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fix: #6742 calcula hasIncoterm a partir de la serie
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This commit is contained in:
Javi Gallego 2024-01-23 12:30:04 +01:00
parent 371cc1e0a9
commit 540c76005e
2 changed files with 6 additions and 9 deletions
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2
print/templates/reports/invoice-incoterms/sql/incoterms.sql
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@ -15,7 +15,7 @@ SELECT pack.packages,
LEFT JOIN vn.province p ON p.id = c.provinceFk LEFT JOIN vn.province p ON p.id = c.provinceFk
JOIN vn.ticket t ON t.refFk = io.ref JOIN vn.ticket t ON t.refFk = io.ref
JOIN vn.address a ON a.id = t.addressFk JOIN vn.address a ON a.id = t.addressFk
LEFT JOIN vn.incoterms ic ON ic.code = a.incotermsFk JOIN vn.incoterms ic ON ic.code = a.incotermsFk
LEFT JOIN vn.customsAgent ca ON ca.id = a.customsAgentFk LEFT JOIN vn.customsAgent ca ON ca.id = a.customsAgentFk
JOIN vn.sale s ON s.ticketFk = t.id JOIN vn.sale s ON s.ticketFk = t.id
JOIN ( JOIN (

13
print/templates/reports/invoice/sql/hasIncoterms.sql
View File

@ -1,8 +1,5 @@
SELECT IF(incotermsFk IS NULL, FALSE, TRUE) AS hasIncoterms SELECT COUNT(*) AS hasIncoterms
FROM ticket t FROM invoiceOut io
JOIN invoiceOut io ON io.ref = t.refFk JOIN vn.invoiceOutSerial ios ON ios.code = io.serial
JOIN client c ON c.id = t.clientFk AND ios.taxAreaFk = 'WORLD'
JOIN address a ON a.id = t.addressFk WHERE io.ref = ?
WHERE t.refFk = ?
AND IF(c.hasToinvoiceByAddress = FALSE, c.defaultAddressFk, TRUE)
LIMIT 1