What Coupon Code do my subscriber use?

How do I find out what pro coupon code my subscriber used? (Non affiliate)

Hi Edward,

s2Member does not store a record of the coupon code that was used. Applying the coupon code to a Pro-Form during checkout simply recalculates the checkout total based on what discount should be applied for that coupon. If you need to figure out which coupon code was used, you’ll need to calculate the difference in the total that was paid and go from there.

That seems like a significant oversite given that coupons are usually as much for tracking (i.e. what ad did this customer come from) as they are for discounts. Calculating backwards won’t help as we have multiple codes that apply the same discount.

I noticed that you do pass the codes for the affiliates. I am in the process of creating my own faux affiliate system where all will simply have “-1” appended. I wrote my php script that will take be called by silent-php and will pass the coupon code and the user_ip address (as opposed to %%subscr_id%% which does not appear to be available). Once in my script I plan to query the usermeta and match up the user_ip and then insert the coupon code as a usermeta record. Unfortunately the user_ip is not guaranteed unique (like %%subscr_id%%) as all my tests have the same one but I am hoping it will be in practical application.

This all seems extremely convoluted. Is there an easier way? It would seem that it might be theoretically possible to have the coupon code copied to another custom field on the registration form but I am not sure how one might go about this.

Are there plans to add the ability to capture coupon codes in the near future. I have been very impressed with the product so far but this seems a big hole.

Hi Edward.

That’s a good idea to implement this in the current s2Member.

Yes, we have plans to improve the coupon codes, including their tracking, vastly. Jason is working on the rewrite and it should be ready for the new major version of s2member in -probably- a couple of months.