I have a table called primeWeek. Im trying to get weekly avg depends on dates.
Example of my table
id | count | date
1 | 70 | 2020-08-29
2 | 67 | 2020-08-30
3 | 69 | 2020-08-31
4 | 82 | 2020-09-01
5 | 73 | 2020-09-02
I tried few things but results are not correct.
count and date are both keywords in SQL, so you should surround them with backticks.
SELECT
AVG(`count`) AS primeCount,
CONCAT(`date`, '-', `date` + INTERVAL 6 DAY) AS week
FROM primeWeek
GROUP BY WEEK(`date`)
ORDER BY WEEK(`date`);
Related
I have a table like this
userID time NoOfVisits
1 2014 50
2 2015 60
3 2016 70
4 2017 80
5 2018 90
6 2019 100
I need to write a sql query which will print time and average of past 3 years NoOfVisits for a particular site.
output should be as
userID time NoOfVisits
1 2014 50.0000
2 2015 55.0000
3 2016 60.0000
4 2017 70.0000
5 2018 80.0000
6 2019 90.0000
Explanation :
For user Id 6 (80+90+100)/3=90.0000
Please help me to solve this problem.
You can use a cumulative average, available in MySQL 8+:
select t.*,
avg(visits) over (order by time rows between 2 preceding and current row) as avg_visits_3
from t;
Assuming there are no gaps between the years (like your sample data), you can self join the table and group by userid, time to get the average:
select
t.userid, t.time, avg(tt.noofvisits) NoOfVisits
from tablename t inner join tablename tt
on tt.time between t.time - 2 and t.time
group by t.userid, t.time
See the demo.
Results:
| userid | time | NoOfVisits |
| ------ | ---- | ---------- |
| 1 | 2014 | 50 |
| 2 | 2015 | 55 |
| 3 | 2016 | 60 |
| 4 | 2017 | 70 |
| 5 | 2018 | 80 |
| 6 | 2019 | 90 |
I have this query to extract total_hours, start_date and end_date:
select proj.start_date, proj.end_date, sum(ifnull(work.hours_estimate,0)) as total_hours
from project_table proj
left outer join project_task work on
work.project_id = proj.id
where proj.id = 3
This query gives me a single row of result:
start_date | end_date | total_hour
----------------------------------------
2017-04-24 | 2017-05-15 | 119
What I want is to generate a daily interval of rows, constantly decreasing the total_hours by a certain amount, say 19 hours, and the day increasing by 1 day.
Expected results:
day | hours_left
------------------------
2017-04-24 | 119
2017-04-25 | 100
2017-04-26 | 81
2017-04-27 | 62
2017-04-28 | 43
2017-04-29 | 24
... and so on and so forth until it reaches 2017-05-15 (of course, no negative for hours_left, just zero if negative)
can't seem to figure out how to do this.
QUESTIONS:
1.) Is this possible in MySQL?
2.) If this is possible in MySQL, is it efficient/convinient?
If not, I could just do it in application, as state in the comments
I don't know if this is possible, but it'd be really awesome. I have a table of sign-ins for people who are logging time on different projects and I need to compile a report of time logged for each project for a given time period.
My table looks something like this:
id | project | time_in | time_out | break
----------------------------------------------------------------
1 | 1 | 2014-12-07 05:00:00 | 2014-12-07 10:00:00 | 30
2 | 2 | 2014-12-07 06:00:00 | 2014-12-07 13:00:00 | 15
3 | 1 | 2014-12-07 14:00:00 | 2014-12-07 18:00:00 | 0
4 | 3 | 2014-12-07 08:30:00 | 2014-12-07 18:45:00 | 75
5 | 2 | 2014-12-07 12:00:00 | 2014-12-07 16:30:00 | 0
What I'd like to be able to do is get a report of the time logged for each project given a date range, i.e. the total time, probably in seconds, logged for each project.
time_in and time_out are fields of type TIMESTAMP; break is an integer representing the number of minutes the person was on break. I need to get the sum of time_out - time_in - break for each project, e.g. for December 7:
project | time
---------------
1 | 34200
2 | 40500
3 | 34200
This is all I have so far:
SELECT DISTINCT
`project`
FROM `sign_ins`
WHERE
`time_in` >= '2014-12-07 00:00:00' AND
`time_out` <= '2014-12-08 00:00:00';
I appreciate your help on this, SO community. You guys are so brilliant.
You can get the difference in seconds by converting the date/time values to Unix time stamps. Then, just aggregate the differences using sum():
SELECT project,
SUM(UNIX_TIMESTAMP(time_out) - UNIX_TIMESTAMP(time_in) - (break * 60)) as DiffSecs
FROM `sign_ins`
WHERE `time_in` >= '2014-12-07 00:00:00' AND
`time_out` <= '2014-12-08 00:00:00'
GROUP BY project;
I have table something like this (there's also "device_id" and "timestamp" columns)
day | interval | value
----------------------------
1 | 14 | 63 // start of a day
1 | 14 | 83
1 | 14 | 73
1 | 15 | 23
1 | 15 | 33
1 | 15 | 50
2 | 16 | 23 // start of a day
2 | 16 | 33
2 | 16 | 50
I want to select all intervals in a day. That is simple.
However, an interval can start a bit before a day flips, or end a bit past:
day | interval | value
----------------------------
7 | 14 | 63
7 | 14 | 83
8 | 14 | 73 // start of a day
8 | 15 | 23
8 | 15 | 33
8 | 15 | 50
8 | 16 | 23
8 | 16 | 33
9 | 16 | 50 // start of a day
Now I'd like to select all three intervals - or even better intervals that are mostly in that day.
SELECT ... WHERE day = 8
Gives me only parts of the start/end intervals (14, 16). That's useless, I need the complete intervals.
If there's no solution, I'll just do three queries, but there might be some SQL trick I'm not aware of?
It's MySQL, called from PHP.
More visually:
day 7 | day 8 | day 9
------------------+-------------------+---------------
###13### ###14### ###15### ###16### ###17###
... 63 83 73 23 33 50 23 33 50 ...
I want all values in day 8 -> intervals 14, 15, 16
I think you are looking for this:
SELECT * FROM intervals
WHERE interval IN (
SELECT DISTINCT interval FROM intervals WHERE day = 8)
This selects all interval data where at least one of the entries for that interval occurs in day 8. The subquery determines which unique intervals happen in the day, which is then used by the outer query to select their specifics.
SELECT DISTINCT y.*
FROM my_table x
JOIN my_table y
ON y.some_column = x.some_column
WHERE x.some_other_column = 8;
I have a mysql db which I use to return amounts of orders by hour in a specific day. I use this SELECT statement for that.
select
hour(datains),sum(valore)
from
ordini
where (stato=10 or stato = 1 ) and DATE(datains) = DATE_SUB(CONCAT(CURDATE(), ' 00:00:00'), INTERVAL 0 DAY)
group by hour(datains)
order by
id DESC
It returns:
+--------------+---------------+
| hour datains | valore |
| 12 | 34 |
| 11 | 56 |
| 10 | 134 |
+-------------------------------
Now I need to have columns for a certain number of days, like this.
+--------------+---------------+--------------+--------------+
| hour datains | 01-01-2014 | 02-01-2014 | 03-01-2014 |
| 12 | 34 | 34 | 77 |
| 11 | 56 | 0 | 128 |
| 10 | 134 | 66 | 12 |
+------------------------------+-----------------------------+
Is this possible?
It seems you have a table ordini with columns datains, valore, and stato.
Perhaps you can try this query to generate hour-by-hour aggregates for a three days' worth of recent sales, but not including today.
SELECT DATE_FORMAT(datains, '%Y-%m-%d %H:00') AS hour,
SUM(valore) AS valore
FROM ordini
WHERE (stato = 1 OR stato = 10)
AND datains >= CURRENT_DATE() - INTERVAL 3 DAY
AND datains < CURRENT_DATE
GROUP BY DATE_FORMAT(datains, '%Y-%m-%d %H:00')
ORDER BY DATE_FORMAT(datains, '%Y-%m-%d %H:00')
This will give you a result set with one row for each hour of the three days, for example:
2014-01-01 10:00 456
2014-01-01 11:00 123
2014-01-02 10:00 28
2014-01-02 11:00 350
2014-01-02 12:00 100
2014-01-02 13:00 17
2014-01-03 10:00 321
2014-01-03 11:00 432
2014-01-03 12:00 88
2014-01-03 13:00 12
That's the data summary you have requested, but formatted row-by-row. Your next step is to figure out an appropriate technique to pivot that result set, formatting it so some rows become columns.
It happens that I have just written a post on this very topic. It is here:
http://www.plumislandmedia.net/mysql/sql-reporting-time-intervals/