我正在做一个维护项目,其中所有都是sql查询,但是我想将其转换为laravel。这是下面的查询:-
$members = Member::select('members.id','members.first_name',
'members.surname','members.username','members.password',
'members.email','user_access.active',
DB::raw('SUM( IF (user_access.active = "y", 1, 0) ) as acount'))
->join('user_access','user_access.member_id','=','members.id')
->where('special_access','=','n')
->groupby('user_access.member_id')
->having('acount','>','0')
->orderby('members.id','desc')
->orderby('members.username','ASC')
->orderby('user_access.note','DESC')
->paginate(30);
当我执行查询时,这给了我错误。出现在有子句的错误下面
SQLSTATE[42S22]: Column not found: 1054 Unknown column 'acount' in 'having clause' (SQL: select count(*) as aggregate from `members` inner join `user_access` on `user_access`.`member_id` = `members`.`id` where `special_access` = n group by `user_access`.`member_id` having `acount` > 0)
答案 0 :(得分:1)
我想类似的东西会对您有所帮助。
$userAccess = DB::table('user_access')
->where('special_access','=','n')
->where('active','y')
->groupby('member_id');
Member::select([
'members.id',
'members.first_name',
'members.surname',
'members.username',
'members.password',
'members.email',
'user_access.active'
])->joinSub($userAccess, 'user_access', function($join){
$join->on('user_access.member_id', '=', 'members.id');
})->orderby('members.id','desc')
->orderby('members.username','ASC')
->orderBy('user_access.note','asc')
->paginate(30);
OR
$members = Member::select([
'members.id',
'members.first_name',
'members.surname',
'members.username',
'members.password',
'members.email',
'user_access.active'
])->join('user_access', function($join){
$join->on('user_access.member_id', '=', 'members.id')->on('user_access.active', '=', DB::raw('"y"'));
})
->where('special_access','=','n')
->groupby('user_access.member_id')
->orderby('members.id','desc')
->orderby('members.username','ASC')
->orderby('user_access.note','DESC')
->paginate(30);
如apokryfos在评论中所述:
分页器将尝试获取查询的计数,并在执行此操作时将其从所有选择中删除,从而导致此错误。
如果您只需要使用user_access.active = "y"
的记录,则无需首先选择它们,然后尝试用HAVING
过滤掉它们