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运行此查询时，COUNT 函数不会按“check_if_new_customer”标志进行过滤。我在这篇文章中读到：http ://dev.mysql.com/tech-resources/articles/wizard/page4.html 在某些情况下可以使用 SUM 而不是 COUNT 来获得更准确的结果，但是当我尝试这样做时，我得到了一些非常不同的东西，它似乎显示出大量的数字翻倍。我认为这可能是因为我正在对 id 字段中的 UUID 求和，而不是在该点进行计数。关于我可以放什么来计算所有现有客户与新客户的任何建议？

SELECT
  YEAR(so.date_entered),
  so.technical_address_country,
  so.technical_address_state,
  COUNT(so.id) as all_sales,
  COUNT(mf.id) as all_jobs,
  SUM(so.total_value) as all_value,
    COUNT(IF(so.check_if_new_customer=1,so.id,0)) as sales_order_new,
  SUM(IF(so.check_if_new_customer = 1,so.total_value,0)) as total_value_new,
    COUNT(IF(so.check_if_new_customer=1,mf.id,0)) as jobs_new,
    COUNT(IF(so.check_if_new_customer=0,so.id,0)) as sales_order_existing,
  SUM(IF(so.check_if_new_customer = 0,so.total_value,0)) as total_value_existing,
    COUNT(IF(so.check_if_new_customer=0,mf.id,0)) as jobs_existing,
    SUM(IF(so.check_if_new_customer=0,mf.id,0)) as jobs_existing_t
FROM 
  sugarcrm2.so_order so 
LEFT JOIN 
  sugarcrm2.mf_job mf on so.id = mf.sales_order_id 
WHERE 
  so.date_entered > "2011-10-30" AND 
    so.technical_address_country IS NOT NULL AND  
  so.technical_address_state IS NOT NULL AND 
  so.deleted = 0 AND 
  so.has_been_promoted = 1 
GROUP BY 
    YEAR(so.date_entered),
  so.technical_address_country, 
  so.technical_address_state
ORDER BY 
  so.technical_address_country, so.technical_address_state
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3 回答 3

1

COUNT()返回其参数（如果指定）为 non 的记录数NULL。由于在这种情况下它的参数是一个IF()表达式的结果（如果为真，如果为假，则计算某个列的值0），几乎每条记录都将被计算在内，而与测试条件无关。

SUM()，顾名思义，将其参数的值相加。在这种情况下，只要测试条件为真，它就会对引用列的值求和。

显然，您所追求的也不是，尽管您的问题对于您到底想要什么是相当模棱两可的。猜测一下，你可能想要这样的东西：

SUM(so.check_if_new_customer)
于 2012-11-07T21:12:38.713 回答
1

If you want to use SUM() like COUNT() you will need to pass it either a 1 or 0 so that all of the 1's will sum up to your desired count. So in your example, if you want a sum of all the new jobs you would do this:

SUM(IF(so.check_if_new_customer=1,1,0)) as jobs_new

or if so.check_if_new_customer always returns a 1 or 0 you could alternatively do this:

SUM(so.check_if_new_customer) as jobs_new

于 2012-11-07T21:23:28.303 回答
0

每次连接表的 id 字段不为空时，我想求和 +1。

SUM(IF(so.check_if_new_customer=0 AND mf.id IS NOT NULL,1,0)) as jobs_existing

于 2012-11-07T21:21:29.827 回答