修正錯誤

Question

我有3個表，其中我問的第一個表（用戶）得到這樣的結果：

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

，我想通過其他兩個朋友來過濾和freinds_request和刪除IDS在這些數字讓我CUD做一些事情以後，但我不太瞭解它返回此錯誤：

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

然後我試圖運行從SQL命令查詢 - phpMyAdmin來看看什麼是錯的：

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')) 
              ) 
              ) 

但MySQL表示：

#1241 - Operand should contain 1 column(s) 

下面

是我的PHP代碼也：

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); 

    } 



} 



} 

下面

存儲表和數據：

用戶

朋友

friend_request

我該如何解決這個問題？謝謝。

Answer 1

修訂： OK，那麼您的SQL語句應該是這樣的：

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 
) 

附：而且，當然，請使用準備好的語句。我建議使用PDO。如果您沒有可能，請使用mysqli::prepare方法。 它會使你的代碼看起來像這樣（只是改變:loggedInId到?）：

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

Answer 2

有喜歡刪除single quote裏面（一期），否則將考慮一個字符串：

(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')))) 

請如下更新：

(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

您的查詢不正確的邏輯，當你做where子句+蘇b選擇像：

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

您應該只選擇1個操作數/列，即在子查詢中的columnX。 請參見下面的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

在抓我的頭讀什麼Dimitiri貼我重新思考和重新編寫下面的SQL語句是我的回答：

$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）） 或 （user_one IN（$ string_users）AND user_two ='$ IsLoggIn'））AND ID NOT IN（SELECT user_two FROM $ friends_table WHERE（user_one ='' $ IsLoggIn'和user_two IN（$ string_users））或 （user_one IN（$ string_users）AND user_t WO = '$ IsLoggIn'））

和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）和to_user = '$ IsLoggIn'））

和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）和to_user ='$ IsLoggIn'））「;

這將過濾並刪除這兩個表中的所有id。 以前在我的朋友表我有5個IDS表示一旦運行這些ID上面這個查詢被刪除的朋友：

下面

是照片來證明這一點：

$ 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

解釋：

這意味着我已經是朋友IDS：21,22,26,30（友表）

和

我不得不這暗示着ID 33（friend_requst_table 1個朋友請求）