I have a table that has start_date and end_date timestamps, the rows contain data from a radius accounting table.
Shortly once a user logs in, a row is inserted with the start_date timestamp, and once the user logs off, the end_date is populated with an UPDATE.
Here are couple of rows:
ID | start_date | end_date
22 2013-11-19 12:00:22 2013-11-20 14:20:22 (*)
23 2013-11-19 12:02:22 2013-11-20 15:20:22 (*)
23 2013-11-19 17:02:22 2013-11-20 20:20:22
24 2013-11-20 12:06:22 2013-11-20 15:20:22 *
25 2013-11-20 12:40:22 2013-11-20 15:23:22 *
26 2013-11-20 12:50:22 2013-11-20 17:23:22 *
27 2013-11-20 16:56:22 2013-11-20 17:29:22
28 2013-11-20 17:58:22 2013-11-20 20:24:22
So in this case, for 2013-11-19 the max number of concurrent user is 2 (marked with (*) )(their times between start and end overlaps), for 2013-11-20 it is 3 (marked with *)
I am trying to write an SQL query to get the number of most concurrent users in a day (based on the start and end date), so a short result would be that on 2013-08-12, the most online at the same time is xx number.
I could do this in PHP by analyzing row by row, but I would like to keep it as an SQL query.
Try
select d, count(*) as user_count
from
(
select start_date as d from your_table
union all
select end_date as d from your_table
) x
group by d
order by user_count desc
You could calculate count of concurrent users on each datapoint you have (start_date / end_date) and then calculate the max out of that:
select max(cnt)
from (
select q.t, count(*) as 'cnt'
from (
select start_date as 't'
from log
where start_date between YOUR_START_DATE and YOUR_END_DATE
union
select end_date
from log
where end_date between YOUR_START_DATE and YOUR_END_DATE
) as q
join log l on q.t between l.start_date and l.end_date
group by q.t
) a
Related
This is my table, i want to find concurrent user per hour for a given week
I am trying to calculate number of concurrent users in a time range. The input looks something like the below
Table
id user_id login_time
1 23 2016-06-08 09:10:00
2 24 2016-06-08 08:55:00
3 25 2016-06-08 09:29:00
4 26 2016-06-08 09:40:00
5 27 2016-06-08 09:08:00
6 28 2016-06-09 13:40:00
7 31 2016-06-09 14:04:00
How to get the concurrent users in time range ?
Expected Output Table
Date
Hour
User
2014-08-04
0
3
2014-08-04
1
2
2014-08-04
2
0
2014-08-05
0
1
Similar question
concurrent users sql
I created a DBFIDDLE
first I entered the data from your question
half-way I changed data to what was given here: http://sqlfiddle.com/#!9/67356f/2
first the cte1 contains the first and last date from users.
cte2 contains all the dates between StartDate and EndDate
cte3 contains all (24) hours for the dates.
After this is is just counting to see if a user is logged in.
WITH RECURSIVE cte1 AS (
SELECT
DATE(MIN(login_time)) StartDate,
DATE(MAX(login_time)) EndDate
FROm users),
cte2 AS (
SELECT cte1.StartDate
from cte1
union all
select DATE_ADD(cte2.StartDate, INTERVAL 1 DAY)
from cte2
cross join cte1 where cte2.StartDate < cte1.EndDate
),
cte3 AS (
SELECT StartDate, 0 as H
FROM cte2
UNION ALL
SELECT StartDate, H+1 FROM cte3 WHERE H<24
)
select * from (
select
StartDate as `Date`,
H as `hour`,
(SELECT count(*) from users
WHERE login_time BETWEEN DATE_ADD(StartDate, interval H HOUR) AND DATE_ADD(StartDate, interval (H+1) HOUR)
) as `Count`
from cte3) x
where x.`Count` <>0
order by 1,2;
You can begin with this, but (from my opinion) it has no sense the result you are trying to get because you need to calculate the time:
If a user enters 9:30 and left 9:35 and re-enter 9:45 is not a concurrent user but you get this in the SQL.
If a user enters 9:59 and enter 10:01 you have a concurrent user but you won't see this with this logic of "hour"
Concurrent user with different day (23:59 and 00:01 logins)
In any case, the SQL you are asking for:
SQL Fiddle
SELECT
up.user_id,
up.diff as TimeDiff,
FROM
(
SELECT TIMESTAMPDIFF(HOUR,u1.login,u2.login) as diff, u1.user_id FROM users u1
JOIN users u2
ON u1.user_id = u2.user_id
AND u1.login < u2.login ) up
WHERE up.diff < 1
And without DIFF time (as you requested):
SELECT
g.id,
g.hour,
g.datelogin,
COUNT(*) as times
FROM
(SELECT HOUR(login) as hour, DATE(login) as datelogin, id FROM users) g
GROUP BY datelogin, hour, id
HAVING COUNT(*) > 1 -- This will show only counts is bigger than 1
The table provided shows all new users signing up on a specific date in the format YYYY-MM-DD.
Your query should output the change from one month to the next. Because the first month has no preceding month, your output should skip that row. Your output should look like the following table.
My table data
Table data:
ID DateJoined
1 2017-01-06
2 2017-01-12
3 2017-01-16
4 2017-01-25
5 2017-02-05
6 2017-02-07
7 2017-02-21
8 2017-03-05
9 2017-03-07
10 2017-03-14
11 2017-03-16
12 2017-03-25
13 2017-03-25
14 2017-03-25
15 2017-03-25
16 2017-03-26
17 2017-04-05
18 2017-04-14
19 2017-04-21
20 2017-05-07
23 2017-05-14
24 2017-05-16
25 2017-05-25
26 2017-05-25
27 2017-05-25
28 2017-05-25
Enter image description here
I want this output:
count all records from every month and subtract it from the next month record.
This is my query:
SELECT
MONTH(L.joindate),
COUNT(L.joindate) - COUNT(R.joindate),
MONTH(R.joindate),
COUNT(R.joindate)
FROM
userlog AS L
LEFT JOIN
userlog AS R
ON MONTH(R.joindate)= (SELECT MIN(MONTH(joindate)) FROM userlog WHERE MONTH(joindate) < MONTH(L.joindate))
GROUP BY (MONTH(L.joindate)),(MONTH(R.joindate));
Use lag(), available in MySQL 8.0:
select date_format(joindate, '%Y-%m-01') joinmonth,
count(*) - lag(count(*), 1, 0) over(order by date_format(joindate, '%Y-%m-01')) m2m
from userlog
group by joinmonth
Note that I changed the logic to truncate dates to the first of month to use date_format().
In earlier versions, you can use a correlated subquery:
select date_format(joindate, '%Y-%m-01') joinmonth,
count(*) - (
select count(*)
from userlog l1
where l1.joindate >= date_format(l.joindate, '%Y-%m-01') - interval 1 month
and l1.joindate < date_format(l.joindate, '%Y-%m-01')
) m2m
from userlog l
group by joinmonth
LIMIT 12 OFFSET 1
You need to use Lag. Also, since it says you need to skip the first row so I have used the not null condition. I believe this query should work.
select
Month,
MonthToMonthChange
from
(
select
m_name as Month,
(total_id - diff) as MonthToMonthChange
from
(
select
total_id,
m_name,
Lag(total_id, 1) OVER(
ORDER BY
m_num ASC
) AS diff
from
(
select
MonthNAME(DateJoined) m_name,
Month(DateJoined) m_num,
count(*) total_id
from
maintable
Group by
m_name,
m_num
) as first_subquery
) as second_subquery
) as final_query
where
MonthToMonthChange IS NOT NULL;
select
MONTHNAME(UL1.DateJoined) as MONTH,
count(UL1.DateJoined) -
(
select count(UL2.DateJoined)
from tablename UL2
where MONTH(UL2.DateJoined )=MONTH(UL1.DateJoined) -1
) as MonthToMonthChange
from tablename UL1
where Month(UL1.DateJoined)!=1
Group by MONTHNAME(UL1.DateJoined)
order by UL1.DateJoined ASC;
https://i.stack.imgur.com/BXXDb.png
I tried this and it worked
select date_format(DateJoined, CONCAT('%M')) as Month,
count(*) - lag(count(*), 1, 0) over(order by date_format(DateJoined, CONCAT('%m'))) MonthToMonthChange
from maintable_OKLOT
group by Month
limit 12 offset 1
I have the following query:
SELECT
DATE(`timeStamp`),COUNT(*)
FROM
`wf`.sh`
WHERE
(DATE(`timeStamp`) >= curdate()- INTERVAL 31 DAY)
GROUP BY
DATE(`timeStamp`)
HAVING
COUNT(DATE(`timeStamp`)) > 0
ORDER BY
DATE(`timeStamp`) ASC;
The purpose of this query is to retrieve the amount of users online in my system per day, in the space of a month.
Example dataset:
uID timeStamp
1 2016-11-28 00:27:01
1 2016-11-28 01:10:15
1234 2016-11-28 02:50:00
2 2016-11-28 06:11:09
47 2016-11-28 08:32:48
1246 2016-11-28 09:51:47
In its current format, this query returns the count of rows with duplicate dates, for example:
timeStamp COUNT(*)
2017-01-29 256
2017-01-30 224
2017-01-31 240
2017-02-01 95
2017-02-02 136
I have another field uID; I need to modify my query so that GROUP also ignores rows with a duplicate uID field for each day. I tried creating another GROUP BY but was given an error that 'incorrect GROUP BY clause' (or something of that nature).
Can this be done via pure MySQL?
You can use a subselect
SELECT
visitDate,COUNT(*)
FROM
(SELECT DISTINCT DATE(`timeStamp`) as visitDate, uID FROM `wf`.sh`) alias_t
WHERE
(visitDate >= curdate()- INTERVAL 31 DAY)
GROUP BY
visitDate
HAVING
COUNT(visitDate) > 0
ORDER BY
visitDate ASC;
I have an MySQL table, similar to this example:
c_id date value
66 2015-07-01 1
66 2015-07-02 777
66 2015-08-01 33
66 2015-08-20 200
66 2015-08-21 11
66 2015-09-14 202
66 2015-09-15 204
66 2015-09-16 23
66 2015-09-17 0
66 2015-09-18 231
What I need to get is count of periods where dates are in row. I don't have fixed start or end date, there can be any.
For example: 2015-07-01 - 2015-07-02 is one priod, 2015-08-01 is second period, 2015-08-20 - 2015-08-21 is third period and 2015-09-14 - 2015-09-18 as fourth period. So in this example there is four periods.
SELECT
SUM(value) as value_sum,
... as period_count
FROM my_table
WHERE cid = 66
Cant figure this out all day long.. Thx.
I don't have enough reputation to comment to the above answer.
If all you need is the NUMBER of splits, then you can simply reword your question: "How many entries have a date D, such that the date D - 1 DAY does not have an entry?"
In which case, this is all you need:
SELECT
COUNT(*) as PeriodCount
FROM
`periods`
WHERE
DATE_ADD(`date`, INTERVAL - 1 DAY) NOT IN (SELECT `date` from `periods`);
In your PHP, just select the "PeriodCount" column from the first row.
You had me working on some crazy stored procedure approach until that clarification :P
I should get deservedly flamed for this, but anyway, consider the following...
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(date DATE NOT NULL PRIMARY KEY
,value INT NOT NULL
);
INSERT INTO my_table VALUES
('2015-07-01',1),
('2015-07-02',777),
('2015-08-01',33),
('2015-08-20',200),
('2015-08-21',11),
('2015-09-14',202),
('2015-09-15',204),
('2015-09-16',23),
('2015-09-17',0),
('2015-09-18',231);
SELECT x.*
, SUM(y.value) total
FROM
( SELECT a.date start
, MIN(c.date) end
FROM my_table a
LEFT
JOIN my_table b
ON b.date = a.date - INTERVAL 1 DAY
LEFT
JOIN my_table c
ON c.date >= a.date
LEFT
JOIN my_table d
ON d.date = c.date + INTERVAL 1 DAY
WHERE b.date IS NULL
AND c.date IS NOT NULL
AND d.date IS NULL
GROUP
BY a.date
) x
JOIN my_table y
ON y.date BETWEEN x.start AND x.end
GROUP
BY x.start;
+------------+------------+-------+
| start | end | total |
+------------+------------+-------+
| 2015-07-01 | 2015-07-02 | 778 |
| 2015-08-01 | 2015-08-01 | 33 |
| 2015-08-20 | 2015-08-21 | 211 |
| 2015-09-14 | 2015-09-18 | 660 |
+------------+------------+-------+
4 rows in set (0.00 sec) -- <-- This is the number of periods
there is a simpler way of doing this, see here SQLfiddle:
SELECT min(date) start,max(date) end,sum(value) total FROM
(SELECT #i:=#i+1 i,
ROUND(Unix_timestamp(date)/(24*60*60))-#i diff,
date,value
FROM tbl, (SELECT #i:=0)n WHERE c_id=66 ORDER BY date) t
GROUP BY diff
This select groups over the same difference between sequential number and date value.
Edit
As Strawberry remarked quite rightly, there was a flaw in my apporach, when a period spans a month change or indeed a change into the next year. The unix_timestamp() function can cure this though: It returns the seconds since 1970-1-1, so by dividing this number by 24*60*60 you get the days since that particular date. The rest is simple ...
If you only need the count, as your last comment stated, you can do it even simpler:
SELECT count(distinct diff) period_count FROM
(SELECT #i:=#i+1 i,
ROUND(Unix_timestamp(date)/(24*60*60))-#i diff,
date,value
FROM tbl,(SELECT #i:=0)n WHERE c_id=66 ORDER BY date) t
Tnx. #cars10 solution worked in MySQL, but could not manage to get period count to echo in PHP. It returned 0. Got it working tnx to #jarkinstall. So my final select looks something like this:
SELECT
sum(coalesce(count_tmp,coalesce(count_reserved,0))) as sum
,(SELECT COUNT(*) FROM my_table WHERE cid='.$cid.' AND DATE_ADD(date, INTERVAL - 1 DAY) NOT IN (SELECT date from my_table WHERE cid='.$cid.' AND coalesce(count_tmp,coalesce(count_reserved,0))>0)) as periods
,count(*) as count
,(min(date)) as min_date
,(max(date)) as max_date
FROM my_table WHERE cid=66
AND coalesce(count_tmp,coalesce(count_reserved,0))>0
ORDER BY date;
Let's say I have a table that says how many items of something are valid between two dates.
Additionally, there may be multiple such periods.
For example, given a table:
itemtype | count | start | end
A | 10 | 2014-01-01 | 2014-01-10
A | 10 | 2014-01-05 | 2014-01-08
This means that there are 10 items of type A valid 2014-01-01 - 2014-01-10 and additionally, there are 10 valid 2014-01-05 - 2014-01-08.
So for example, the sum of valid items at 2014-01-06 are 20.
How can I query the table to get the sum per day? I would like a result such as
2014-01-01 10
2014-01-02 10
2014-01-03 10
2014-01-04 10
2014-01-05 20
2014-01-06 20
2014-01-07 20
2014-01-08 20
2014-01-09 10
2014-01-10 10
Can this be done with SQL? Either Oracle or MySQL would be fine
The basic syntax you are looking for is as follows:
For my example below I've defined a new table called DateTimePeriods which has a column for StartDate and EndDate both of which are DATE columns.
SELECT
SUM(NumericColumnName)
, DateTimePeriods.StartDate
, DateTimePeriods.EndDate
FROM
TableName
INNER JOIN DateTimePeriods ON TableName.dateColumnName BETWEEN DateTimePeriods.StartDate and DateTimePeriods.EndDate
GROUP BY
DateTimePeriods.StartDate
, DateTimePeriods.EndDate
Obviously the above code won't work on your database but should give you a reasonable place to start. You should look into GROUP BY and Aggregate Functions. I'm also not certain of how universal BETWEEN is for each database type, but you could do it using other comparisons such as <= and >=.
There are several ways to go about this. First, you need a list of dense dates to query. Using a row generator statement can provide that:
select date '2014-01-01' + level -1 d
from dual
connect by level <= 15;
Then for each date, select the sum of inventory:
with
sample_data as
(select 'A' itemtype, 10 item_count, date '2014-01-01' start_date, date '2014-01-10' end_date from dual union all
select 'A', 10, date '2014-01-05', date '2014-01-08' from dual),
periods as (select date '2014-01-01' + level -1 d from dual connect by level <= 15)
select
periods.d,
(select sum(item_count) from sample_data where periods.d between start_date and end_date) available
from periods
where periods.d = date '2014-01-06';
You would need to dynamically set the number of date rows to generate.
If you only needed a single row, then a query like this would work:
with
sample_data as
(select 'A' itemtype, 10 item_count, date '2014-01-01' start_date, date '2014-01-10' end_date from dual union all
select 'A', 10, date '2014-01-05', date '2014-01-08' from dual)
select sum(item_count)
from sample_data
where date '2014-01-06' between start_date and end_date;