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Hidden "return" value does not override the default return url-2

Contributor

Hidden "return" value does not override the default return url-2

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="'$successful_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

1 REPLY 1
New Community Member

Re: Hidden "return" value does not override the default return url-2

We experienced this issue appearing in August 2020 as well.

 

Looks like PayPal changed the Button API and no longer honors the legacy "return" URL:

https://www.paypal-community.com/t5/REST-APIs/SmartButtons-Return-URL-s-not-working/td-p/1793386#

 

It seems an upgrade to the JavaScript-rendered "SmartButtons" is required: