0

我有这个查询：

    SELECT ava_users.*, 0 AS ord
    FROM ava_friend_requests
    LEFT JOIN ava_users 
    ON ava_friend_requests.from_user = ava_users.id
    WHERE ava_friend_requests.to_user = $user[id]
    UNION ALL
    SELECT ava_users.*, 1 AS ord
    FROM ava_friends
    LEFT JOIN ava_users 
    ON ava_friends.user2 = ava_users.id
    WHERE ava_friends.user1 = $user[id]
    ORDER BY ord
    LIMIT $from, $display_num

如您所见，有两个带有 UNION ALL 的查询。

现在我的问题是：我如何计算每个查询？如何检索每个计数值？我想要查询 1 的计数结果和查询 2 的计数结果。

更新：我想要一个这样的计数结果：count 1: 34, count 2: 45

见这里：从多个表中选择计数

4

2 回答 2

2

如果要计算两个子查询返回的行数，如何：

SELECT 'Query1' as which, count(*) as cnt
FROM ava_friend_requests
LEFT JOIN ava_users 
ON ava_friend_requests.from_user = ava_users.id
WHERE ava_friend_requests.to_user = $user[id]
UNION ALL
SELECT 'Query2', count(*)
FROM ava_friends
LEFT JOIN ava_users 
ON ava_friends.user2 = ava_users.id
WHERE ava_friends.user1 = $user[id];

编辑：

要在限制之后执行此操作，请将当前查询设为子查询并进行计数：

select ord, count(*) as cnt
from (SELECT ava_users.*, 0 AS ord
      FROM ava_friend_requests
      LEFT JOIN ava_users 
      ON ava_friend_requests.from_user = ava_users.id
      WHERE ava_friend_requests.to_user = $user[id]
      UNION ALL
      SELECT ava_users.*, 1 AS ord
      FROM ava_friends
      LEFT JOIN ava_users 
      ON ava_friends.user2 = ava_users.id
      WHERE ava_friends.user1 = $user[id]
      ORDER BY ord
      LIMIT $from, $display_num
     ) t
group by ord;

ord仅在应用程序级别计算列也可能更容易。

于 2013-08-11T15:50:49.663 回答
1

你可能想试试这个：

select count(*) from (SELECT ava_users.*, 0 AS ord
FROM ava_friend_requests
LEFT JOIN ava_users 
ON ava_friend_requests.from_user = ava_users.id
WHERE ava_friend_requests.to_user = $user[id]
UNION ALL
SELECT ava_users.*, 1 AS ord
FROM ava_friends
LEFT JOIN ava_users 
ON ava_friends.user2 = ava_users.id
WHERE ava_friends.user1 = $user[id]
ORDER BY ord
LIMIT $from, $display_num) as users;

这意味着要像一张表一样处理你的 union all 并计算所有元素。

于 2013-08-11T15:52:04.990 回答