MySQL Query GROUP BY day / month / year

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Is it possible to make a simple query to count how many records I have in a determined period of time like a year, month, or day, having a TIMESTAMP field, like:

SELECT COUNT(id)
FROM stats
WHERE record_date.YEAR = 2009
GROUP BY record_date.YEAR

Or even:

SELECT COUNT(id)
FROM stats
GROUP BY record_date.YEAR, record_date.MONTH

To have a monthly statistic.

Thanks!

GROUP BY YEAR(record_date), MONTH(record_date)

Check out the date and time functions in MySQL.

MySQL Query GROUP BY day / month / year, Try this code: GROUP BY YEAR(Record_date), MONTH(Record_date). OR. You can use DATE_FORMAT operator. If you are using this you  Using the DATE_FORMAT operator, you can easily group the timestamp, date or datetime column using any format you want. For example, I needed to group rows that were added on the same day. Here is my query:

GROUP BY DATE_FORMAT(record_date, '%Y%m')

Note (primarily, to potential downvoters). Presently, this may not be as efficient as other suggestions. Still, I leave it as an alternative, and a one, too, that can serve in seeing how faster other solutions are. (For you can't really tell fast from slow until you see the difference.) Also, as time goes on, changes could be made to MySQL's engine with regard to optimisation so as to make this solution, at some (perhaps, not so distant) point in future, to become quite comparable in efficiency with most others.

MySQL group by day, month or year, For example, I needed to group rows that were added on the same day. Here is my query: ? 1. select count(*), DATE_FORMAT(created_at  Is it possible I make a simple query to count how many records I have in a determined period of time like a Year, month or day, having a TIMESTAMP field, like: SELECT COUNT(id) FROM stats. WHERE record_date.YEAR = 2009. GROUP BY record_date.YEAR. Or even: SELECT COUNT(id) FROM stats. GROUP BY record_date.YEAR, record_date.MONTH. To have a

try this one

SELECT COUNT(id)
FROM stats
GROUP BY EXTRACT(YEAR_MONTH FROM record_date)

EXTRACT(unit FROM date) function is better as less grouping is used and the function return a number value.

Comparison condition when grouping will be faster than DATE_FORMAT function (which return a string value). Try using function|field that return non-string value for SQL comparison condition (WHERE, HAVING, ORDER BY, GROUP BY).

MySQL: GROUP BY Day, Month and Year, In the following example, the date column is "dat". Here is how to get the number grouped by years: SELECT COUNT (id)  MySQL: GROUP BY Day, Month and Year Question by Compi | 24/04/2016 at 19:59 I have some data sets with a timestamp column in which the date of the record is specified.

I tried using the 'WHERE' statement above, I thought its correct since nobody corrected it but I was wrong; after some searches I found out that this is the right formula for the WHERE statement so the code becomes like this:

SELECT COUNT(id)  
FROM stats  
WHERE YEAR(record_date) = 2009  
GROUP BY MONTH(record_date)

Group month and year in MySQL?, The GROUP BY clause is also used. The syntax is as follows − SELECT DATE_FORMAT(yourColumnName, '%m-%Y') from yourTableName  The right way to GROUP BY MONTH or DAY is to add YEAR to the GROUP BY clause. This way data will not be consolidated. For eg: now July 2007 and July 2008 are treated separate as a GROUP BY YEAR has been added. Here’s the same query with the GROUP BY YEAR clause

If your search is over several years, and you still want to group monthly, I suggest:

version #1:

SELECT SQL_NO_CACHE YEAR(record_date), MONTH(record_date), COUNT(*)
FROM stats
GROUP BY DATE_FORMAT(record_date, '%Y%m')

version #2 (more efficient):

SELECT SQL_NO_CACHE YEAR(record_date), MONTH(record_date), COUNT(*)
FROM stats
GROUP BY YEAR(record_date)*100 + MONTH(record_date)

I compared these versions on a big table with 1,357,918 rows (innodb), and the 2nd version appears to have better results.

version1 (average of 10 executes): 1.404 seconds version2 (average of 10 executes): 0.780 seconds

(SQL_NO_CACHE key added to prevent MySQL from CACHING to queries.)

Group by command in date field to list number of records in day wise, Displaying number of records against each day by using group by & count command in date field. Year- Month : 2015-Apr, 2016-Jan; Month wise ( in all years ): Jan , Feb; Financial Year: SELECT YEAR(payment_dt),COUNT(*) as Nos,SUM(amount) as total Date & Time functions used in Query for MySQL Table GROUP BY year_month_registered If you run the SQL query above, you will see that it returns the following result: As you can see, the Year and Month has been stored in a YYYYMM format in an alias called “year_month_registered.” Using DATE_FORMAT to get the year and month.

How to select from mysql that group by month in one year, Well, a quick search on Google seems to result in an answer to your question. This is for the derived table to be used with the main query - which will use MAX (firstcount) etc. THe main query will also have to group by the year and month fields. (SELECT YEAR(date1) AS yr, MONTH(date1) AS m, COUNT(date1) AS firstcount, 0 AS secondcount, 0 AS firstsum

MySQL - How to group data by Day and Hour, I have some problem in getting the query for this sample format. My idea is select hours on that day and group it by day or select all the 7 days  Use the below MySQL query for fetch the last month’s records from date wise. SELECT COUNT(id) as Count, DAY(created_at) as 'Day', DAYNAME(created_at) as 'Day Name', MONTHNAME(created_at) as 'Month Name' FROM employees WHERE MONTH(created_at) = MONTH(CURDATE() - INTERVAL 1 MONTH) AND YEAR(created_at) = YEAR(CURDATE()) GROUP BY DAY(created_at)

Laravel Group Query Result by Day/Month/Year, Quick tip for those who want to list some events, grouping them by some date value – by day, month, year etc. Let's take a look at example – we  One option would be to just run this query thirty times and increment the date manually each time. Obviously there is a better way. First, lets expand the scope of our query to include the entire month of August. We also need to select the created_at attribute, which we will call 'date', so that our query will return two columns - date and total.

Comments
  • I guess it's supposed to be GROUP BY record_date.MONTH in your first code snippet?
  • You may want to add an extra column for added clarity in some cases such as where records span several years. SELECT COUNT(event_id), DATE_FORMAT(event_start, '%Y/%m')
  • Simple complete example: SELECT count(*), record_date FROM anytable WHERE anytable.anycolumn = 'anycondition' GROUP BY YEAR(record_date), month(record_date); note: record_date is a date type TIMESTAMP
  • Probably worth mentioning this didn't run on my MySQL 5.7 with a COUNT aliased column (no error, I got zero results). When I changed to select those fields with alias, I could then group by the alias. This is standard MySQL 5.7 docker image running in a local environment so I've no idea why it didn't error or return results.
  • Oh god, if I knew this earlier...so many lines of PHP to do something mysql can do in one line.
  • I have a feeling that this would not perform well because a format function wouldn't be able to use an index on the date column.
  • @Stv: You might want to consider @fu-chi's answer then. As far as I can tell, the grouping expressions in both that answer and mine evaluate to the same thing but EXTRACT() may be more efficient than DATE_FORMAT(). (I don't have a MySQL for proper testing, though.)
  • Consider including @fu-chi's suggestion into your tests, it may prove even more efficient. Also, you tested GROUP BY YEAR(record_date)*100 + MONTH(record_date), but why not test GROUP BY YEAR(record_date), MONTH(record_date) as well?
  • If you use COUNT(1) insteed COUNT(*) it will be even faster, and result data are same.
  • What is that *100 on the versión #2? Thanks in advance.
  • *100 to YEAR(record_date)*100 + MONTH(record_date) == DATE_FORMAT(record_date, '%Y%m')
  • It's important to note that you'd also need to group by MONTH(record_date) as well to account for multiple months.
  • This would be very very slow.