Hidden "return" value does not override the default return url

ThinkPad123
Contributor
Contributor

Hi all,

 

I just set up a nice pay now button for my website and want to redirect the user to a specific page after checkout. I am attempting to do this via the hidden input named "return", just as paypal documentation specifies:

 

<INPUT TYPE="hidden" NAME="return" value="'.$return_url.'">

the $return_url, in my case, is dynamically generated via php and outputs something like: https://example.com/confirm_transaction.php?product_id=1

 

My problem is, that, despite using the correct syntax in my HTML form, paypal always redirects the customer (which of course is me as I am testing my implementation) to the default return URL specified in my (business accounts') account settings.

Any help with this issue would be greatly appreciated.

Cheers

 

Login to Me Too
1 REPLY 1

angelleye
Advisor
Advisor

My assumption is that your button is a hosted button, in which case most variables included in the button code directly will be ignored.  Hosted buttons needs to be updated in the PayPal account button manager.  If you need to dynamically set your return URL you'll need to use a non-hosted button.

 

Another option would be to switch to Express Checkout, which is recommended, but would require more development experience.  That said, our PayPal PHP SDK makes setting up these API calls very quick and easy for you.

Angell EYE - www.angelleye.com
PayPal Partner and Certified Developer - Kudos are Greatly Appreciated!
Login to Me Too

Haven't Found your Answer?

It happens. Hit the "Login to Ask the community" button to create a question for the PayPal community.