使用自引用查询更新mysql

我有一份调查表，其中包括（除其他外）以下各栏

survey_id  - unique id
user_id    - the id of the person the survey relates to
created    - datetime
ip_address - of the submission
ip_count   - the number of duplicates

由于记录集很大，动态运行此查询是不切实际的，因此尝试创建更新语句，定期将“缓存”结果存储在ip_计数中

ip_计数的目的是显示同一用户id在12个月内（创建日期+/-6个月）收到的重复ip_地址调查提交的数量

使用以下数据集，这是预期的结果

survey_id   user_id    created    ip_address     ip_count  #counted duplicates survey_id
  1            1      01-Jan-12   123.132.123       1      # 2
  2            1      01-Apr-12   123.132.123       2      # 1, 3
  3            2      01-Jul-12   123.132.123       0      # 
  4            1      01-Aug-12   123.132.123       3      # 2, 6
  6            1      01-Dec-12   123.132.123       1      # 4

这是迄今为止我提出的最接近的解决方案，但是这个查询没有考虑到日期限制，并且很难找到替代方法

UPDATE surveys
JOIN(
  SELECT ip_address, created, user_id, COUNT(*) AS total
  FROM surveys  
  WHERE surveys.state IN (1, 3) # survey is marked as completed and confirmed
  GROUP BY ip_address, user_id
) AS ipCount 
  ON (
    ipCount.ip_address = surveys.ip_address
    AND ipCount.user_id = surveys.user_id
    AND ipCount.created BETWEEN (surveys.created - INTERVAL 6 MONTH) AND (surveys.created + INTERVAL 6 MONTH)
  )
SET surveys.ip_count = ipCount.total - 1 # minus 1 as this query will match on its own id.
WHERE surveys.ip_address IS NOT NULL # ignore surveys where we have no ip_address

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提前感谢您的帮助：）

我没有您的表，因此我很难形成正确的sql，但我可以尝试一下，希望能够帮助您

首先，我需要将调查的笛卡尔乘积与自身进行对比，并过滤掉我不想要的行

select s1.survey_id x, s2.survey_id y from surveys s1, surveys s2 where s1.survey_id != s2.survey_id and s1.ip_address = s2.ip_address and (s1.created and s2.created fall 6 months within each other)

此项的输出应包含匹配的每对调查（根据您的规则）两次（第一个位置的每个id一次，第二个位置的每个id一次）

然后我们可以对这个输出进行

分组

，得到一个表，该表基本上为每个调查id提供了正确的ip计数

(select x, count(*) c from (select s1.survey_id x, s2.survey_id y from surveys s1, surveys s2 where s1.survey_id != s2.survey_id and s1.ip_address = s2.ip_address and (s1.created and s2.created fall 6 months within each other)) group by x)

现在我们有了一个表，将每个调查id映射到其正确的ip计数。要更新原始表，我们需要将其与此关联，并将值复制到

所以看起来应该是这样的

UPDATE surveys SET s.ip_count = n.c from surveys s inner join (ABOVE QUERY) n on s.survey_id = n.x

这里有一些伪代码，但我认为总体思路应该是可行的

我以前从未根据另一个查询的输出更新过表。。试图从这个问题中猜出正确的语法-

另外，如果我需要为自己的工作做类似的事情，我不会试图在一个查询中完成。。这将是一个难以维护的问题，并且可能存在内存/性能问题。最好让脚本逐行遍历表，在事务中的一行上更新，然后再移动到下一行。速度要慢得多，但更容易理解，而且数据库可能更轻松。

对上面显示的内容进行了一些（非常）小的调整。再次感谢你

UPDATE surveys AS s
INNER JOIN (
  SELECT x, count(*) c
  FROM (
    SELECT s1.id AS x, s2.id AS y
    FROM surveys AS s1, surveys AS s2
    WHERE s1.state IN (1, 3) # completed and verified
      AND s1.id != s2.id # dont self join
      AND s1.ip_address != "" AND s1.ip_address IS NOT NULL # not interested in blank entries
      AND s1.ip_address = s2.ip_address
      AND (s2.created BETWEEN (s1.created - INTERVAL 6 MONTH) AND (s1.created + INTERVAL 6 MONTH))
      AND s1.user_id = s2.user_id # where completed for the same user
  ) AS ipCount
  GROUP BY x
) n on s.id = n.x
SET s.ip_count = n.c

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非常感谢你！在我看到这个答案之前，我开始摆弄临时桌子等。请参阅下面最后完成的查询，了解处于类似情况下的其他人。