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。)
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