修复SQL语句中的错误

Question

I have 3 tables of which i queried the first table (users) to get this result:

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

and I want to filter through two other tables friends and freinds_request and remove ids in the numbers so that I cud do something with later but to my poor understanding it returns this error:

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\cebs\include\functions.php on line 825

I then tried to run query from sql command - phpmyadmin to see what was wrong:

 SELECT id FROM users WHERE id NOT IN
(SELECT user_one,user_two FROM friends WHERE (user_one='18' AND 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') AND user_two='18')) 
     AND id NOT IN( SELECT to_user,from_user FROM friend_reqest WHERE (to_user='18' AND 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') AND from_user='18'))
                                           )
                                           )

but MYSQL Said:

#1241 - Operand should contain 1 column(s) 

below is my php code also:

function somepeopleyoumayknow(){


global $dbc_conn, $IsLoggIn,$table_name,$friend_request_table,$friends_table ; 
$cu_school = getuser($IsLoggIn,'cell_group');

//assuming $IsLoggIn is equal  to 18...

$peopleids= mysqli_query($dbc_conn,"SELECT id FROM $table_name WHERE id !='$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 FROM $table_name WHERE id NOT IN
 (SELECT user_one,user_two FROM $friends_table WHERE (user_one='$IsLoggIn' AND user_two IN($string_users)) 
 OR (user_one IN($string_users) AND user_two='$IsLoggIn')) 
 AND id NOT IN( SELECT to_user,from_user FROM $friend_request_table WHERE (to_user='$IsLoggIn' AND from_user IN($string_users) OR (to_user IN($string_users) AND 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 are the tables and data stored:

users

friends

friend_request

How can I fix this problem? Thank you.

Answer 1

UPDATED : OK, then your sql-statement should look like this:

SELECT * FROM (
    SELECT user_one as id
    FROM friends 
    WHERE user_one != :loggedInId
    UNION 
    SELECT user_two as id
    FROM friends
    WHERE user_two != :loggedInId
)
WHERE id NOT IN (
    SELECT from_user as id
    FROM friend_request 
    WHERE to_user = :loggedInId
    UNION
    SELECT to_user as id
    FROM friend_request
    WHERE from_user = :loggedInId
)

PS And, of course, use prepared statements instead. I suggest using PDO . If you don't have a possibility, use mysqli::prepare method. It will make you code look like this (just change :loggedInId to ? ):

if ($stmt = $mysqli->prepare($sql)) {
    $stmt->bind_param('i', $loggedInId);
    $stmt->execute();
    $stmt->fetch();
    $stmt->close();
}

Answer 2

There is a issue like remove single quote inside in() otherwise it will consider a single string:

(SELECT user_one,user_two FROM friends WHERE (user_one='18' AND 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') AND user_two='18')) AND id NOT IN( SELECT to_user,from_user FROM friend_reqest WHERE (to_user='18' AND 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') AND from_user='18')) ) )

Please update as below:

(SELECT user_one,user_two FROM friends WHERE (user_one='18' AND 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) AND user_two='18')) AND id NOT IN( SELECT to_user,from_user FROM friend_reqest WHERE (to_user='18' AND 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) AND from_user='18')) ) )

Answer 3

Your query is not logically correct, when you do where clause + sub select like:

WHERE <columnX> NOT IN (SELECT <columnX from table x)

You should select only 1 operand/column ie columnX in the sub query. See below the boldded :

$sql = "SELECT id FROM $table_name WHERE **id** NOT IN
 (SELECT **id** FROM $friends_table WHERE (user_one='$IsLoggIn' AND user_two IN($string_users)) 
 OR (user_one IN($string_users) AND user_two='$IsLoggIn')) 
AND id NOT IN( SELECT **id** FROM $friend_request_table WHERE   (to_user='$IsLoggIn' AND from_user IN($string_users) OR (to_user  IN($string_users) AND from_user='$IsLoggIn'))

Answer 4

Upon scratching my head reading what Dimitiri posted I re-think and re-write the SQL Statement below is my answer:

$sql = "SELECT id FROM $table_name WHERE id NOT IN

( SELECT user_one FROM $friends_table WHERE (user_one='$IsLoggIn' AND user_two IN($string_users)) OR (user_one IN($string_users) AND user_two='$IsLoggIn')) AND id NOT IN(SELECT user_two FROM $friends_table WHERE (user_one='$IsLoggIn' AND user_two IN($string_users)) OR ( user_one IN($string_users) AND user_two='$IsLoggIn'))

AND id NOT IN (SELECT from_user FROM $friend_request_table WHERE(from_user='$IsLoggIn' AND to_user IN($string_users)) OR (from_user IN($string_users) AND to_user='$IsLoggIn'))

AND id NOT IN (SELECT to_user FROM $friend_request_table WHERE(from_user='$IsLoggIn' AND to_user IN($string_users)) OR (from_user IN($string_users) AND to_user='$IsLoggIn'))";

This filters and remove all the ids in both tables. Previously in my friends table i had 5 ids representing friends upon running this query above these ids are removed:

below is a photo to prove it:

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

$filtered_id_users= 19,25,27,28,29,31,32,34

EXPLAINING:

This means i was already friends with ids: 21,22,26,30 (Friends table)

and

I had 1 friend request which connotes to id 33 (friend_requst_table)