Mysql 推进自定义标准

我正在尝试使用propel critiera构建以下查询：

SELECT *
FROM mytable
WHERE facility_id IN (4,7)
GROUP BY housing_id
HAVING COUNT(DISTINCT facility_id) = 2;

推进标准：

$criteria->add(HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID, $facility_filter, Criteria::IN);

$criteria->addGroupByColumn(HousingsHousingFacilitiesPeer::HOUSING_ID);

$criteria->addHaving(
  $criteria->getNewCriterion(
    HousingsHousingFacilitiesPeer::HOUSING_ID, 
    'COUNT(DISTINCT housings_housing_facilities.HOUSING_FACILITY_ID) = 2',
    Criteria::CUSTOM
  )
);

但是查询不会返回任何结果

---

我是否正确使用了自定义条件？

在第一次检查时，您的

计数不一致

似乎有误：

'COUNT(DISTINCT HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID) = 2'

如果只是一个坏的c/p，关于，您可以：

$c = new Criteria();
$c->add(HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID, $facility_filter, Criteria::IN);
$c->addGroupByColumn(HousingsHousingFacilitiesPeer::HOUSING_ID);
$c->addAsColumn('numFacilities', 'COUNT(DISTINCT '.HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID.')');
$c->addHaving($c->getNewCriterion(HousingsHousingFacilitiesPeer::ID, 'numArticles=2', Criteria::CUSTOM));

---

第一次检查时，您的

计数不重复

似乎有误：

'COUNT(DISTINCT HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID) = 2'

如果只是一个坏的c/p，关于，您可以：

$c = new Criteria();
$c->add(HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID, $facility_filter, Criteria::IN);
$c->addGroupByColumn(HousingsHousingFacilitiesPeer::HOUSING_ID);
$c->addAsColumn('numFacilities', 'COUNT(DISTINCT '.HousingsHousingFacilitiesPeer::HOUSING_FACILITY_ID.')');
$c->addHaving($c->getNewCriterion(HousingsHousingFacilitiesPeer::ID, 'numArticles=2', Criteria::CUSTOM));