Question

我有3个表，我查询了第一个表（用户）以获得此结果：

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

我希望过滤其他两个表朋友和freinds_request并删除数字中的ID，以便我稍后再做一些事情，但是由于我的理解不好，它会返回此错误：

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

然后我尝试从sql命令运行查询 - phpmyadmin来查看错误：

 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'))
                                           )
                                           )

但是MYSQL说：

#1241 - Operand should contain 1 column(s)

下面是我的PHP代码：

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);

   }



}



}

以下是存储的表格和数据：

用户

朋友

friend_request

如何解决此问题？谢谢。

Answer 1

<强>已更新：好的，那么你的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
)

P.S。当然，请使用预备语句。我建议使用PDO。如果您没有可能，请使用mysqli::prepare方法。它会使您的代码看起来像这样（只需将:loggedInId更改为?）：

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

Answer 2

有一个问题，比如在（）中删除single quote，否则会考虑单个字符串：

(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')) ) )

请更新如下：

(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

您的查询在逻辑上不正确，当您执行where子句时，子选择如下：

var re = /@(\w+)/; 
var arr = ['@one', 'two', '@three'];

arr.forEach(function(str) {
  if (str.match(re))
  document.writeln(str.match(re)[1] + '<br>');
});

您应该只选择1个操作数/列，即子查询中的columnX。见下面的粗体：

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

Answer 4

读到Dimitiri发布的内容时，我正在思考并重新编写下面的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））要么（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'））“;

这会过滤并删除两个表中的所有ID。以前在我的朋友表中我有5个ID代表朋友，在上面运行此查询时，这些ID被删除：

下面是一张证明它的照片：

$ 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

说明：

这意味着我已经是ids的朋友：21,22,26,30（朋友表）

和

我有一个朋友请求，意味着id 33（friend_requst_table）

修复sql语句中的错误

4 个答案: