php - Fixing errors in sql statement -

i have 3 tables of queried first table (users) result:

$string_users='19,20,21,22,25,26,27,28,29,30,31,32,33,34'; 

and want filter through 2 other tables friends , freinds_request , remove ids in numbers cud later poor understanding returns error:

warning: mysqli_num_rows() expects parameter 1 mysqli_result, boolean given in c:\xampp\htdocs\cebs\include\functions.php on line 825 

i tried run query sql command - phpmyadmin see wrong:

 select id users id not in (select user_one,user_two friends (user_one='18' , user_two in('19,20,21,22,25,26,27,28,29,30,31,32,33,34'))       or (user_one in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') , user_two='18'))       , id not in( select to_user,from_user friend_reqest (to_user='18' , from_user in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') or (to_user in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') , from_user='18'))                                            )                                            ) 

but mysql said:

#1241 - operand should contain 1 column(s)  

below php code also:

function somepeopleyoumayknow(){   global $dbc_conn, $isloggin,$table_name,$friend_request_table,$friends_table ;  $cu_school = getuser($isloggin,'cell_group');  //assuming $isloggin equal  18...  $peopleids= mysqli_query($dbc_conn,"select id $table_name id !='$isloggin'"); $sql_num_rows = mysqli_num_rows($peopleids); if($sql_num_rows > 0){ while($run_peopleids= mysqli_fetch_array($peopleids)){      $users_ids[] = $run_peopleids['id'];  }    $string_users = implode(',',$users_ids);   $sql = "select id $table_name id not in  (select user_one,user_two $friends_table (user_one='$isloggin' , user_two in($string_users))   or (user_one in($string_users) , user_two='$isloggin'))   , id not in( select to_user,from_user $friend_request_table (to_user='$isloggin' , from_user in($string_users) or (to_user in($string_users) , from_user='$isloggin'))                                        )";    $filter_id_query = mysqli_query($dbc_conn,$sql);    $fnrows = mysqli_num_rows($filter_id_query);     if($fnrows > 0){        while($run_fiq=mysqli_fetch_array($filter_id_query)){            $uid[] = $run_fiq['id'];        }        echo $filtered_id_users = implode(',',$uid);     }    }    } 

below tables , data stored:

users

friends

friend_request

how can fix problem? thank you.

there issue remove single quote inside in() otherwise consider single string:

(select user_one,user_two friends (user_one='18' , user_two in('19,20,21,22,25,26,27,28,29,30,31,32,33,34')) or (user_one in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') , user_two='18')) , id not in( select to_user,from_user friend_reqest (to_user='18' , from_user in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') or (to_user in('19,20,21,22,25,26,27,28,29,30,31,32,33,34') , from_user='18')) ) ) 

please update below:

(select user_one,user_two friends (user_one='18' , user_two in(19,20,21,22,25,26,27,28,29,30,31,32,33,34)) or (user_one in(19,20,21,22,25,26,27,28,29,30,31,32,33,34) , user_two='18')) , id not in( select to_user,from_user friend_reqest (to_user='18' , from_user in(19,20,21,22,25,26,27,28,29,30,31,32,33,34) or (to_user in(19,20,21,22,25,26,27,28,29,30,31,32,33,34) , from_user='18')) ) )