Check Status of the Duplicate Records - mysql

Question

Lets say we have a table named record with 4 fields
id (INT 11 AUTO_INC)
email (VAR 50)
timestamp (INT 11)
status (INT 1)
And the table contains following data
Now we can see that the email address test#xample.com was duplicated 4 times (the record with the lowest timestamp is the original one and all copies after that are duplicates). I can easily count the number of unique records using
SELECT COUNT(DISTINCT email) FROM record
I can also easily find out which email address was duplicated how many times using
SELECT email, count(id) FROM record GROUP BY email HAVING COUNT(id)>1
But now the business question is
How many times STATUS was 1 on all the Duplicate Records?
For example:
For test#example.com there was no duplicate record having status 1
For second#example.com there was 1 duplicate record having status 1
For third#example.com there was 1 duplicate record having status 1
For four#example.com there was no duplicate record having status 1
For five#example.com there were 2 duplicate record having status 1
So the sum of all the numbers is 0 + 1 + 1 + 0 + 2 = 4
Which means there were 4 Duplicate records which had status = 1 In table
Question
How many Duplicate records have status = 1 ?

Answer 1

This is a new solution that works better. It removes the first entry for each email and then counts the rest. It's not easy to read, if possible I would write this in a stored procedure but this works.
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
original answer below
Your own query to find "which email address was duplicated how many times using"
SELECT email,
count(id) as duplicates
FROM record
GROUP BY email
HAVING COUNT(id)>1
can easily be modified to answer "How many Duplicate records have status = 1"
SELECT email,
count(id) as duplicates_status_sum
FROM record
GROUP BY email
WHERE status = 1
HAVING COUNT(id)>1
Both these queries will answer including the original line so it's actually "duplicates including the original one". You can subtract 1 from the sums if the original one always have status 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

If I am not wrong in understanding then your query should be
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
First we need to get the minimum timestamp and then find duplicate records that are inserted after this timestamp and having status 1.
If you want the total sum then the query is
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
Hope this is what you want

Answer 3

You can get the count of Duplicate records have status = 1 by
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
The following query will return the duplicate email with status 1 count and timestamp
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