我试图在laravel中获取以下mysql查询。
SELECT * FROM (SELECT `tb`.owner_id,`tb`.property_id, `tb`.tenant_id FROM
`tbl_booking` tb join`tbl_ticket` tt WHERE `tb`.tenant_id = `tt`.tenant_id and
request_status = '4') AS T where T.tenant_id = 25 limit 1;
我已经能够为内部查询编写laravel等效项,如下所示:
DB::table('tbl_booking as tb')
->join('tbl_ticket as tt ', 'tb.tenant_id', '=', 'tt.tenant_id' )
->where('tb.request_status', '=', '4')
->get(['tb.owner_id', 'tb.property_id', 'tb.tenant_id']);
有人可以提供有关如何进行的建议吗?
PS:我正在使用Laravel 5.2
最佳答案
要仅选择一个原始用途take(1)
:
DB::table('tbl_booking as tb')
->join('tbl_ticket as tt', function($join) {
$join->on('tb.tenant_id', '=', 'tt.tenant_id')
->where('request_status', DB::Raw(4));
})
->where('tb.tenant_id', '=', '25')
->take(1)
->get(['tb.owner_id', 'tb.property_id', 'tb.tenant_id']);
关于php - laravel查询mysql等效项,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48964306/