I have such a schema and queries:
http://sqlfiddle.com/#!2/7b032/3
Seperately I have these queries:
SELECT COUNT(*) AS 'times', userid, name
FROM main
WHERE comedate <= DATE_SUB(CURDATE(),
INTERVAL 5 DAY)
GROUP BY userid ORDER BY times DESC LIMIT 0,2;
SELECT * FROM details WHERE 1;
By comparing userid columns of both table I need to join them.
I need an output having these columns:
"times, userid, name, age, location"
Also order, group and limits should be considered.
I would be happy if you can write one query with JOIN and one query with subquery.
I have a 60k table and I will compare the performances.
How about this:
select x.times,
x.userid,
x.name,
d.age,
d.location
from
(
SELECT COUNT(*) AS 'times', userid, name
FROM main
WHERE comedate <= DATE_SUB(CURDATE(),
INTERVAL 5 DAY)
GROUP BY userid
) x
left join details d
on x.userid = d.userid
see SQL Fiddle with Demo
edit:
select x.times,
x.userid,
x.name,
d.age,
d.location
from
(
SELECT COUNT(*) AS 'times', userid, name
FROM main
WHERE comedate <= DATE_SUB(CURDATE(),
INTERVAL 5 DAY)
GROUP BY userid
ORDER BY times DESC
LIMIT 0,2
) x
left join details d
on x.userid = d.userid
see SQL Fiddle with demo
Related
How can I combine my two queries in MySQL 5.7:
(windows functions doesn't work)
1: This query is finding all dialogues where more than 10 messages and spaced at least an hour apart.
SELECT `dialog_id`
FROM `messages`
GROUP BY `dialog_id`
HAVING COUNT(*) >= 10
AND MIN(`timestamp`) + INTERVAL 1 HOUR < MAX(`timestamp`)
2: The second query selects two last rows for each dialogue.
SELECT * FROM messages tbl WHERE
(SELECT COUNT(*) FROM messages tbl1 WHERE tbl1.dialog_id = tbl.dialog_id AND tbl1.id >= tbl.id AND
(user_from = :user_from OR user_to = :user_to)) <= 2 ORDER BY dialog_id DESC
So, what I want is to select last two rows for each dialogue that lasted more than an hour and where more than 10 messages were sent.
Though I didn't got you properly is this what you are looking for:
SELECT * FROM messages tbl WHERE
(SELECT COUNT(*) FROM messages tbl1 WHERE tbl1.dialog_id = tbl.dialog_id AND tbl1.id >= tbl.id AND
(user_from = :user_from OR user_to = :user_to)) <= 2
and dialog_id in (SELECT `dialog_id`
FROM `messages`
GROUP BY `dialog_id`
HAVING COUNT(*) >= 10
AND MIN(`timestamp`) + INTERVAL 1 HOUR < MAX(`timestamp`))
ORDER BY dialog_id DESC
One way you can accomplish this with a subquery:
SELECT *
FROM messages tbl
WHERE (
SELECT COUNT(*)
messages tbl1
tbl1.dialog_id = tbl.dialog_id
tbl1.id >= tbl.id
(user_from = :user_from OR user_to = :user_to)) <= 2
and dialog_id in (
SELECT `dialog_id`
FROM `messages`
GROUP BY `dialog_id`
HAVING COUNT(*) >= 10
AND MIN(`timestamp`) + INTERVAL 1 HOUR < MAX(`timestamp`))
ORDER BY dialog_id DESC
You might have to adjust it slightly, since I don't have the full structure of your tables, but the principle would go as follows: get all the dialogue_ids matching your 10 and over an hour criteria, then use it to limit the messages returned from your "Get the most recent two" logic.
Could you help me to calculate percent of users, which made payments?
I've got two tables:
activity
user_id login_time
201 01.01.2017
202 01.01.2017
255 04.01.2017
255 05.01.2017
256 05.01.2017
260 15.03.2017
2
payments
user_id payment_date
200 01.01.2017
202 01.01.2017
255 05.01.2017
I try to use this query, but it calculates wrong percent:
SELECT activity.login_time, (select COUNT(distinct payments.user_id)
from payments where payments.payment_time between '2017-01-01' and
'2017-01-05') / COUNT(distinct activity.user_id) * 100
AS percent
FROM payments INNER JOIN activity ON
activity.user_id = payments.user_id and activity.login_time between
'2017-01-01' and '2017-01-05'
GROUP BY activity.login_time;
I need a result
01.01.2017 100 %
02.01.2017 0%
03.01.2017 0%
04.01.2017 0%
05.01.2017 - 50%
If you want the ratio of users who have made payments to those with activity, just summarize each table individually:
select p.cnt / a.cnt
from (select count(distinct user_id) as cnt from activity a) a cross join
(select count(distinct user_id) as cnt from payment) p;
EDIT:
You need a table with all dates in the range. That is the biggest problem.
Then I would recommend:
SELECT d.dte,
( ( SELECT COUNT(DISTINCT p.user_id)
FROM payments p
WHERE p.payment_date >= d.dte and p.payment_date < d.dte + INTERVAL 1 DAY
) /
NULLIF( (SELECT COUNT(DISTINCT a.user_id)
FROM activity a
WHERE a.login_time >= d.dte and p.login_time < d.dte + INTERVAL 1 DAY
), 0
) as ratio
FROM (SELECT date('2017-01-01') dte UNION ALL
SELECT date('2017-01-02') dte UNION ALL
SELECT date('2017-01-03') dte UNION ALL
SELECT date('2017-01-04') dte UNION ALL
SELECT date('2017-01-05') dte
) d;
Notes:
This returns NULL on days where there is no activity. That makes more sense to me than 0.
This uses logic on the dates that works for both dates and date/time values.
The logic for dates can make use of an index, which can be important for this type of query.
I don't recommend using LEFT JOINs. That will multiply the data which can make the query expensive.
First you need a table with all days in the range. Since the range is small you can build an ad hoc derived table using UNION ALL. Then left join the payments and activities. Group by the day and calculate the percentage using the count()s.
SELECT x.day,
concat(CASE count(DISTINCT a.user_id)
WHEN 0 THEN
1
ELSE
count(DISTINCT p.user_id)
/
count(DISTINCT a.user_id)
END
*
100,
'%')
FROM (SELECT cast('2017-01-01' AS date) day
UNION ALL
SELECT cast('2017-01-02' AS date) day
UNION ALL
SELECT cast('2017-01-03' AS date) day
UNION ALL
SELECT cast('2017-01-04' AS date) day
UNION ALL
SELECT cast('2017-01-05' AS date) day) x
LEFT JOIN payments p
ON p.payment_date = x.day
LEFT JOIN activity a
ON a.login_time = x.day
GROUP BY x.day;
I have a food selling website in which there is order table which record the order of every user.It column for user id ,user name,orderid ,timestamp of order.I want to know the maximum number of order that has been made in any one hour span through out the day.Give me any formula for this,or any algorithm or any sql queries for these.
SQL server:
with CTE as
(
select cast(t1.timestamp as date) as o_date, datepart(hh, t1.timestamp) as o_hour, count(*) as orders
from MyTable t1
group by cast(t1.timestamp as date), datepart(hh, t1.timestamp)
)
select o_date, o_hour, orders
from CTE
where orders = (select max(orders) from CTE)
Oracle
with CTE as
(
select to_char(t1.timestamp, 'YYYYMMDD') as o_date, to_char(t1.timestamp, 'HH24') as o_hour, count(*)
from MyTable t1
group by to_char(t1.timestamp, 'YYYYMMDD'), to_char(t1.timestamp, 'HH24')
)
select o_date, o_hour, orders
from CTE
where orders = (select max(orders) from CTE)
You can get count by day and hour like this
For SQL
SELECT TOP 1
COUNT(*)
FROM myTable
GROUP BY DATEPART(day, [column_date]), DATEPART(hour, [column_date])
ORDER BY COUNT(*) DESC;
For MySQL
SELECT
COUNT(*)
FROM myTable
GROUP BY HOUR(column_date), DAY(column_date)
ORDER BY COUNT(*) DESC
LIMIT 1;
I have MySQL queries both of which work fine independantly which I would like to combine together so I get three values returned.
Query 1 checks how many accounts have been deleted:
SELECT
COUNT(1) AS deleted_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date
FROM
exit_reasons e
WHERE
e.timestamp>='$sixmonths'
GROUP BY
WEEKOFYEAR(e.timestamp)
ORDER BY
display_date ASC
LIMIT 26
This returns a date and the number who deleted in that week
Query 2 checks how many of these have subsequently signed up again:
SELECT
COUNT(1) AS date_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date
FROM
exit_reasons e
LEFT JOIN
companies c on e.email=c.email
WHERE
e.timestamp>='$sixmonths' AND c.email IS NOT NULL
GROUP BY
WEEKOFYEAR(e.timestamp)
ORDER BY
display_date ASC
LIMIT 26
This returns a date and the number of that weeks deleted who now have a new account
I would like it to return a date and then the number deleted and number rejoined in one query so I tried:
SELECT
COUNT(1) AS date_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date,
date_count as rejoined_count from
(SELECT
COUNT(1) AS date_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date
FROM
exit_reasons e2
LEFT JOIN
companies c on e.email=c.email
LEFT JOIN
companies_users cu on e.email=cu.email
WHERE
e2.timestamp>='$sixmonths' AND c.email IS NOT NULL
GROUP BY
WEEKOFYEAR(e.timestamp)
ORDER BY
display_date ASC
LIMIT 26)
FROM
exit_reasons e
WHERE
e.timestamp>='$sixmonths'
GROUP BY
WEEKOFYEAR(e.timestamp)
ORDER BY
display_date ASC
LIMIT 26
but I am getting a syntax error - how can I combine these queries together into one query?
You should be able to combine the two queries into a single query by using an aggregate function along with some conditional logic like a CASE expression:
SELECT
COUNT(1) AS deleted_count,
SUM(CASE WHEN c.email IS NOT NULL THEN 1 ELSE 0 END) as date_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date
FROM exit_reasons e
LEFT JOIN companies c
on e.email=c.email
WHERE e.timestamp>='$sixmonths'
GROUP BY WEEKOFYEAR(e.timestamp)
ORDER BY display_date ASC
LIMIT 26;
See Demo. Your check on the second query if the c.email IS NOT NULL is moved into the SUM(CASE.. which allows you to get a total of the rows that are not null.
I think the following will do what you want:
SELECT COUNT(*) AS deleted_count,
COUNT(c.email) as date_count,
SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date
FROM exit_reasons e LEFT JOIN
companies c
on e.email = c.email
WHERE e.timestamp >= '$sixmonths'
GROUP BY WEEKOFYEAR(e.timestamp)
ORDER BY display_date ASC
LIMIT 26;
In the event that someone can sign up more than once with the same email, you should change the count() to use distinct:
COUNT(DISTINCT e.email) as deleted_count,
COUNT(DISTINCT c.email) as date_count
I have a pretty flat table - tbl_values which has userids as well as netAmounts in a given row. In the example below, 2280 has no records in the past 30 days based on the timestamp.
I'd expect this to return 3 rows, with 2280 as "0" - but I'm only getting 2 back? Am I missing something obvious here?
SELECT userid, (COALESCE(SUM(netAmount),0)) as Sum FROM `tbl_values` where userid in (2280, 399, 2282) and date > (select DATE_SUB(NOW(), INTERVAL 30 day)) GROUP BY userid
Assuming you always want to return the user, regardless of rather they have a matching record in tbl_values, what you're looking for is an outer join:
SELECT u.userid, COALESCE(SUM(v.netAmount),0) as Sum
FROM (
SELECT 2280 userid UNION ALL
SELECT 399 UNION ALL
SELECT 2282
) u
LEFT JOIN `tbl_values` v ON u.userid = v.userid AND
v.date > DATE_SUB(NOW(), INTERVAL 30 day)
GROUP BY u.userid
If you perhaps have a Users table, then you can use it instead of the subquery.
SELECT u.userid, COALESCE(SUM(v.netAmount),0) as Sum
FROM users u
LEFT JOIN `tbl_values` v ON u.userid = v.userid AND
v.date > DATE_SUB(NOW(), INTERVAL 30 day)
WHERE u.userid in (2280, 399, 2282)
GROUP BY u.userid
This is your query:
SELECT userid, (COALESCE(SUM(netAmount),0)) as Sum
FROM `tbl_values`
where userid in (2280, 399, 2282) and
date > (select DATE_SUB(NOW(), INTERVAL 30 day))
GROUP BY userid;
The filter in the where clause finds no rows that match for user id 2280. Assuming that at least one row exists somewhere, you can get what you want by moving the date comparison to a conditional aggregation:
SELECT userid,
sum(case when date > DATE_SUB(NOW(), INTERVAL 30 day)
then netAmount else 0
end) as Sum
FROM `tbl_values`
WHERE userid in (2280, 399, 2282)
GROUP BY userid;
EDIT:
If you really want all three results, then use a left join:
SELECT u.userid,
coalesce(sum(netAmount), 0) as Sum
FROM (select 2280 as userid union all
select 399 union all
select 2282
) u left join
tbl_values t
on u.userid = t.userid and
t.date > DATE_SUB(NOW(), INTERVAL 30 day)
GROUP BY u.userid;