php - 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所有相关的 ID Table A，另一个查询获得了所有必要ids的Table B.

现在我想通过在两个查询上方不匹配的那些 id 进行第三次查询。

例子：

1. 查询返回：1、5、10、15。
2. 查询返回：5。
3. 查询要按：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

查看子查询

我认为应该是这样的：

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

我认为您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..