修正错误

我有3个表，其中我问的第一个表（用户）得到这样的结果：修正错误

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

，我想通过其他两个朋友来过滤和freinds_request和删除IDS在这些数字让我CUD做一些事情以后，但我不太了解它返回此错误：

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

然后我试图运行从SQL命令查询 - phpMyAdmin来看看什么是错的：

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代码也：

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

存储表和数据：

用户

朋友

friend_request

我该如何解决这个问题？谢谢。

来源

2016-04-25 James Favour

回应您的查询并直接运行到phpmyadmin并检查您的查询错误 – Saty

也许您需要使用'left join'或'inner join'来获取结果...您可以显示结构数据库。 – Naumov

**警告**：使用'mysqli'时，您应该使用参数化查询和['bind_param']（http://php.net/manual/en/mysqli-stmt.bind-param.php）来添加用户数据到您的查询。 **不要**使用手动转义和字符串插值或串联来实现此目的，因为如果您忘记正确地转义某些内容，您将创建严重的[SQL注入漏洞]（http://bobby-tables.com/）。 – tadman

修订： OK，那么您的SQL语句应该是这样的：

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，当然，请使用准备好的语句。我建议使用PDO。如果您没有可能，请使用mysqli::prepare方法。它会使你的代码看起来像这样（只是改变:loggedInId到?）：

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

来源

2016-04-25 07:52:08 freeek

感谢您的回答，但它不适合我。有另一种方法可以达到这个目的吗？ –

请给我们所有的表格说明，以了解存储的字段。似乎你需要这样的东西：select_to_user from friend_request UNION select from_user FROM friend_request WHERE ... – freeek

请问我已经更新了表格和描述的问题。 @freeek –

有喜欢删除single quote里面（一期），否则将考虑一个字符串：

(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'))))

请如下更新：

(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'))))

来源

2016-04-25 07:54:32

您的查询不正确的逻辑，当你做where子句+苏b选择像：

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

您应该只选择1个操作数/列，即在子查询中的columnX。请参见下面的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'))

来源

2016-04-25 07:57:42 Dimitrii

什么是操作数？...试过了你的帮助，但它没有奏效。我仍然有相同的错误信息。 @Dimitrii –

请参阅本手册页面：http://dev.mysql.com/doc/refman/5.7/en/any-in-some-subqueries。HTML 你在哪里（子查询）条款是不正确的，应该像其中，操作数IN（从选择操作数...） – Dimitrii

在抓我的头读什么Dimitiri贴我重新思考和重新编写下面的SQL语句是我的回答：

$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'））“;

这将过滤并删除这两个表中的所有id。以前在我的朋友表我有5个IDS表示一旦运行这些ID上面这个查询被删除的朋友：

下面

是照片来证明这一点：

$ 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

解释：

这意味着我已经是朋友IDS：21,22,26,30（友表）

和

我不得不这暗示着ID 33（friend_requst_table 1个朋友请求）

来源

2016-04-25 10:40:06

回答

相关问题