Correct table

This commit is contained in:
aspangaro 2016-03-28 21:33:49 +02:00
parent a67b9afb6c
commit a1d95e11ed

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@ -4,7 +4,7 @@
* Copyright (C) 2011 Juanjo Menent <jmenent@2byte.es>
* Copyright (C) 2012 Regis Houssin <regis@dolibarr.fr>
* Copyright (C) 2013-2015 Alexandre Spangaro <aspangaro.dolibarr@gmail.com>
* Copyright (C) 2013-2014 Olivier Geffroy <jeff@jeffinfo.com>
* Copyright (C) 2013-2016 Olivier Geffroy <jeff@jeffinfo.com>
* Copyright (C) 2013-2016 Florian Henry <florian.henry@open-concept.pro>
*
* This program is free software; you can redistribute it and/or modify
@ -90,7 +90,7 @@ $sql .= " s.code_compta_fournisseur, p.accountancy_code_buy , ct.accountancy_cod
$sql .= " FROM " . MAIN_DB_PREFIX . "facture_fourn_det as fd";
$sql .= " LEFT JOIN " . MAIN_DB_PREFIX . "c_tva as ct ON fd.tva_tx = ct.taux AND ct.fk_pays = '" . $idpays . "'";
$sql .= " LEFT JOIN " . MAIN_DB_PREFIX . "product as p ON p.rowid = fd.fk_product";
$sql .= " LEFT JOIN " . MAIN_DB_PREFIX . "accountingaccount as aa ON aa.rowid = fd.fk_code_ventilation";
$sql .= " LEFT JOIN " . MAIN_DB_PREFIX . "accounting_account as aa ON aa.rowid = fd.fk_code_ventilation";
$sql .= " JOIN " . MAIN_DB_PREFIX . "facture_fourn as f ON f.rowid = fd.fk_facture_fourn";
$sql .= " JOIN " . MAIN_DB_PREFIX . "societe as s ON s.rowid = f.fk_soc";
$sql .= " WHERE f.fk_statut > 0 ";