我已在查询一个范围雄辩复杂的连接
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);
});
}
记录此范围增加:
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
我没发现一个很好的解决方案,JoinClause有函数:On,orOn,Where,或者Where。
但非所有可以采取功能作为输入,并且组查询......
别人的理想?
谢谢,这样做的工作) – 2014-09-11 13:27:46