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check sort field when searching

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If the field they want to search is the one being used to sort, this saves us
having to go the full search route
This commit is contained in:
Damien Elmes 2011-12-08 09:53:27 +09:00
parent 08660f6bf8
commit 95106908dd
2 changed files with 12 additions and 5 deletions
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5
anki/find.py
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@ -159,8 +159,9 @@ and c.nid=n.id %s""" % (q, order)
extra = "not" if neg else ""
if not self.full:
self.lims['args']["_text_%d"%c] = "%"+val+"%"
self.lims['preds'].append("flds %s like :_text_%d escape '\\'" % (
extra, c))
self.lims['preds'].append("""\
(sfld %s like :_text_%d escape '\\' or
flds %s like :_text_%d escape '\\')""" % (extra, c, extra, c))
else:
# in the future we may want to apply this at the end to speed up
# the case where there are other limits

12
tests/test_find.py
View file

@ -111,12 +111,18 @@ def test_findCards():
# full search
f = deck.newNote()
f['Front'] = u'hello<b>world</b>'
f['Back'] = u''
f['Back'] = u'abc'
deck.addNote(f)
# as it's the sort field, it matches
assert len(deck.findCards("helloworld")) == 2
assert len(deck.findCards("helloworld", full=True)) == 2
# if we put it on the back, it won't
(f['Front'], f['Back']) = (f['Back'], f['Front'])
f.flush()
assert len(deck.findCards("helloworld")) == 0
assert len(deck.findCards("helloworld", full=True)) == 1
assert len(deck.findCards("helloworld", full=True)) == 2
assert len(deck.findCards("front:helloworld")) == 0
assert len(deck.findCards("front:helloworld", full=True)) == 1
assert len(deck.findCards("back:helloworld", full=True)) == 2
# searching for an invalid special tag should not error
assert len(deck.findCards("is:invalid")) == 0