如何避免一个查询的匹配ID与另一个查询

Question

我的第一个查询获得了Table A的所有相关ID，另一个查询从ids获得了所有必要的Table B。

现在我想通过上面两个查询都不匹配的ID进行第三次查询。

示例：

1. 查询返回：1,5,10,15。
2. 查询返回：5。
3. 查询想要通过：1,10,15。

这是我的工作：

//1st query:
foreach((array)$f2_ids as $indx => $value) {
    $g = mysqli_query($dbh,"SELECT id FROM update WHERE 
    `to_id`='".$pg_ids[$indx]."' AND `from_id`='".$f2_ids[$indx]."' GROUP BY id") or die(mysqli_error($dbh));
    while ($rows = mysqli_fetch_assoc($g)) {
    $ids[] = $rows['id'];
    }
}

//2st query:
foreach((array)$ids as $id) {
    $p = mysqli_query($dbh,"SELECT post_id FROM view WHERE `post_id`='".$id."' AND `user_id` ='".$session->id."'") or die(mysqli_error($dbh));
    while ($rows = mysqli_fetch_assoc($p)) {
    $Vids[] = $rows['post_id'];
    }
}

//3st query: I tried which not get proper result.
foreach((array)$ids as $index => $value) {
    if($ids[$index] !== $Vids[$index]){  // avoid match ids
    //echo $ids[$index];
    $j = mysqli_query($dbh,"SELECT * FROM update WHERE `id`='".$ids[$index]."' ORDER BY created DESC");
    }
}

更新：

foreach((array)$ids as $id) {
    $p = mysqli_query($dbh,"SELECT * 
    FROM update 
    WHERE `id` = '$id'
    AND `id` NOT IN (SELECT post_id
            FROM view
            WHERE `post_id`='$id'
            AND `user_id`='$myid')
    ORDER BY created DESC") or die(mysqli_error($dbh));
            $rows = mysqli_fetch_assoc($p);
            $id = $rows['id'];
            echo $id;
}

Answer 1

查看subqueries

我认为它应该是这样的：

SELECT * 
FROM update 
WHERE `id` NOT IN (SELECT id 
                  FROM update 
                  WHERE `to_id`=".$pg_ids[$indx]."
                  AND `from_id`=".$f2_ids[$indx]."
                  GROUP BY id )
AND `id` NOT IN (SELECT post_id
                FROM view
                WHERE `post_id`=".$id."
                AND `user_id`=".$session->id.")
ORDER BY created DESC

Answer 2

我认为你应该使用单独的第三个查询，不要从第一个和第二个查询中获取值..