From 3f30f1d46f407d145d0c51a0d435c54172e9a6ce Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Frederik=20=E2=80=9CFreso=E2=80=9D=20S=2E=20Olesen?= Date: Sat, 29 Apr 2017 13:54:43 +0200 Subject: [PATCH] Always disambiguate in the release title part of the template Fixes https://github.com/JoeLametta/whipper/issues/127 --- whipper/common/program.py | 11 +++++++---- 1 file changed, 7 insertions(+), 4 deletions(-) diff --git a/whipper/common/program.py b/whipper/common/program.py index 5f084b8..62e9ec8 100644 --- a/whipper/common/program.py +++ b/whipper/common/program.py @@ -243,10 +243,13 @@ class Program: # when disambiguating, use catalogNumber then barcode if disambiguate: templateParts = list(os.path.split(template)) - if self.metadata.catalogNumber: - templateParts[-2] += ' (%s)' % self.metadata.catalogNumber - elif self.metadata.barcode: - templateParts[-2] += ' (%s)' % self.metadata.barcode + # Find the section of the template with the release name + for i, part in enumerate(templateParts): + if "%d" in part: + if self.metadata.catalogNumber: + templateParts[i] += ' (%s)' % self.metadata.catalogNumber + elif self.metadata.barcode: + templateParts[i] += ' (%s)' % self.metadata.barcode template = os.path.join(*templateParts) logger.debug('Disambiguated template to %r' % template)