dropdown question

PaulaRun

I am displaying products from a db. The user chooses which product they want from a dropdown list of prices. I have verified the user has indeed chosen one of the 2 choices of prices through javascript. Now, I need to send the appropriate price to paypal. Can somebody tell me what is wrong with this if statement??

By the way, I want to go to paypal on submit - not to php_self.

THANKS.

            echo "<select size=\"1\" name=\"dd\">";
          echo "<option value=0 selected>Select</option>";
          echo "<option value=1>$".stripslashes($row["price1"])."</option>";
          echo "<option value=2>$".stripslashes($row["price2"])."</option></select>&nbsp;&nbsp;&nbsp;";
          echo '<br><br>';
          echo "<input type=\"hidden\" name=\"item_name\" value=".stripslashes($row["prod_name"]).">";
          if(dd == '1'){
			echo "<input type=\"hidden\" name=\"amount\" value=".stripslashes($row["price1"]).">";
		  }
			elseif(dd == '2')
			{
			echo "<input type=\"hidden\" name=\"amount\" value=".stripslashes($row["price2"]).">";
		  }

jpmoriarty

dd is a variable, so it sould be:

if($dd == '1'){
            echo "<input type=\"hidden\" name=\"amount\" value=".stripslashes($row["price1"]).">";
          }
elseif($dd == '2')
            {
            echo "<input type=\"hidden\" name=\"amount\" value=".stripslashes($row["price2"]).">";
          }

a provisional word of warning though - javascript is client side, but php is server side: when you select a value from the drop down list you have to send it back to the php engine before it gets registered.

Think about how that script runs off the server - first time it's called $dd hasnt been set, so it wont be = to either 1 or 2, it will be blank. Then when you select one of them, you have to run the php again to see whether to do $dd==1 or $dd==2 : the php wont be run again until you send the value to it, so it will remain unset until you submit the form back to the php page.

PaulaRun

Thanks jpmoriarty. I tried using $dd and it didn't work. The "dd" is coming from the name in the select statement. How can I find out which selection the use has made? The if statement isn't being executed at all.

THANKS!

jpmoriarty

i can see very well where the dd is coming from, but you've not taken on board what i posted previously:

Think about how that script runs off the server - first time it's called $dd hasnt been set, so it wont be = to either 1 or 2, it will be blank. Then when you select one of them, you have to run the php again to see whether to do $dd==1 or $dd==2 : the php wont be run again until you send the value to it, so it will remain unset until you submit the form back to the php page.

PHP WILL NOT update any of its variables or values until they have been sent back to the server. You changing a value on the screen will have no effect on the variable that is stored on the server that is running your php script - how can it!!??

The easiest way to tell what selection is made is to make proper use of the form you're trying to use: send the variable using a post method to another page, and then use the variable then.
eg

<form method="POST" action="check_variable.php">
.... your select bits ....
<input type="submit" value="Send this selection">
</form>

Alternatively you can use javascript to send the value automatically when the selection is changed, but you cant get away from having to send the information back to the server (since that's where all the back end gubbins is happening), and to do that you'll need some sort of refresh. THEN you can test to see what the value of $dd is, and code around it accordingly.

echo "<form method=\"POST\" action=\"check_variable.php\">\n";
echo "<select size=\"1\" name=\"dd\">\n";
echo "<option value=0 selected>Select</option>\n";
echo "<option value=1>$".stripslashes($row["price1"])."</option>\n";
echo "<option value=2>$".stripslashes($row["price2"])."</option></select>\n";
echo "<input type=\"submit\" value=\"Send this selection\">\n";
echo "</form>\n";

and then test.php:

echo "<p>On the last form you selected $_POST['dd']</p>\n";
....here you can put if's, but's and whatever's...

It seems odd that you'rre trying to send the name and value and so on anyway. From your code these are already in a database - why risk people fiddling (believe me, they will) with data that's much safer to get to by leaving it on the server anyway?

PaulaRun

Thank you John for your patience! I'm just soooo confused.

I must send the product name and price to paypal so they can process the credit card. Is there another way other than what I'm doing??

Anyway, I have decided to rip out the quantity and drop down box for the 2 prices. Instead, I'll just put an add to cart button next to each price. Even that isn't working!!

Can you tell me why the add to cart button isn't next to the price? It is on the next line - ugh!

echo "<tr>";
          echo "<td width=\"1\" height=\"1\" colspan=\"2\"></td>";
          echo "<td width=\"464\" height=\"1\"><strong><font size=\"2\">".stripslashes($row["price1"])."</strong></font>";
          include "order_form1.php";
          echo "</td>"; 
     echo "<tr>";

order_form1.php
<html>

<head>

<title>Order Form</title>
</head>

<body>
<?
echo '<form target="open" action="https://www.paypal.com/cgi-bin/webscr" method="POST"> ';
echo "<font size=\"1\"><input type=\"submit\" value=\"Add to Cart\"></font>";
echo "<input type=\"hidden\" name=\"item_name\" value=".stripslashes($row["prod_name"]).">";
echo "<input type=\"hidden\" name=\"amount\" value=".stripslashes($row["price1"]).">";
echo '<input type="hidden" name="cmd" value="cart">';
echo '<input type="hidden" name="business" value="me@mysite.com">';
echo '<input type="hidden" name="cancel_return" value="http://mysite.com/home.htm">';
echo '<input type="hidden" name="return" value="http://mysite.com/confirm.htm">';
echo '<input type="hidden" name="add" value="1">';

echo '</form>';

?>
</body>

</html>

dalecosp

originally posted by PaulaRun
Can you tell me why the add to cart button isn't next to the price? It is on the next line - ugh!

This is a case for "View Source" in your browser...your HTML code determines where the objects appears on the page (in the absence of any CSS stylesheets).

On the larger subject, jpmoriarty has got the right idea; surely you need to wrap up the order on your server before submitting it to PayPal...at least two steps (in the bigger picture) which may requre a half dozen or more scripts, assuming you follow the recommended practice of seperating logic from content, etc...

Display products and prices. Provide a means for a surfer to indicate that he/she would like product X. Provide a means for your website to remember what is in his/her cart....
Provide a way for the surfer to indicate that he'd like to check out now. When he/she does this, total up the cart and send the total price, along with your company name, etc. to PayPal for their verification process.

It almost looks as if you're trying to submit each action of cart addition to PayPal...if you do that, what happens when the surfer changes his/her mind about an item?

Don't get too discouraged. Sit down and think things through. It has taken two months to get my inventory control/shopping cart/catalog display/purchase system even close to working. Granted, I'm not putting in 40 hours a week, but it's an extremely large project to tackle. Good news: it's much easier in PHP than, say, Perl, or most other CGI's...