计数联盟或加入Sum - MySQL

时间:2009-07-08 16:06:15

标签: mysql join count union sum

我想在查询中组合三个表 - 日期,潜在客户和点击。

表格如下:

日期:

|date|

铅:

id|time|commission

点击:

id|time|commission

表格日期只是存储日期,用于获取没有点击或潜在客户的日期。

因此,如果我们在表格中有以下数据:

日期:

2009-06-01
2009-06-02
2009-06-03

铅:

1|2009-06-01|400
2|2009-06-01|300
3|2009-06-03|350

点击:

1|2009-06-01|1
2|2009-06-03|2
3|2009-06-03|2
4|2009-06-03|0

我想获得日期,点击次数,点击产生的佣金(有不提供佣金的点击次数),潜在客户数量,潜在客户产生的佣金和总佣金。因此,根据上面的表格,我想得到:

2009-06-01|1|1|2|700|701|
2009-06-02|0|0|0|0|0
2009-06-03|3|4|1|350|354|

我尝试过以下联盟:

 SELECT  
    campaign_id, 
    commission_date,  
    SUM( click_commission ) AS click_commission,
    click,
    SUM( lead_commission ) AS lead_commission ,  
    lead,
    SUM( total_commission ) as total_commission
    FROM(
        SELECT  
            click.campaign_id AS campaign_id, 
            DATE( click.time ) AS commission_date, 
            click.commission AS click_commission, 
            (SELECT count(click.id) from click GROUP BY date(click.time)) as click,
            0 as lead_commission,
            0 as lead,
            click.commission AS total_commission
        FROM click
        UNION ALL
        SELECT 
            lead.campaign_id AS campaign_id, 
            DATE( lead.time ) AS commission_date, 
            0 as click_commission,
            0 as click,
            lead.commission AS lead_commission, 
            lead.id as lead,
            lead.commission AS total_commission
        FROM lead
        UNION ALL
        SELECT 
            0 AS campaign_id, 
            date.date AS commission_date, 
            0 AS click_commission, 
            0 as click,
            0 AS lead_commission, 
            0 as lead,
            0 AS total_commission
        FROM date 
    ) AS foo 
    WHERE commission_date BETWEEN '2009-06-01' AND '2009-07-25' 
    GROUP BY  commission_date 
    ORDER BY commission_date LIMIT 0, 10

但这不能同时计算点击次数和潜在客户数量,上面的代码可以在所有潜在客户上获得正确的点击量0。如果我移动代码并从引导表中选择选择,我会在所有点击中获得引导权限0。我无法找到从查询中获取两个计数的方法。

所以我尝试了左连接:

SELECT
    date.date as date, 
    count( DISTINCT click.id ) AS clicks, 
    sum(click.commission) AS click_commission, 
    count( lead.id ) AS leads, 
    sum(lead.commission) AS lead_commission
FROM date
LEFT JOIN click ON ( date.date = date( click.time ) )
LEFT JOIN lead ON ( date.date = date( lead.time ) )
GROUP BY date.date
LIMIT 0 , 30 

此查询的问题是,如果日期上有多个点击或潜在客户,它将返回预期值* 2.因此,在2009-06-01,它将返回1400而不是预期的700佣金佣金。

所以在UNION中我有计数问题,在左连接中它是SUM无法正常工作。

如果可能的话,我真的很想坚持UNION,但我还没有办法从中得到两个数据。

(这是this之前问题的后续问题,但由于我没有要求计数,因此我发布了一个新问题。)

2 个答案:

答案 0 :(得分:2)

SELECT  date,
        COALESCE(lcomm, 0), COALESCE(lcnt, 0),
        COALESCE(ccomm, 0), COALESCE(ccnt, 0),
        COALESCE(ccomm, 0) + COALESCE(lcomm, 0),
        COALESCE(ccnt, 0) + COALESCE(lcnt, 0)
LEFT JOIN
        (
        SELECT  date, SUM(commission) AS lcomm, COUNT(*) AS lcnt
        FROM    leads
        GROUP BY
                date
        ) l
ON      l.date = d.date
LEFT JOIN
        (
        SELECT  date, SUM(commission) AS ccomm, COUNT(*) AS ccnt
        FROM    clicks
        GROUP BY
                date
        ) с
ON      c.date = d.date
FROM    date d

答案 1 :(得分:0)

我使用的代码,是根据Quassnoi的建议构建的:

SELECT  date,
        COALESCE(ccomm, 0) AS click_commission, COALESCE(ccnt, 0) AS click_count,
        COALESCE(lcomm, 0) AS lead_commision, COALESCE(lcnt, 0) AS lead_count,
        COALESCE(ccomm, 0) + COALESCE(lcomm, 0) as total_commission
FROM    date d
LEFT JOIN
        (
        SELECT  DATE(time) AS lead_date, SUM(commission) AS lcomm, COUNT(*) AS lcnt
        FROM    lead
        GROUP BY
                lead_date
        ) l
ON     lead_date = date
LEFT JOIN
        (
        SELECT  DATE(time) AS click_date, SUM(commission) AS ccomm, COUNT(*) AS ccnt
        FROM    click
        GROUP BY
                click_date
        ) с
ON      click_date =  date