联合选择列不匹配答案

【问题标题】：Union Select Column Mismatch联合选择列不匹配
【发布时间】：2011-08-17 08:19:21
【问题描述】：

这是我的查询：

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
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

这是我的表格数据：

用户列表

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 表中 formatted_date 的收入相加，作为输出查询的第 6 列。我不知道从哪里开始。如果我将 sum(revenue) 添加到 union select 我会收到“列不匹配”错误。

【问题讨论】：

标签： mysql select union

【解决方案1】：

列revenue 属于ads_served，但您从不存在revenue 的子查询中进行选择。将其添加到子查询中：

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

【讨论】：

感谢您的回答。可悲的是，你迟到了 2 分钟。为您的帮助点赞。

【解决方案2】：

以这种方式尝试。你为什么使用 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

【讨论】：

顺便说一句，“as total”后面应该有一个逗号
为了这个例子，我忘了编辑 WHERE 1=1。在代码中，每个 SELECT 语句都应用了过滤器。我使用“1=1”来确保以下过滤器前面有“和”，以便更轻松地应用过滤器。
好的，我猜你正在创建动态查询。
啊，好的。当我发布我的信息时，我已经把它作为儿子修复了。我怕我忘记了别的东西。 :)
@user396449：manji 做了同样的事情，但他使用 0 而不是空字符串的解决方案更正确。 :) 将其应用于您的最终查询。