检查重复的记录

Question

的状态可以说，我们有4个领域

id (INT 11 AUTO_INC) 

email (VAR 50) 

timestamp (INT 11) 

status (INT 1) 

命名record表和表包含以下数据

现在我们可以看到的电子邮件地址test@xample.com被复制了4次（时间戳最低的记录是原始记录，之后的所有副本都是重复记录）。多少次使用

SELECT email, count(id) FROM record GROUP BY email HAVING COUNT(id)>1 

我可以很容易地计算使用

SELECT COUNT(DISTINCT email) FROM record 

了独特的记录数我也可以很容易找到的电子邮件地址被复制但现在的商业问题是

如何所有重复记录中有多次STATUS是1？

例如：

• 对于test@example.com有具有状态1
• 没有重复记录对于second@example.com有具有1个重复的记录状态1
• 对于第三@ example.com有1个重复的记录，状态为1
• 对于four@example.com，没有重复记录状态为1
• 对于five@example.com，有2个重复记录状态为1

因此，所有的数字之和是0 + 1 + 1 + 0 + 2 = 4

这意味着有其中有status = 1在表4点中重复的记录

问题

多少重复的记录有状态= 1？

Answer 1

这是一个更好的新解决方案。它会删除每封电子邮件的第一个条目，然后计算其余条目。阅读起来并不容易，如果可能的话，我会在存储过程中编写它，但这是有效的。

select sum(status) 
    from dude d1 
    join (select email, 
       min(ts) as ts 
      from dude 
     group by email) mins 
using (email) 
where d1.ts != mins.ts; 

sqlfiddle

下面

自己的查询找到原来的答案 “的电子邮件地址被重复多少次使用”

SELECT email, 
     count(id) as duplicates 
    FROM record 
GROUP BY email 
HAVING COUNT(id)>1 

可以很容易地修改答案“有多少条重复记录的状态= 1”

SELECT email, 
     count(id) as duplicates_status_sum 
    FROM record 
GROUP BY email 
WHERE status = 1 
HAVING COUNT(id)>1 

这两个查询都会回答，包括原始行，所以它实际上是“包括原始行的重复”。您可以从金额减去1，如果原来总是有状态1.

SELECT email, 
     count(id) -1 as true_duplicates 
    FROM record 
GROUP BY email 
HAVING COUNT(id)>1 

SELECT email, 
     count(id) -1 as true_duplicates_status_sum 
    FROM record 
GROUP BY email 
WHERE status = 1 
HAVING COUNT(id)>1 

Answer 2

如果我的理解没有错的，然后你的查询应该是

SELECT `email` , COUNT( `id`) AS `tot` 
FROM `record` , (
SELECT `email` AS `emt` , MIN( `timestamp`) AS `mtm` 
FROM `record` 
GROUP BY `email` 
) AS `temp` 
WHERE `email` = `emt` 
AND `timestamp` > `mtm` 
AND `status` =1 
GROUP BY `email` 
HAVING COUNT( `id`) >=1 

首先，我们需要得到最小的时间戳然后发现如果你想总和那些这个时间戳和有状态1.

后插入重复记录则查询

SELECT SUM( `tot`) AS `duplicatesWithStatus1` 
FROM (
SELECT `email` , COUNT( `id`) AS `tot` 
FROM `record` , (
SELECT `email` AS `emt` , MIN( `timestamp`) AS `mtm` 
FROM `record` 
GROUP BY `email` 
) AS `temp` 
WHERE `email` = `emt` 
AND `timestamp` > `mtm` 
AND `status` =1 
GROUP BY `email` 
HAVING COUNT( `id`) >=1 
) AS t 

希望这是你想要的

Answer 3

你可以得到重复记录的计数有状态= 1通过

select count(*) as Duplicate_Record_Count 
from (select * 
from record r 
where r.status=1 
group by r.email,r.status 
having count(r.email)>1) t1 

下面的查询将返回重复的电子邮件与状态1的计数和时间戳

select r.email,count(*)-1 as Duplicate_Count,min(r.timestamp) as timestamp 
from record r 
where r.status=1 
group by r.email 
having count(r.email)>1