I have a table with 4 columns:
1. customerID
2. dateAdded
3. productID
4. quantity
The format of dateAdded is like this: 20180730 (YearMonthDay).
I want to group the rows in the table by year and month.
I tried the code below but it doesn't work. I still see rows with the same year and month repeated, but different day.
SELECT dateAdded
, SUM(quantity)
FROM testTable
GROUP
BY DATE_FORMAT(dateAdded, '%Y%m')
, dateAdded
ORDER
BY dateAdded DESC
Any idea how to fix this?
Thank you
Do not group by what you do not want the data to be grouped. Since you provide dateAdded as a parameter to be grouped with, it splits the data by dateAdded and you gain nothing out of year-month grouping. Remove the columns from select/groupby that you use for grouping like this:
SELECT DATE_FORMAT(dateAdded, '%Y%m')
, SUM(quantity)
FROM testTable
GROUP BY
DATE_FORMAT(dateAdded, '%Y%m')
If you store dateAdded as 20180730, looks like it's a string, so it wont work with DATE_FORMAT function, you can use SUBSTRING instead
something like
SELECT dateAdded, SUM(quantity), SUBSTRING(dateAdded, 1, 6) as d
FROM testTable
GROUP BY d
ORDER BY dateAdded DESC
use year and month function
SELECT Year(dateAdded) as YearOfDate,Month(dateAdded) as MonthOfdate,
, SUM(quantity) as Qty
FROM testTable
GROUP
BY Year(dateAdded), Month(dateAdded)
ORDER BY dateAdded DESC
Related
I expect this query to give me the avg value from daily active users up to date and grouped by month (from Oct to December). But the result is 164K aprox when it should be 128K. Why avg is not working? Avg should be SUM of values / number of current month days up to today.
SELECT sq.month_year AS 'month_year', AVG(number)
FROM
(
SELECT CONCAT(MONTHNAME(date), "-", YEAR(DATE)) AS 'month_year', count(distinct id_user) AS number
FROM table1
WHERE date between '2020-10-01' and '2020-12-31 23:59:59'
GROUP BY EXTRACT(year_month FROM date)
) sq
GROUP BY 1
Ok guys thanks for your help. The problem was that on the subquery I was pulling the info by month and not by day. So I should pull the info by day there and group by month in the outer query. This finally worked:
SELECT sq.day_month, AVG(number)
FROM (SELECT date(date) AS day_month,
count(distinct id_user) AS number
FROM table_1
WHERE date >= '2020-10-01' AND
date < '2021-01-01'
GROUP BY 1
) sq
GROUP BY EXTRACT(year_month FROM day_month)
Do not use single quotes for column aliases!
SELECT sq.month_year, AVG(number)
FROM (SELECT CONCAT(MONTHNAME(date), '-', YEAR(DATE)) AS month_year,
count(distinct id_user) AS number
FROM table1
WHERE date >= '2020-10-01' AND
date < '2021-01-01'
GROUP BY month_year
) sq
GROUP BY 1;
Note the fixes to the query:
The GROUP BY uses the same columns as the SELECT. Your query should return an error (although it works in older versions of MySQL).
The date comparisons have been simplified.
No single quotes on column aliases.
Note that the outer query is not needed. I assume it is there just to illustrate the issue you are having.
I have a table where I need to get the last 12 records which is grouped by year() month() ASC. I try to use the query below but the result is not as expected.
SELECT * FROM
(
SELECT
id,
tanggal,
date_format(tanggal,'%b-%Y') as bulan,
sum(sisa_pokok) as jumlah
FROM transaksi_detail
GROUP BY date_format(tanggal,'%b-%Y')
ORDER BY id DESC LIMIT 12
) sub
ORDER BY id ASC
the query result is as below
My expected result is sort by bulan column order by year(), month() as follows
Bulan jumlah
Mar-2018 26600000
Oct-2017 1000000
Sept-2017 4500000
and so on....
EXTRACT(YEAR FROM bulan) as year
SELECT EXTRACT(YEAR FROM tanggal) as year , EXTRACT(MONTH FROM tanggal) as month, id FROM table_name group by year order by month
you can get year same like you can get month after that put group by and order i hope it will help you
This works for my situation
SELECT * FROM
(
SELECT
id,
tanggal,
month(tanggal),
year(tanggal),
date_format(tanggal,'%b-%Y') as bulan,
sum(sisa_pokok) as jumlah
FROM transaksi_detail
GROUP BY date_format(tanggal,'%b-%Y')
ORDER BY id DESC LIMIT 12
) sub
ORDER BY year(tanggal), month(tanggal) ASC
Basically I have a table like this:
Table Time:
ID.......Date
1......08/26/2016
1......08/26/2016
2......05/29/2016
3......06/22/2016
4......08/26/2015
5......05/23/2015
5......05/23/2015
6......08/26/2014
7......04/26/2014
8......08/26/2013
9......03/26/2013
The query should return like this
Year........CountNum
2016........4
2015........3
To find out which year does its value tend to increase in. I notice that I want to display the years that have more values (number of row in this case) than the previous year.
What I've done so far
SELECT Year, count(*) as CountNum
FROM Time
GROUP BY Year
ORDER BY CountNum DESC;
I don't know how to get the year from date format. I tried year(Date) function, but I got Null data.
Please help!
It should works fine.
select year(date), count(*) as countNum
from time
group by year(date)
order by countNum
Join the grouped data to itself with 1 year offset:
select
a.*
from
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) a
left join
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) b
on a._year = b._year-1
where a._n > b._n
order by 1
I have the following table structure: Value (stores random integer values), Datetime` (stores purchased orders datetimes).
How would I get the average value from all Value rows across a full day?
I'm assuming the query would be something like the following
SELECT count(*) / 1
FROM mytable
WHERE DateTime = date(now(), -1 DAY)
You can GROUP BY the DATE part of DATETIME and use AVG aggregate function to find an average value for each group :
SELECT AVG(`Value`)
, DATE(`Datetime`)
FROM `mytable`
GROUP BY DATE(`Datetime`)
Looks like a simple AVG task:
SELECT `datetime`,AVG(`Value`) as AvgValue
FROM TableName
GROUP BY `datetime`
To find average of a specific day:
SELECT `datetime`,AVG(`Value`) as AvgValue
FROM TableName
WHERE `datetime`=#MyDate
GROUP BY `datetime`
Or Simply:
SELECT AVG(`Value`) as AvgValue
FROM TableName
WHERE `datetime`=#MyDate
Explanation:
AVG is an aggregate function used to find the average of a column. Read more here.
The following query will give you what u want..
SELECT DATE_FORMAT(thedate_field, '%Y-%m-%d') as theday, avg (value)
FROM mytable group by
DATE_FORMAT(thedate_field, '%Y-%m-%d')
Try It its work...
Select AVG(value),convert(nvarchar,DateTime,103) from table group by convert(nvarchar,DateTime,103)
I have a MySQL table with 5 rows:
email
message_id
date
time
And I would like to count the number of emails there is per days.
So far I have this
SELECT Date,COUNT(*) AS Num FROM mail2_mailing_log WHERE Id_message=#Id GROUP BY Date ORDER BY Date DESC
But we found out that there was a problem with a script and many data got multiplied (Which have since been fixed), but I would like to be able to use the data I have.
So basically I would want to "merge" all the rows that match per email, date and time and then group by date and count the number of items.
SELECT q.date, COUNT(*)
FROM (SELECT DISTINCT email, date, time
FROM mail2_mailing_log
WHERE Id_Message = #Id) q
GROUP BY q.date
ORDER BY q.date DESC
Use the distinct keyword:
SELECT DISTINCT Date,COUNT(*) AS Num FROM mail2_mailing_log WHERE Id_message=#Id GROUP BY Date ORDER BY Date DESC
If the Date column contains the time too, then you will need to format the Date column using the DATE_FORMAT() function.
SELECT DATE_FORMAT(Date, '%W %M %Y'),COUNT(*) AS Num FROM mail2_mailing_log
WHERE Id_message=#Id GROUP BY DATE_FORMAT(Date, '%W %M %Y')
ORDER BY DATE_FORMAT(Date, '%W %M %Y') DESC