我已在查詢一個範圍雄辯複雜的連接
public function scopeCollaborative($query){
return $query->leftJoin('collaborative', function($join){
$join->on('imms.phone2', '=', 'collaborative.phone')
->orOn('imms.phone', '=', 'collaborative.phone')
->where('collaborative.user_id', '=', App('CURUSER')->id);
});
}
記錄此範圍增加:
left join `cs_collaborative` on
`cs_imms`.`phone2` = `cs_collaborative`.`phone` or
`cs_imms`.`phone` = `cs_collaborative`.`phone` and
`cs_collaborative`.`user_id` = 3
,但我需要有:
left join `cs_collaborative` on
(`cs_imms`.`phone2` = `cs_collaborative`.`phone` or
`cs_imms`.`phone` = `cs_collaborative`.`phone`) and
`cs_collaborative`.`user_id` = 3
我沒發現一個很好的解決方案,JoinClause有函數:On,orOn,Where,或者Where。
但非所有可以採取功能作爲輸入,並且組查詢......
別人的理想?
謝謝,這樣做的工作) – 2014-09-11 13:27:46