如何在mysql中选择不同值的计数总和?

时间:2015-12-24 06:29:07

标签: php mysql distinct

这是我的查询:

SELECT 
    p.PRO_Id, p.PRO_Name,
    COUNT(DISTINCT c.COM_Id) as commentCount,
    COUNT(DISTINCT d.DIS_Id) AS disCount, 
    d.DIS_ThreadDesc,
    dev.DEV_Name,
    loc.LOC_Name,
    d.USE_Id,
    d.DIS_Date
FROM projects p
LEFT JOIN discussions d ON p.PRO_Id = d.PRO_Id
LEFT JOIN comments c ON d.DIS_Id = c.DIS_Id
LEFT JOIN developer dev ON p.DEV_Id=dev.DEV_Id
LEFT JOIN locality loc ON p.LOC_Id=loc.LOC_Id
WHERE p.PRO_Status=1 
  and d.DIS_Status=1 
  and c.COM_Status=2
GROUP BY p.PRO_Id
ORDER  BY 3 desc LIMIT 3

我想要得到的是commenCount(总评论)和disCount(总讨论)的总和。
我这样做了; -

 sum(COUNT(DISTINCT c.COM_Id) + COUNT(DISTINCT d.DIS_Id)) AS totalResponse
但是没机会。

1 个答案:

答案 0 :(得分:4)

您可以添加COUNT的结果:

SELECT 
    p.PRO_Id, p.PRO_Name,
    COUNT(DISTINCT c.COM_Id) AS commentCount,
    COUNT(DISTINCT d.DIS_Id) AS disCount, 
    COUNT(DISTINCT c.COM_Id) + COUNT(DISTINCT d.DIS_Id) AS totalResponse
    d.DIS_ThreadDesc, dev.DEV_Name, loc.LOC_Name, d.USE_Id, d.DIS_Date
FROM projects p
...

或者使用子查询:

SELECT sub.*,
       commentCount + disCount AS totalResponse
FROM (
  SELECT 
    p.PRO_Id, p.PRO_Name,
    COUNT(DISTINCT c.COM_Id) as commentCount,
    COUNT(DISTINCT d.DIS_Id) AS disCount, 
    d.DIS_ThreadDesc,
    dev.DEV_Name,
    loc.LOC_Name,
    d.USE_Id,
    d.DIS_Date
  FROM projects p
  LEFT JOIN discussions d ON p.PRO_Id = d.PRO_Id
  LEFT JOIN comments c ON d.DIS_Id = c.DIS_Id
  LEFT JOIN developer dev ON p.DEV_Id=dev.DEV_Id
  LEFT JOIN locality loc ON p.LOC_Id=loc.LOC_Id
  WHERE p.PRO_Status=1 
   and d.DIS_Status=1 
   and c.COM_Status=2
  GROUP BY p.PRO_Id
) AS sub
ORDER BY commentCount LIMIT 3

请注意,您的SELECT列表与GROUP BY不匹配。这适用于MySQL,但不符合ANSI标准。您应该在两个子句中使用相同的列表,或者对GROUP BY中未指定的列使用聚合函数。更多信息here