3

我将这两个查询单独列出，但现在我需要将结果放在同一个列表中。

现在我通过使用javascript按id排序项目来“修复”它。但是从服务器合并这些数据会很棒，这样我就可以对结果进行分页了..

数据库架构

桌友：

在此处输入图像描述

表帮助

在此处输入图像描述

查询 A（返回当前用户的结果）

   $sql = 'SELECT * FROM helps WHERE id_user ='.$value;

查询 B（返回当前用户好友的结果）

$sql = 'SELECT
    h.*,
    f.* 
    FROM (
            SELECT
                     id  AS friendsId,
                     CASE followerid WHEN '.$value.' THEN followingid ELSE followerid END AS friend_id
                        FROM friends
                        WHERE acepted = 1
                        AND (followerid  = '.$value.' OR followingid = '.$value.')
                    ) AS f
                        INNER JOIN helps AS h ON h.id_user = f.friend_id
                        ORDER BY h.id DESC';

有没有办法合并这些查询？老实说，我不知道该怎么做。

-编辑-

考虑联合，但我不知道如何处理订单..

$sql = '(SELECT * FROM helps WHERE id_user = '.$value.')
                    UNION
                    (SELECT
                                h.*,
                                f.* 
                            FROM (
                                SELECT
                                    id  AS friendsId,
                                    CASE followerid WHEN '.$value.' THEN followingid ELSE followerid END AS friend_id
                                FROM friends
                                WHERE acepted = 1
                                AND (followerid  = '.$value.' OR followingid = '.$value.')
                            ) AS f
                                INNER JOIN helps AS h ON h.id_user = f.friend_id
                                ORDER BY h.id DESC)';

正如您在第二个查询中看到的，帮助表已重命名为 h。我怎样才能在第一次做同样的事情？（两个查询都以相同的格式返回数据）

4

4 回答 4

4

要合并这两个查询和订单记录的结果，您需要将UNION ALL与GROUP BY. 试试这个查询：

SELECT h.id AS id, h.title, h.content, id_user, h.id_group,
       h.id_type, h.id_lic, h.avatar, h.attached, h.attached,
       h.fetcha, h.likes, h.lan, h.needsCount, h.receivedCount,
       MAX(h.fid) AS fid, MAX(h.friend_id) AS friend_id
FROM (   
      SELECT h.id AS id, h.title, h.content, id_user, h.id_group,
             h.id_type, h.id_lic, h.avatar, h.attached, h.attached,
             h.fetcha, h.likes, h.lan, h.needsCount, h.receivedCount,
             0 AS fid, 0 AS friend_id
      FROM helps h
      WHERE h.id_user ='.$value.'

      UNION ALL

      SELECT h.id, h.title, h.content, id_user, h.id_group,
             h.id_type, h.id_lic, h.avatar, h.attached, h.attached,
             h.fetcha, h.likes, h.lan, h.needsCount, h.receivedCount,
             f.id AS fid, f.friend_id
      FROM (
             SELECT id  AS friendsId,
                    CASE followerid WHEN '.$value.' THEN followingid
                                    ELSE followerid
                    END AS friend_id
             FROM friends
             WHERE acepted = 1
                   AND (followerid  = '.$value.' OR followingid = '.$value.')
            ) AS f
            INNER JOIN helps AS h
                ON h.id_user = f.friend_id
     ) h
GROUP BY h.id
ORDER BY h.id DESC;
于 2012-09-11T13:50:31.030 回答
1

的基本规则UNION是

  • 每个查询返回的列数必须相等，并且
  • 数据类型必须相互匹配。

如您所见，我没有修改您制定的查询。只是我为每个查询添加了一个虚拟列。此列将用于对所有记录进行排序。查询被另一个包装，select statment因此虚拟列不会显示在结果列表中，但它将是对记录进行排序的基础。下面的查询结果包含最顶部的用户记录（因为ORDER BY OrderBy ASC），然后是按好友ID排序的用户好友记录。如果您想要任何澄清，请通知我。还有一件事，我f.*在您的第二个查询中删除了，因为它不是必需的，并且可能由于列（number）不匹配而导致您的语法错误。

SELECT  ID, Title, Content, ID_User, ID_Group, 
        ID_Type, ID_Loc, Avatar, Attached, Fecha, Likes, 
        Lan, NeedsCount, RecivedCount
FROM
    (
        SELECT *, 1 as OrderBy           -- first query you formulated
        FROM helps 
        WHERE id_user = '.$value.'
        UNION
        SELECT  h.*, 2 as OrderBy        -- second query you formulated
        FROM    (
                    SELECT  id  AS friendsId,
                            CASE followerid 
                                WHEN '.$value.' THEN followingid 
                                ELSE followerid 
                            END AS friend_id
                    FROM    friends
                    WHERE   acepted = 1 AND 
                            (followerid  = '.$value.' OR followingid = '.$value.')
                ) AS f
                    INNER JOIN helps AS h 
                        ON h.id_user = f.friend_id
    ) x
ORDER BY OrderBy ASC, ID DESC
于 2012-09-13T09:19:05.880 回答
0

UNION如果两个查询的结果集相似，请尝试运算符

于 2012-09-09T12:49:36.557 回答
0

就像 Omesh 描述的那样做一个 UNION，然后处理排序，这样做：

SELECT * FROM
(
    ....Omesh's code....
) T
ORDER BY T.id DESC
于 2012-09-11T15:36:12.680 回答