如何计算where子句的结果以计算MYSQL select子句的比例？答案

【问题标题】：How to count the results of a where clause in order to calculate a proportion a MYSQL select clause?如何计算where子句的结果以计算MYSQL select子句的比例？
【发布时间】：2016-04-28 21:03:45
【问题描述】：

我从这个查询开始，它返回了 8 条具有“已声明”状态的记录。我正在查看invites-from-address 列中的任何地址是否与moves-from-address 列中的地址不同：

SELECT i.id, i.company_id, i.status,  
  ia_f.base_street as "invites-from-address", a_f.base_street as "moves-from-address", 
  ia_t.base_street as "invites-to-address", a_t.base_street as "moves-to-address", i.`mover_first_name`, 
  i.mover_last_name, i.`to_address_id`

FROM invites i
JOIN moves m ON i.id = m.`claimed_invite_id`
JOIN `invite_addresses` ia_f ON ia_f.id = i.`from_address_id`
JOIN addresses a_f ON a_f.id = m.from_address_id

JOIN `invite_addresses` ia_t ON ia_t.id = i.to_address_id
JOIN addresses a_t ON a_t.id = m.to_address_id

WHERE i.`company_id` = 1040345
GROUP BY id

我在下面的这个查询中试图做的是动态创建一个average_discrepancy 列，显示invites-from-address 和moves-from-address 之间不同地址的比例。我能够通过使用WHERE 子句来成功检查地址差异，该子句检查ia_f.base_street 不等于a_f.base_street（分别别名为invites-from-address 和moves-from-address 列）但是当我把这个我的SELECT 中的count 函数内的WHERE 子句导致它不起作用。是因为我不能将WHERE 子句放在SELECT 或count 函数或两者中？在我的SELECT 子句中尝试划分对count 函数的两次调用的结果是否也存在问题？

SELECT i.id, i.company_id, i.status, 
  count(WHERE ia_f.base_street != a_f.base_street)/count(i.status="claimed") as "average_discrepancy", 
  ia_f.base_street as "invites-from-address", a_f.base_street as "moves-from-address", 
  ia_t.base_street as "invites-to-address", a_t.base_street as "moves-to-address",
  i.`mover_first_name`, 
  i.mover_last_name, i.`to_address_id`

FROM invites i
JOIN moves m ON i.id = m.`claimed_invite_id`
JOIN `invite_addresses` ia_f ON ia_f.id = i.`from_address_id`
JOIN addresses a_f ON a_f.id = m.from_address_id

JOIN `invite_addresses` ia_t ON ia_t.id = i.to_address_id
JOIN addresses a_t ON a_t.id = m.to_address_id

WHERE i.`company_id` = 1040345

AND i.status = "claimed"

【问题讨论】：

标签： mysql select subquery

【解决方案1】：

您需要将其放入 SUM 而不是 COUNT。像这样的东西可以解决问题：

 SELECT i.id, i.company_id, i.status, 
  SUM(CASE WHEN ia_f.base_street != a_f.base_street THEN 1 ELSE 0 END)/ SUM(CASE WHEN i.status='claimed' THEN 1 ELSE 0 END) as 'average_discrepancy', 
  ia_f.base_street as 'invites-from-address', 
  a_f.base_street as 'moves-from-address', 
  ia_t.base_street as 'invites-to-address', 
  a_t.base_street as 'moves-to-address',
  i.mover_first_name, 
  i.mover_last_name, 
  i.to_address_id

FROM invites i
JOIN moves m ON i.id = m.claimed_invite_id
JOIN invite_addresses ia_f ON ia_f.id = i.from_address_id
JOIN addresses a_f ON a_f.id = m.from_address_id

JOIN invite_addresses ia_t ON ia_t.id = i.to_address_id
JOIN addresses a_t ON a_t.id = m.to_address_id

WHERE i.company_id = 1040345

AND i.status = 'claimed'

【讨论】：

做到了！后续问题，如果我只想显示具有地址差异的记录，我必须使用哪种子句？有没有办法在这个条件的基础上实现这个期望的结果？
您可以将其包装在另一个 SELECT 语句中，以仅提取您想要显示的记录。将此查询更改为仅显示这些记录的问题是，如果我们更改查询返回的内容，您的 SUM 操作将会更改。如果它被包装在另一个查询中，它看起来像这样：'SELECT sub.id, sub.company_id, sub.status, sub.average_discrepancy, sub.invites_from_address, move_from_address, move_to_address, sub.mover_first_name, sub.mover_last_name, sub。 to_address_id FROM ( -- Original Query go here ) AS sub WHERE sub.base_street != sub_f.base