I have table shown below :
login
date user
2016-11-23 1
2016-11-23 2
2016-11-23 3
2016-11-25 2
2016-11-25 5
2016-11-27 1
from above table what I want to get is like this:
date count(*)
2016-11-21 0
2016-11-22 0
2016-11-23 3
2016-11-24 0
2016-11-25 2
2016-11-26 0
2016-11-27 1
But, because there are only dates 2016-11-23 and 2016-11-25 and 2016-11-27, when I query like this :
select date, count(*)
from login
where date between (current_date()-interval 7 day) and current_date()
group by date
order by date asc
It can't get result like what I really want to get. Is that result possible from my login table?
One way is to generate all days before JOIN
select GenDate, count(Date)
from login
right join
(select a.GenDate
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as GenDate
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a
where a.GenDate between (current_date()-interval 7 day) and current_date())x
ON x.GenDate=login.Date
group by GenDate
order by GenDate asc
Use a derived table with the wanted dates :
SELECT t.date, count(s.date)
FROM (SELECT '2016-11-21' as `date` UNION ALL
SELECT '2016-11-22' as `date` UNION ALL
...) t
LEFT JOIN login s
ON(t.date = s.date)
WHERE
t.date between (current_date()-interval 7 day) and current_date()
GROUP BY t.date
ORDER BY t.date
This is a very well known problem in programming. There are several solutions.
Go over the result with PHP, and fill the missing days in the resulting array.
AS sagi proposed, create a separate table that contains all the dates in the range of days your application works with, then you can JOIN that table with your query. One of the issues is that from time to time you have to add more days to this table, if you suddenly have missing days in future or in past.
Related
Im running an Openhab2 instance and collect data aswell as timestamps for important changes in datacollection. Inside my table item45 I store 2 colums Time and Value witch looks like this.
TABLE (item45) Design Screenshot 1 Screenshot 2
Time (datetime,primary) Value (datetime)
... ...
2018-10-17 03:08:30 2018-10-17 03:08:30
2018-10-19 00:13:13 2018-10-19 00:13:13
2018-10-19 00:27:58 2018-10-19 00:27:57
Its kind of the design of how Openhab stores data so nothing i can do about that.
i now try to use these values in Grafana just as ones like this:
Example of sucessful use
My Problem lies within Grafana im using to plot the querys. It disconnects the graphs if the next older point is outside the viewport. I tried to cope with that with virutally filling up the gaps in data inside the query. Like this:
Currently used Query
SELECT
1 as value,
'Net Reset' as metric,
UNIX_TIMESTAMP(v.gen_date) AS time_sec
from
(select DATE_SUB( FROM_UNIXTIME(1539707286), INTERVAL t3*1000 + t2*100 + t1*10 + t0 HOUR) gen_date from
(select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0
CROSS JOIN (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1
CROSS JOIN (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2
CROSS JOIN (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3
) v
LEFT JOIN item45
ON DATE_FORMAT( Time, '%Y-%m-%dT%H:00:00') = v.gen_date
where $__timeFilter(v.gen_date)
ORDER BY v.gen_date DESC;
AS u can see im using one of the given "hacks" here to generate hourly fillingdata just set to ones.
I need help to implement this query:
Query to replace FROM_UNIXTIME(...)
SELECT * FROM item45 ORDER BY Time DESC LIMIT 1
into the query 1 where FROM_UNIXTIME(1539707286) to set the actual stop of filling up the future data.
Im running MySQL57-server and currently cant upgrade to version 8 to use the with clause.
Desired Output
Time Value
2018-10-18 21:00:00 1 <- Inserted
2018-10-18 22:00:00 1 <- Inserted
2018-10-18 23:00:00 1 <- Inserted
2018-10-19 00:00:00 1 <- Inserted
2018-10-19 00:13:13 1
2018-10-19 00:27:58 1
I have a table with reservations in it. Each row is a reservation and has a start & end datetime field.
I want to construct a query which gives me the count of reservations on each day in a certain time interval, eg april 2018.
Selecting all the reservations within the given interval is fairly simple:
SELECT * FROM reservation
WHERE start <= '2018-05-01 00:00:00'
AND end >= '2018-04-01 00:00:00'
But then the 'trouble' starts.
I want to display a 'count' of reservations on each day in the interval. But a reservation could span multiple days. So grouping them on DAY(start) is not correct.
I don't want to query each day in the interval seperately as this would be very server-intensive.
Is there a way to do this through a MySQL query?
Sample data:
id | start | end
2 | 2018-04-01 12:00:00 | 2018-04-03 09:00:00
3 | 2018-04-01 09:00:00 | 2018-04-01 11:00:00
4 | 2018-04-06 13:00:00 | 2018-05-20 09:00:00
Result for 2018-04-01 to 2018-04-06:
2018-04-01 | 2 (2/3)
2018-04-02 | 1 (2)
2018-04-03 | 1 (2)
2018-04-04 | 0
2018-04-05 | 0
2018-04-06 | 1 (4)
in a sqlfiddle: http://sqlfiddle.com/#!9/e62ffa/2/0
First we will reuse the answer from DBA StackExchange. (You can use the accepted answer if you want, you would just need to create a dedicated table for that).
We will just modify the query a bit by using the condition that you need.
Your condition:
SELECT * FROM reservation
WHERE start <= '2018-05-01 00:00:00'
AND end >= '2018-04-01 00:00:00'
Modified answer from DBA Stackexchange:
SELECT date_field
FROM
(
SELECT
MAKEDATE(YEAR(NOW()),1) +
INTERVAL (MONTH(NOW())-1) MONTH +
INTERVAL daynum DAY date_field
FROM
(
SELECT t * 10 + u daynum
FROM
(SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) B
ORDER BY daynum
) AA
) AAA
/*WHERE MONTH(date_field) = MONTH(NOW())*/
WHERE date_field BETWEEN '2018-04-01' AND '2018-05-01'
Take note that I only changed the WHERE Clause.
Now using that query as a DERIVED TABLE, we will include your Reservations table using LEFT JOIN.
SELECT D.date_field
, COUNT(R.Id)
FROM (
/* The query from above goes here */
) D
LEFT JOIN Reservations R ON D.date_field BETWEEN DATE(R.StartDate) AND DATE(R.EndDate)
GROUP BY D.date_field
Notice again that we used the DATE function to truncate the TIME part of our StartDate and EndDate because for example, 2018-04-01 denotes the whole day and it cannot be in between 2018-04-01 09:00:00 and 2018-04-01 11:00:00 for some under the hood reason I am not completely familiar of.
Here is a SQL Fiddle Demo of the result.
If someone could help me on this one. SELECT '2018-04-02' BETWEEN '2018-04-01 23:59:59' AND '2018-04-02 00:00:00' will result to 1 (TRUE). It seems that by default DATE will have a TIMESTAMP of 00:00:00.
Update for More Flexible Date Range (2018-04-11)
The query above from DBA StackExchange only lists down the days of the current month. I tried to search a bit and found this another good answer here in StackOverflow. Here is a part of the query:
SELECT CURDATE() - INTERVAL (A.A+ (10 * B.A)) DAY AS Date
FROM (
SELECT 0 AS A UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) AS A
CROSS JOIN (
SELECT 0 AS A UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) AS B
The query above will generate numbers (1 to 100) using CROSS JOIN and then subtracting it to the Current Date, then you will have dates from now up to 100 days back. You can add another CROSS JOIN of numbers to generate 1000 numbers if necessary.
I assume you will have StartDate and EndDate in your stored procedure or somewhere. We can replace the CURDATE with EndDate and then we will have 100 days back up to our EndDate. We will just add a WHERE clause to filter only the dates that we need using subquery/derived table.
SELECT D.Date
FROM (
SELECT CURDATE() - INTERVAL (A.A+ (10 * B.A)) DAY AS Date
FROM (
SELECT 0 AS A UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) AS A
CROSS JOIN (
SELECT 0 AS A UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) AS B
) AS D
WHERE D.Date BETWEEN #startDate AND #endDate
We can now use LEFT JOIN to include the Reservations table.
Here is another SQL Fiddle Demo for that. This also includes the Start and End Date variables, and a sample date range spanning from a previous year to the current year.
Again if you need more than 100 days of range, we will just need to add another CROSS JOIN of numbers, let's name that as C:
CROSS JOIN (
SELECT 0 AS A UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) AS C
And then add it to the calculation of past days in the SELECT statement.
SELECT CURDATE() - INTERVAL (A.A + (10 * B.A) + (100 * C.A)) DAY AS Date
So, where I work, my employees work alternating Saturdays: some employees are assigned to work the 1st and 3rd Saturdays of each month, some are assigned to work the 2nd and 4th Saturdays of each month.
One small problem arises: there are four months in the year which have five Saturdays. Which is easy enough to work around: 1st/3rd Employees work the 1st and 3rd Fifth Saturdays in the year and so on.
Some years have five months with fifth Saturdays, but we're not talking about that right now.
Anyway, to generate schedules for my employees, I first use the following code to generate a list of dates:
select
curdate() - interval (a.a + (10 * b.a) + (100 * c.a)) day as Date
from (
select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) as a
Then, I select the dates from the subquery, and join them to my Scheduling table to create a list of what each employee works on each date.
All of this works fine until Fifth Saturdays rear their ugly heads. At the moment, I use the following code to mathematically figure out which fifth saturday of the year it is:
mod(((datediff(date_add(a.date, interval (7-dayofweek(a.date)) day),date_add(subdate(a.Date,dayofyear(a.date-1)), interval (1-dayofweek(subdate(a.Date,dayofyear(a.date-1)))) day))+1)/7),4)
Or, to summarize, I find the number of days that have elapsed from the current first Sunday of the year to the current Saturday of this week, divide it by seven to figure out how many Saturdays have passed since then, and then find the modulo of 4 to figure out how many "extra" fifth Saturdays have passed since then.
...and this code will work perfectly fine unless the Fifth Saturday happens to fall on a week which is a multiple of four. Thankfully, that doesn't happen at all this year, but next year I'll need to figure out how to deal with this.
Is there a better way to mathematically figure out which fifth Saturday of the month a given fifth Saturday is?
Oh man. I was way overthinking the SQL piece of this. It was actually pretty easy. Create three temporary tables: one to hold the completed date table, one to hold all of the dates of interest in your period, and a third table specifically to find Fifth Saturdays:
drop temporary table if exists DATES;
create temporary table DATES (AdherenceDates date, WhichSaturday int,
WhichFifthSaturday int);
create temporary table DATES1 (AdherenceDates date, WhichSaturday int);
create temporary table DATES2 (FifthSaturdayDates date, WhichFifthSaturday
int auto_increment, primary key (WhichFifthSaturday));
insert into DATES1 (AdherenceDates, WhichSaturday)
select
a.Date as AdherenceDates,
case
when day(adddate(a.date,(7-dayofweek(a.date)))) <= 7
then 1
when day(adddate(a.date,(7-dayofweek(a.date)))) <= 14
then 2
when day(adddate(a.date,(7-dayofweek(a.date)))) <= 21
then 3
when day(adddate(a.date,(7-dayofweek(a.date)))) <= 28
then 4
when day(adddate(a.date,(7-dayofweek(a.date)))) > 28
then 5
end as WhichSaturday
from
(select
curdate() - interval (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a)) day
as Date
from (
select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all select 7
union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as d
) as a
where dayofweek(a.Date) <> 1
and a.Date >= '2017-01-01'
order by a.Date asc
;
insert into DATES2 (FifthSaturdayDates)
select
a.Date as AdherenceDates
from
(select
curdate() - interval (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a)) day
as Date
from (
select 0 as a union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6 union all select 7
union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union
all select 3 union all select 4 union all select 5 union all select 6 union
all select 7 union all select 8 union all select 9) as d
) as a
where dayofweek(a.Date) = 7
and a.Date >= '2017-01-01'
and day(a.Date) > 28
order by a.date asc;
insert into DATES (AdherenceDates, WhichSaturday, WhichFifthSaturday)
select
DATES1.AdherenceDates,
DATES1.WhichSaturday,
DATES2.WhichFifthSaturday
from
DATES1
left join
DATES2
on DATES1.AdherenceDates = DATES2.FifthSaturdayDates;
drop temporary tables DATES1, DATES2;
select * from DATES
order by AdherenceDates asc;
I'm not very good when it comes to using joins - so I have a single table where I'm counting the number of records that meet certain conditions, and returns those counts by week. The problem is, I need the weeks that have a zero count too....I tried to get this to work with a left join, but I'm struggling...any help appreciated: (Stamp is a datetime field)
Query:
SELECT week(stamp), count(*) AS mycount, YEAR(stamp) as theyear
FROM merges
WHERE completed = 1
AND stamp BETWEEN '2017/4/1 00:00:00' AND '2017/6/1 00:00:00' GROUP BY week(stamp)
This returns:
week(stamp) | mycount | theyear
15 | 21 |2017
17 | 10 |2017
18 | 62 |2017
19 | 13 |2017
20 | 76 |2017
21 | 22 |2017
Notice week 16 is missing? I need to have this result included in the above, like:
16 | 0 |2017
I appreciate any help - I know this isn't too difficult, but I'm pulling my hair out trying to understand how to do this while I read other posts....
select weekValue, yearValue, coalesce(mycount,0)
from
( SELECT distinct week(#startDate := #startDate + Interval 1 day) as weekValue,
year(#startDate := #startDate + Interval 1 day) as yearValue
FROM
(select 0 union all select 1 union all select 3 union all select 4
union all select 5 union all select 6 union all select 6 union all select 7
union all select 8 union all select 9) t,
(select 0 union all select 1 union all select 3
union all select 4 union all select 5 union all select 6
union all select 6 union all select 7 union all select 8 union all select 9) t2,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t3,
(select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) t4,
(SELECT #startDate := '2017-03-31 00:00:00' ) as g
where
#startDate < '2017-06-01 00:00:00' ) as generateWeekYear left join
(SELECT week(stamp) as theweek, count(*) AS mycount, YEAR(stamp) as theyear
FROM merges
WHERE completed = 1
AND stamp BETWEEN '2017/4/1 00:00:00' AND '2017/6/1 00:00:00' GROUP BY week(stamp) ) as actualQuery
on generateWeekYear.weekValue = actualQuery.theweek
and generateWeekYear.yearValue = actualQuery.theyear
Let me explain the above query,
Sub Query generateWeekYear = This is used to genearate distinct week and year based on two inputs
lets say startDate and endDate. startDate should be 1 day less to actual startDate. Because if you do not
subtract 1 day then there might chance to loose one week.
Now you have all week and year which needs to be displayed.
Now you are thinking generateWeekYear is going to be more time to execute but this is not case. You can
check this generate an integer sequence in MySQL.
After that you simply join your table with above table and you can get your required result.
I'm trying to count entries grouped per hour.
I've found some useful info inform on different sites and here on: MySQL Group By Hours
But the result is not what I've expected.
With the following code I get:
SELECT CONCAT(Hour, ':00-', Hour+1, ':00') AS Hours,
COUNT(`usage_time`) AS `usage` FROM `usage`
RIGHT JOIN (
SELECT 0 AS Hour
UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3
UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6
UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18
UNION ALL SELECT 19 UNION ALL SELECT 20 UNION ALL SELECT 21
UNION ALL SELECT 22 UNION ALL SELECT 23
) AS AllHours ON HOUR(`usage_time`) = Hour
WHERE `usage_function` LIKE 'PlayedWholeSong' AND `usage_date` = DATE_SUB(CURDATE(), INTERVAL 0 DAY) OR `usage_time` IS NULL
GROUP BY Hour
ORDER BY Hour
Result:
Hours usage
2:00-3:00 0
4:00-5:00 6
6:00-7:00 2
8:00-9:00 3
9:00-10:00 20
10:00-11:00 1
14:00-15:00 14
15:00-16:00 1
16:00-17:00 32
17:00-18:00 10
As these are entry's from today, I don't have any entries after 19:00.
Also I don't see an entry from 00:00 - 01:00, 03:00 - 04:00 and several others are missing.
But I do want to show a list with every 24 hour and the result, even if there's nothing.
String thing is the result shows a 0 between 02:00 - 03:00.
I've learned a lot today about mysql, but nothing that solves my issue.
I hope you can learn me something, doesn't have to be code, a direction would be great.
I prefer LEFT JOIN over RIGHT JOIN personally. That way you can add your WHERE criteria in your JOIN and it won't constrict your results. Try this:
SELECT CONCAT(Hour, ':00-', Hour+1, ':00') AS Hours,
COUNT(`usage_time`) AS `usage`
FROM
(
SELECT 0 AS Hour
UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3
UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6
UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18
UNION ALL SELECT 19 UNION ALL SELECT 20 UNION ALL SELECT 21
UNION ALL SELECT 22 UNION ALL SELECT 23
) AS AllHours
LEFT JOIN `usage` ON HOUR(`usage_time`) = Hour
AND `usage_function` LIKE 'PlayedWholeSong'
AND `usage_date` = DATE_SUB(CURDATE(), INTERVAL 0 DAY)
GROUP BY Hour
ORDER BY Hour
Here is a simplified SQL Fiddle.
Good luck.