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use the first set for sorting

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This commit is contained in:
Baris Usakli
2017-11-08 16:32:34 -05:00
parent 4fdc1b99c6
commit 5604c62776
2 changed files with 16 additions and 1 deletions
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5
src/categories/topics.js
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@@ -90,7 +90,10 @@ module.exports = function (Categories) {
var stop = data.stop === -1 ? data.stop : start + normalTidsToGet - 1; var stop = data.stop === -1 ? data.stop : start + normalTidsToGet - 1;
if (Array.isArray(set)) { if (Array.isArray(set)) {
db[direction === 'highest-to-lowest' ? 'getSortedSetRevIntersect' : 'getSortedSetIntersect']({ sets: set, start: start, stop: stop }, next); var weights = set.map(function (s, index) {
return index ? 0 : 1;
});
db[direction === 'highest-to-lowest' ? 'getSortedSetRevIntersect' : 'getSortedSetIntersect']({ sets: set, start: start, stop: stop, weights: weights }, next);
} else { } else {
db[direction === 'highest-to-lowest' ? 'getSortedSetRevRange' : 'getSortedSetRange'](set, start, stop, next); db[direction === 'highest-to-lowest' ? 'getSortedSetRevRange' : 'getSortedSetRange'](set, start, stop, next);
} }

12
test/database/sorted.js
View File

@@ -661,6 +661,18 @@ describe('Sorted Set methods', function () {
}); });
}); });
it('should return the reverse intersection of two sets', function (done) {
db.getSortedSetRevIntersect({
sets: ['interSet1', 'interSet2'],
start: 0,
stop: 2,
}, function (err, data) {
assert.ifError(err);
assert.deepEqual(['value3', 'value2'], data);
done();
});
});
it('should return the intersection of two sets with scores aggregate MIN', function (done) { it('should return the intersection of two sets with scores aggregate MIN', function (done) {
db.getSortedSetIntersect({ db.getSortedSetIntersect({
sets: ['interSet1', 'interSet2'], sets: ['interSet1', 'interSet2'],