Get count on two different date columns and group by date - mysql

Question

I have table containing two DATE columns. TS_customer and TS_verified
I am searching for a way to get a result where in the first column I have dates where either someone created a user (TS_customer) or someone got verified (TS_verified).
In the second column I want count(TS_customer) grouped by the first column.
The third column I want count(TS_verified) grouped by the first column.
It might be 0 customers verified on a sign up date, and in another case 0 signups on a date someone got verified.
I guess it should be an easy one, but I've spent so many hours on it now. Would really appreciate some help. I need this for a graph in excel, so i basicly want how many customers signed up and how many got verified one day without having the hassle to have two selects and combinding them manually.
EDIT: link to SQLfiddle http://sqlfiddle.com/#!2/b14fc/1/0
Thanks

Answer 1

First, we need the list of days.
That looks like this http://sqlfiddle.com/#!2/b14fc/14/0:
SELECT DISTINCT days
FROM (
SELECT DISTINCT DATE(TS_customer) days
FROM customer
UNION
SELECT DISTINCT DATE(TS_verified) days
FROM customer
) AS alldays
WHERE days IS NOT NULL
ORDER BY days
Next we need a summary of customer counts by day. That's pretty easy http://sqlfiddle.com/#!2/b14fc/16/0:
SELECT DATE(TS_customer) days, COUNT(TS_customer)
FROM customer
GROUP BY days
The summary of verifications by day is similarly easy.
Next we need to join these three subqueries together http://sqlfiddle.com/#!2/b14fc/29/0.
SELECT alldays.days, custcount, verifycount
FROM (
SELECT DISTINCT DATE(TS_customer) days
FROM customer
UNION
SELECT DISTINCT DATE(TS_verified) days
FROM customer
) AS alldays
LEFT JOIN (
SELECT DATE(TS_customer) days, COUNT(TS_customer) custcount
FROM customer
GROUP BY days
) AS cust ON alldays.days = cust.days
LEFT JOIN (
SELECT DATE(TS_verified) days, COUNT(TS_verified) verifycount
FROM customer
GROUP BY days
) AS verif ON alldays.days = verif.days
WHERE alldays.days IS NOT NULL
ORDER BY alldays.days
Finally, if you want 0 displayed rather than (null) for days when there weren't any customers and/or verifications, change the SELECT line to this http://sqlfiddle.com/#!2/b14fc/30/0.
SELECT alldays.days,
IFNULL(custcount,0) AS custcount,
IFNULL(verifycount,0) AS verifycount
See how that goes? We build up your result set step by step.

Answer 2

I'm a bit confused on why you created a fiddle that can not hold null values on the TS_Customer and then mention that the field can hold null values.
Having said that, I've modified the solution to work with null values and still be pretty efficient and simple:
SELECT days, sum(custCount) custCount, sum(verifCount) verifCount FROM (
SELECT DATE(TS_customer) days, count(*) custCount, 0 verifCount
FROM customer
WHERE TS_customer IS NOT NULL
GROUP BY days
UNION ALL
SELECT DATE(TS_verified) days, 0, count(*)
FROM customer
WHERE TS_verified IS NOT NULL
GROUP BY days
) s
GROUP BY days
I've also created a different fiddle containing some null values here.