Question

我有这个问题：

    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

正如您所看到的，UNION ALL有两个查询。

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？我想查询1号的计数结果和查询号2的计数结果。

更新：我想得到一个这样的计数结果：数1:34，数为2：45

见这里：select count from multiple tables

Answer 1

如果要计算两个子查询中返回的行，请执行以下操作：

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，请将当前查询作为子查询并执行计数：

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;

在应用程序级别计算ord列可能更容易。

Answer 2

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;

这意味着像对待桌子一样处理你所有的联合并计算所有元素。

如何使用UNION ALL计算两个查询

2 个答案: