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Fix bug 1310523 by not allowing more than 26 equal verse names

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bzr-revno: 2193
Fixes: https://launchpad.net/bugs/1310523
This commit is contained in:
Samuel Mehrbrodt 2014-05-27 22:28:35 +02:00 committed by Raoul Snyman
commit 725be17e7e
2 changed files with 20 additions and 7 deletions
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23
openlp/plugins/songs/forms/editsongform.py
View File

@ -723,19 +723,32 @@ class EditSongForm(QtGui.QDialog, Ui_EditSongDialog):
return False
cnt_errors = 0
error_list = ''
verse_tag = []
verse_num = []
verse_tags = []
wrong_verse_tags = []
wrong_verse_nums = []
for i in range(self.verseListWidget.rowCount()):
item = self.verseListWidget.item(i, 0)
tags = self.find_tags.findall(item.text())
verse_tags.append(unicode(item.data(QtCore.Qt.UserRole).toString()))
if self._validate_tags(tags) == False:
field = unicode(item.data(QtCore.Qt.UserRole).toString())
verse_tag.append(VerseType.translated_name(field[0]))
verse_num.append(field[1:])
wrong_verse_tags.append(VerseType.translated_name(field[0]))
wrong_verse_nums.append(field[1:])
cnt_errors += 1;
for tag in verse_tags:
if verse_tags.count(tag) > 26:
# lp#1310523: OpenLyrics allows only a-z variants of one verse:
# http://openlyrics.info/dataformat.html#verse-name
critical_error_message_box(
message=translate('SongsPlugin.EditSongForm',
'You have %(count)s verses named %(name)s %(number)s. '
'You can have at most 26 verses with the same name' %
{'count': verse_tags.count(tag), 'name': VerseType.translated_name(tag[0]),
'number': tag[1:]}))
return False
if cnt_errors > 0:
for i in range(cnt_errors):
error_list += '%s %s' % (verse_tag[i], verse_num[i])
error_list += '%s %s' % (wrong_verse_tags[i], wrong_verse_nums[i])
if i < cnt_errors-1:
error_list += ', '
critical_error_message_box(

4
openlp/plugins/songs/lib/xml.py
View File

@ -360,9 +360,9 @@ class OpenLyrics(object):
verse_tag = verse[0][u'type'][0].lower()
verse_number = verse[0][u'label']
verse_def = verse_tag + verse_number
verse_tags.append(verse_def)
# Create the letter from the number of duplicates
verse[0][u'suffix'] = chr(96 + verse_tags.count(verse_def))
verse[0][u'suffix'] = chr(97 + (verse_tags.count(verse_def) % 26))
verse_tags.append(verse_def)
# If the verse tag is a duplicate use the suffix letter
for verse in verse_list:
verse_tag = verse[0][u'type'][0].lower()