这是我的疑问:
SELECT
FROM_UNIXTIME( date_added, '%m-%d-%Y' ) AS formatted_date,
SUM( tb =1 ) AS sum_users,
SUM( tb =2 ) AS sum_links,
SUM( tb =3 ) AS sum_ads,
SUM( tb =4 ) AS sum_actions
FROM (
SELECT date_added, 1 AS tb
FROM users_list WHERE 1=1
UNION ALL
SELECT date_added, 2
FROM users_links WHERE 1=1
UNION ALL
SELECT date_served, 3
FROM ads_served WHERE 1=1
UNION ALL
SELECT date_served, 4
FROM actions WHERE 1=1
) AS t
GROUP BY formatted_date
ORDER BY formatted_date DESC
这是我的表数据:
users_list
id date_added
1 1234567890
2 1334567890
3 1434567890
users_links
id date_added
1 1244567890
2 1354567890
3 1464567890
ads_served
id date_served revenue
1 1234567891 0.01
2 1334567892 0.02
3 1434567893 0.02
动作
id date_served
1 1234561890
2 1334562890
3 1434563890
我正在尝试将ads_served表格中formatted_date的收入作为输出查询的第6列。我迷失在哪里开始。如果我将总和(收入)添加到联合选择中,则会出现“列不匹配”错误。
答案 0 :(得分:3)
列revenue
属于ads_served
,但您要从不存在revenue
的子查询中进行选择。将其添加到子查询:
SELECT
FROM_UNIXTIME( date_added, '%m-%d-%Y' ) AS formatted_date,
SUM( tb =1 ) AS sum_users,
SUM( tb =2 ) AS sum_links,
SUM( tb =3 ) AS sum_ads,
SUM( tb =4 ) AS sum_actions,
SUM( revenue ) As sum_revenue
FROM (
SELECT date_added, 1 AS tb, 0 As revenue
FROM users_list WHERE 1=1
UNION ALL
SELECT date_added, 2, 0
FROM users_links WHERE 1=1
UNION ALL
SELECT date_served, 3, revenue
FROM ads_served WHERE 1=1
UNION ALL
SELECT date_served, 4, 0
FROM actions WHERE 1=1
) AS t
GROUP BY formatted_date
ORDER BY formatted_date DESC
答案 1 :(得分:1)
尝试这种方式。你为什么用1 = 1?
SELECT
FROM_UNIXTIME( date_added, '%m-%d-%Y' ) AS formatted_date,
SUM( tb =1 ) AS sum_users,
SUM( tb =2 ) AS sum_links,
SUM( tb =3 ) AS sum_ads,
SUM( tb =4 ) AS sum_actions,
sum(total) as tot_rev
FROM (
SELECT date_added,'' as total, 1 AS tb
FROM users_list WHERE 1=1
UNION ALL
SELECT date_added,'', 2
FROM users_links WHERE 1=1
UNION ALL
SELECT date_served,revenue, 3
FROM ads_served WHERE 1=1
UNION ALL
SELECT date_served,'', 4
FROM actions WHERE 1=1
) AS t
GROUP BY formatted_date
ORDER BY formatted_date DESC