联合计数或加入总和 - MySQL

Question

我想结合查询中的三个表 - 日期，潜在顾客和点击。

的表是这样的：

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导致的日期。

因此，如果我们有表中的以下数据：

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，由点击产生的佣金（有点击不给佣金），销售线索数量，由销售线索和佣金产生的佣金。因此，与上述我的表想获得：

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

我曾尝试用下面的工会：

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 

但是，这并不工作，以计算点击次数和潜在客户的两个号码，代码上面给出了所有线索上适量的点击量0。如果我移动代码并从主表中选择，我会在所有的点击中获得引导权利。我一直无法找到从查询中获得两个计数的方法。

于是，我就不是一个左联接：

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 

与此查询的问题是，如果有一个以上的点击，或者导致一个日期，将返回预期值* 2。所以2009年-06-01它将返回1400，而不是预期的700美元，作为牵头佣金。

所以在UNION中，我遇到了计数问题，在左边的连接中，它是SUM不工作。

如果可能的话，我真的想坚持UNION，但我还没有找到一种方法从它得到两个计数。

（这是后续的this之前的问题，但由于我没有要求计数，因此我发布了一个新问题。）

Answer 1

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 

Answer 2

，我使用的代码，从建议修建从Quassnoi：

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