【发布时间】:2019-04-03 10:59:37
【问题描述】:
大家好,我是 laravel 查询生成器的新手。 我想将此查询转换为 laravel 查询生成器。提前谢谢你。
$bo_facilities = ' SELECT
a.bo_facility_code,
a.bo_facility_groupcode,
a.bo_number,
a.bo_indYesOrNo,
a.bo_indLogic,
b.bo_content_facility_description
FROM
bo_facilities AS a
RIGHT JOIN bo_content_facilities AS b
ON b.bo_content_facility_code = a.bo_facility_code AND b.bo_content_facility_facilityGroupCode = a.bo_facility_groupcode
WHERE a.bo_hotel_code = "1" GROUP BY b.bo_content_facility_description';
我的 groupBy 方法不起作用..
$join->groupBy('b.bo_content_facility_description');
这是我的全部代码 我不知道最好的方法。
DB::table('bo_facilities AS a')
->select("a.bo_facility_code", "a.bo_facility_groupcode", "a.bo_number", "a.bo_indYesOrNo", "a.bo_indLogic", "b.bo_content_facility_description")
->join('bo_content_facilities AS b', function ($join) {
$join->on('b.bo_content_facility_code', '=', 'a.bo_facility_code');
$join->on('b.bo_content_facility_facilityGroupCode', '=', 'a.bo_facility_groupcode');
$join->groupBy('b.bo_content_facility_description');
})
->where('a.bo_hotel_code', $hotelCode)
->get();
【问题讨论】:
标签: laravel query-builder