Appreciate a little guidance here:

Background: I have an invoicing system. Invoices are generated and (invoice and Invoice Items) and Payments are generated (Payments and Payment Items). Payment items are amount of the Payment Total allocated to payoff open invoices. So I may have 3 open invoice for $100 each and I may generate one payment for $300 with 3 payment items for $100 each to pay off those 3 open invoices.

In most cases, clients will pay their own invoices off, but in rare cases another client will pay an invoice for that client (ie... spouse, parent, etc...) My client want me to some how display when the payee (or one of the payees - there can be multiple) of the invoice IS NOT the same client as the invoice being paid.

So I need to display a result that show a comma delineated string of payees OMITTING the invoice's client_id. I only want to show a result if one or more of the payees are different than the invoice's client_id.

So now with the mySQL queries that are working:

First of all, the client_id of the invoice I am querying on is 251719.

query 1 = select group_concat(payment_id) from tl_trans_pmt_items where inv_id = 1033911

This produces a string "1033882,1021630,1021632". These are parent Payment records which have the payee client_ids.

So if I run a query:

query 2 = select group_concat(client_id) FROM tl_transactions WHERE transaction_id IN (1033882,1021630,1021632) AND client_id != 251719

This produces a string "251711,251713". These are the client_ids of the Payment records OMITTING the Invoice's client_id

So far this works fine. Now where I run into issues is where I try to combine these queries together:

query 3 = select group_concat(client_id) FROM tl_transactions WHERE transaction_id IN ((select group_concat(payment_id) from tl_trans_pmt_items where inv_id = 1033911)) AND client_id != 251719

I do not get the same results. Am I missing something? Hopefully something simple ;-)

Don





--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:    http://lists.mysql.com/mysql

Reply via email to