://www.DOMAIN.com/thankyou" />
	<input type="hidden" name="currency_code" value="USD" />
	<input type="hidden" name="undefined_quantity" value="1" />
	<input type="hidden" name="no_note" value="1" />
	<input type="hidden" name="bn" value="PP-BuyNowBF" />
	<input type="hidden" name="add" value="1" />
	<input type="submit" value="Add To Cart" />
</form>

jroryl

Ok, the above code isn't working so I re-wrote it, so now my code is this, which is still submitting the last record in the SELECT statement. How do I send the actual database values that the user picks in the dropdown? ($title AND $price)

anyone?

<?php
include('../../../config.php');

?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>
<body>

Products :
<select name">
<?

$list=mysql_query("SELECT * FROM products");

// Show records by while loop.
while($row_list=mysql_fetch_array($list)){

$title = $row_list[2];
$price = $row_list[4];

print "<option value=$title>$title $ $price</option>";

// End while loop.
}
?>
</select>
<form method="get" action="https://www.paypal.com/cgi-bin/webscr" target="post">
    <input type="hidden" name="cmd" value="_cart" />
    <input type="hidden" name="business" value="name@email.com" />
    <input type="hidden" name="amount" value="<?php echo $price; ?>">
    <input type="hidden" name="item_name" value="<?php echo $title; ?>">
    <input type="hidden" name="item_number" value="<?php echo $title; ?>">
    <input type="hidden" name="return" value="http://www.DOMAIN.com/thankyou" />
    <input type="hidden" name="currency_code" value="USD" />
    <input type="hidden" name="undefined_quantity" value="1" />
    <input type="hidden" name="no_note" value="1" />
    <input type="hidden" name="bn" value="PP-BuyNowBF" />
    <input type="hidden" name="add" value="1" />
    <input type="submit" value="Add To Cart" />

</select>
</form>
<?
// End if statement.


// Close database connection.
mysql_close();
?>
</p>
</body>
</html>

dagon

<?php echo $_POST['$OPTION']; ?>

should be

<?php echo $_POST['select']; ?>

that's the select name

jroryl

thanks .. I'm a little confused 😕 because I have already posted a new update to my first post which isn't showing up yet. I think since I'm new that moderators need to approve my postings ... so thanks for answering but the code you saw has more problems than what you fixed

I'll try to work with what you sent but I think the 'select' value will have both $title and $price when it should only have $title.

This is the newest code ... maybe it will illustrate what I'm trying to accomplish 😕

<?php
include('../../../config.php');

?>
<html>
<body>

Products :
<select name">
<?

$list=mysql_query("SELECT * FROM products");

// Show records by while loop.
while($row_list=mysql_fetch_array($list)){

$title = $row_list[2];
$price = $row_list[4];

print "<option value=$title>$title $ $price</option>";

// End while loop.
}
?>
</select>
<form method="get" action="blah blah" target="post">
    <input type="hidden" name="cmd" value="_cart" />
    <input type="hidden" name="business" value="blah blah" />
    <input type="hidden" name="amount" value="<?php echo $price; ?>">
    <input type="hidden" name="item_name" value="<?php echo $title; ?>">
    <input type="hidden" name="item_number" value="<?php echo $title; ?>">
    <input type="hidden" name="return" value="blah blah" />
    <input type="hidden" name="currency_code" value="USD" />
    <input type="hidden" name="undefined_quantity" value="1" />
    <input type="hidden" name="no_note" value="1" />
    <input type="hidden" name="bn" value="PP-BuyNowBF" />
    <input type="hidden" name="add" value="1" />
    <input type="submit" value="Add To Cart" />

</select>
</form>
<?
// End if statement.


// Close database connection.
mysql_close();
?>
</p>
</body>
</html>

jroryl

great ...now the posting is showing correctly ... I hope anyone reading this thread won't be as confused as I am

dagon

jroryl;10975393 wrote:
I'll try to work with what you sent but I think the 'select' value will have both $title and $price when it should only have $title.

if a value is set, then that is returned, so in your case it will be title, add

print_r($_POST), to see what your form is returning.

jroryl

@, are you looking at the code that I first posted yesterday or the second that I posted today?

jroryl

By the way, if I put print_r($_POST); after </form> in either code I get

Array ( )

To me seems not exactly what I want.

dagon

i don't know where you are at with this, it seems you said it was all fixed, if not post the exact problem

As to using print_r($_POST) it shows you what the form is posted, but you have to post the form first or it will be empty.

jroryl

The reply where I said "the posting is showing correctly" means that this thread was displaying correctly, nothing to do with the code....

Back on track, yeah still having problems. I'm trying to submit 2 values from MySQL, these values are populating a drop down correctly $title and $price but I can't seem to get the selected values sent to the form, the form always picks up the last record. All of this is contained in one php file, not trying to send\get values to other pages.

As to using print_r($_POST) it shows you what the form is posted, but you have to post the form first or it will be empty.

By placing print_r($_POST) after the closing </form> tag, does that mean "post the form first"?

dagon

<form method="get" action="blah blah" target="post">

method get means you get the results in the $GET array not the $POST

action should be the current page ie page.php

drop the target.

jroryl

I made the changes:
<form method="get" action="page.php"> <-- name of php file

and print_r($_POST) still displays

Array ( )

Thanks for helping but isn't the problem in the array or something other than the form?

dagon

as i said its $_GET not post if you chose action get (post i think would be better unless you have a reason for get)

jroryl

Ok, finally got some data...

Array ( [cmd] => _cart [business] => address@email.com [amount] => 1.00 [item_name] => [item_number] => [return] => http://www.DOMAIN.com/thankyou [currency_code] => USD [undefined_quantity] => 1 [no_note] => 1 [bn] => PP-BuyNowBF [add] => 1 )

jroryl

OK, I fixed my code ... moved the select within the form, etc ...

Here's the latest code, it's still submitting the last record instead of the record I choose in the select??

I believe the SELECT name is what is sent to the form but I need 2 values sent from the form or at least from db, how can I name a SELECT with 2 values? or at minimum send 2 values ($title & $price) to the form?

<?php
include('../../../config.php');

?>
<html>
<body>

Products :
<form method="get" action="blah blah" target="post">
<select name"products">
<?

$list=mysql_query("SELECT * FROM products");

// Show records by while loop.
while($row_list=mysql_fetch_array($list)){

$title = $row_list[2];
$price = $row_list[4];

print "<option value=$title>$title $ $price</option>";

// End while loop.
}
?>
    <input type="hidden" name="cmd" value="_cart" />
    <input type="hidden" name="business" value="blah blah" />
    <input type="hidden" name="amount" value="<?php echo $price; ?>">
    <input type="hidden" name="item_name" value="<?php echo $title; ?>">
    <input type="hidden" name="item_number" value="<?php echo $title; ?>">
    <input type="hidden" name="return" value="blah blah" />
    <input type="hidden" name="currency_code" value="USD" />
    <input type="hidden" name="undefined_quantity" value="1" />
    <input type="hidden" name="no_note" value="1" />
    <input type="hidden" name="bn" value="PP-BuyNowBF" />
    <input type="hidden" name="add" value="1" />
    <input type="submit" value="Add To Cart" />

</select>
</form>
<?
// End if statement.


// Close database connection.
mysql_close();
?>
</p>
</body>
</html>

dagon

its actully populating the values from your db query, you are mixing 2 forms\steps in to one

you need to submit select, then take the values from the select and populate the 2nd form to then submit it to paypal.

step one: select your product
step two: confirm and submit

jroryl

Getting closer ...

So looks like I ended up with an array, here is the result from doing a print_r($_GET);

Array ( [item_name] => Product1 1.49 [amount] => [business] => address@email.com [item_number] => [return] => http://www.DOMAIN.com/thankyou [currency_code] => USD [undefined_quantity] => 1 [no_note] => 1 [bn] => PP-BuyNowBF [add] => 1 [submit] => SEND )

The problem, for the item_name array it has "Product1 1.49" how do I get the "1.49" amount into the hidden input [amount]?

Basically the above array should look like this:

Array ( [item_name] => Product1 [amount] => 1.49 [business] => address@email.com [item_number] => [return] => http://www.DOMAIN.com/thankyou [currency_code] => USD [undefined_quantity] => 1 [no_note] => 1 [bn] => PP-BuyNowBF [add] => 1 [submit] => SEND )

dagon

change your select so the value is product|price

use | only if it will never be in the product name.

explode('|',$foo);
will split them apart and you can then use them.

jroryl

You are correct, the price will never be in the products table. (or I have bigger problems)

Currently the SELECT is:

And <option value> is: (notice the |)

<option value="<? echo $row_list['product']; ?>|<? echo $row_list['price']; ?>" { echo "selected"; } ?><? echo $row_list['product']; ?> $ <? echo $row_list['price']; ?></option>

Then do something like this in the input for amount?

$arr = explode('|', $_GET['item_name']);
$arr2 = $arr[1]; //This should be price?

By the way, the changes above didn’t work …

dagon

why would you use $_GET , when you know the value is $arr2 ($arr[1])