这是我的SQL查询PHP laravel未定义的变量:用户ID在控制器
select * from users us join EmailBox em on us.userid = em.suserid
where em.status = "Delete" and (em.suserid="$userid" or em.riemailid="$user")
我要检查任何一个领域,如果只有“suserid”或“riemailid”的结果。 所以我改变了这个laravel querybuilder。
$lists = EmailBox::join('users as us', 'EmailBox.suserid', '=', 'us.id')
->where('status' , 'Delete')
->where(function($query){
$query->where('suserid' , '$userid');
$query->orwhere('riemailid' , '$user');
})->paginate(5);
此查询工作正常时,该值将被直接传递, 用户ID直接传递1表示
$lists = EmailBox::join('users as us', 'EmailBox.suserid', '=', 'us.id')
->where('status' , 'Delete')
->where(function($query){
$query->where('suserid' , 1);
$query->orwhere('riemailid' , '$user');
})->paginate(5);
这是我在laravel代码。 什么是错误?
if(Auth::check()){
//if(Auth::user()->email!="")
$user=Auth::user()->email;
$userid=Auth::user()->id;
$lists = EmailBox::join('users as us', 'EmailBox.suserid', '=', 'us.id')
->where('status' , 'Delete')
->where(function($query){
$query->where('suserid' , '$userid');
$query->orwhere('riemailid' , '$user');
})->paginate(5);
$links = $lists->render();
return view('front.index', compact('lists', 'links'));
}
我得到错误$ userid变量是未定义的。
它会返回值1多数民众赞成没有问题 –