I have a database with a table containing information on some images, each row containing a createdAt date and a viewCount. The data ranges from September 2014 until today (July 2016). I want to get a monthly sum of the amount of views across all images for the month
When I run the query
SELECT YEAR(createdAt), MONTH(createdAt), SUM(viewCount)
FROM Images
GROUP BY MONTH(createdAt);
I'm only returned 12 rows with results between September 2014 and August 2015
Year | Month | Views
-------------------
2014 | 9 | 1452
2014 | 10 | 279
2014 | 11 | 34428
2014 | 12 | 4763
2015 | 1 | 2826
2015 | 2 | 777
2015 | 3 | 568
2015 | 4 | 1309
2015 | 5 | 46744
2015 | 6 | 1541
2015 | 7 | 8160
2015 | 8 | 91
If I add a date restraint it will give me the latest data, but again only 12 rows
SELECT YEAR(createdAt), MONTH(createdAt), SUM(viewCount)
FROM Images WHERE createdAt > DATE('2015-08-01 00:00:00')
GROUP BY MONTH(createdAt);
Year | Month | Views
--------------------
2015 | 8 | 981
2015 | 9 | 1031
2015 | 10 | 2566
2015 | 11 | 3325
2015 | 12 | 411
2016 | 1 | 2140
2016 | 2 | 710
2016 | 3 | 714
2016 | 4 | 1985
2016 | 5 | 426
2016 | 6 | 119
2016 | 7 | 81
I do realise that since it's July the second query stops there as that's where the data ends, but why does the first query not return all the results?
Group by year/month:
SELECT YEAR(createdAt), MONTH(createdAt), SUM(viewCount)
FROM Images
--WHERE createdAt > DATE('2015-08-01 00:00:00')
GROUP BY YEAR(createdAt), MONTH(createdAt);
Related Group by clause in mySQL and postgreSQL, why the error in postgreSQL?
Keep in mind that from MySQL 5.7.6+ your initial query may not even work because of only_full_group_by which is set by default.
You can simply add Year to you group by
SELECT YEAR(createdAt), MONTH(createdAt), SUM(viewCount)
FROM Images
GROUP BY YEAR(createdAt), MONTH(createdAt)
ORDER BY YEAR(createdAt), MONTH(createdAt)
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 |
SELECT COUNT(client_ID) / DAY(LAST_DAY(dateRequested))
FROM `tbl_client`
WHERE dateRequested BETWEEN DATE_FORMAT(dateRequested,'%Y-%m-01') AND LAST_DAY(dateRequested)
I want to show the average of client per day in the month
client_ID | dateRequested
1 | 2018-07-04
2 | 2018-07-05
3 | 2018-07-06
4 | 2018-07-07
5 | 2018-08-04
6 | 2018-08-06
7 | 2018-08-09
i want to show
Average | Month
4 | July 2018
3 | August 2018
Try below query:
SELECT COUNT(client_ID),concat(month(dateRequested),year(dateRequested))
FROM `tbl_client`
WHERE dateRequested BETWEEN DATE_FORMAT(dateRequested,'%Y-%m-01') AND LAST_DAY(dateRequested)
group by concat(month(dateRequested),year(dateRequested))
Hi i have a table regd like this
name | year | month
james| 2017 | 2
stacy| 2017 | 4
josh | 2017 | 8
june | 2017 | 9
mary | 2018 | 1
hawk | 2018 | 2
sim | 2018 | 3
How can I select all the people between month 8 of year 2017 to month 3 of year 2018.
Thank you for your help and suggestions.
You can write condition like below
year * 100 + month between 201708 and 201803
where 201708 = 2017 * 100 + 8
and 201803 = 2018 * 100 + 3
I'm trying to order a mysql query according to the month name but it is ordering in reverse order .I have tried both the ASC and DESC order but not working!This is what i'm getting :
order_amount month_number
370.245 Dec
0.01 Aug
0.02 July
0.01 May
2 Apr
3 Mar
4 Feb
5 Jan
This is the query:
select sum(amnt) as order_amount,month_number from orders where paid =1 GROUP BY month_number ORDER BY MONTH(month_number) ASC
This is a sample table 'order' on which i'm running query
order_amount month_number paid
370.245 Jan 1
0.01 Aug 1
0.02 July 1
0.01 Apr 1
2 May 0
3 Mar 1
4 Feb 0
5 Nov 0
month_number() is a lousy name for a string name for the month. That said, your attempts will not work, because month() takes a date, not a string representation. The best way is probably just to use case:
order by (case when month_number = 'Jan' then 1
when month_number = 'Feb' then 2
. . .
when month_number = 'Dec' then 12
end)
I would strongly suggest you include a date or actual month number in the table. And, rename month_number to something like month_name.
According to your sample, you've got 'Dec', 'Aug' and so on in your month_number field.
How is month(month_number) supposed to work if you don't pass a valid date type? if you pass some random text you'll just get NULL and can't order by NULL
http://sqlfiddle.com/#!2/3a774d/3/0
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_month
SELECT * FROM my_table;
+-------+
| month |
+-------+
| Apr |
| Aug |
| Dec |
| Feb |
| Jan |
| Jul |
| Jun |
| Mar |
| May |
| Nov |
| Oct |
| Sep |
+-------+
SELECT * FROM my_table ORDER BY STR_TO_DATE(CONCAT('2014-',month,'-01'),'%Y-%b-%d');
+-------+
| month |
+-------+
| Jan |
| Feb |
| Mar |
| Apr |
| May |
| Jun |
| Jul |
| Aug |
| Sep |
| Oct |
| Nov |
| Dec |
+-------+
... or something like that
Assuming I have a table like the following:
id | assignment | duedate
1 | Math | 2012-01-01
2 | History | 2012-02-02
3 | Science | 2012-01-01
4 | Government | 2012-02-01
5 | Government | 2013-01-13
6 | History | 2013-03-13
Is it possible to make some sql query such that I get a grouping of all the dates by month and year? Is there some possibility that I could get a sorted result of:
duedatemonth | count
January 2012 | 2
Feburary 2012 | 2
January 2013 | 1
March 2013 | 1
I know you can GROUP BY duedate, but that only groups those with the same month, day, and year instead of just month and year.
Would it be then possible to even further group it such that it factors in "assignment" to obtain a resulting table of
id | duedatemonth | count
1 | January 2012 | 2
3 | January 2012 | 2
2 | Feburary 2012 | 2
4 | Feburary 2012 | 2
5 | January 2013 | 1
6 | March 2013 | 1
try this
SELECT DATE_FORMAT(duedate,'%M %Y') duedatemonth, COUNT(*) count
FROM Table1
GROUP BY year(duedate), MONTH(duedate)
DEMO HERE
will output this:
DUEDATEMONTH COUNT
January 2012 2
February 2012 2
January 2013 1
March 2013 1
Use the string functions YEAR and MONTH.
SELECT YEAR(duedate), MONTH(duedate), COUNT(*)
FROM sparkles
GROUP BY YEAR(duedate), MONTH(duedate)
Use MONTHNAME or DATE_FORMAT to get the name of the month.
You can use this query.
SELECT DATE_FORMAT(duedate, '%M %Y') duedatemonth, COUNT(1) `count`
FROM Tbl
GROUP BY DATE_FORMAT(duedate, '%M %Y')