How to Group by Day Instead of Date

MySQL Query GROUP BY day / month / year

GROUP BY YEAR(record_date), MONTH(record_date)

Check out the date and time functions in MySQL.

Group by day using the timestamp field

Use MySQL DATE() function to extract the date from the timestamp:

SELECT name, count(*) FROM mytable GROUP BY DATE(mytable.modified);

The DATE() function extracts the date value from a date or datetime expression.

How can I group by date time column without taking time into consideration

Cast/Convert the values to a Date type for your group by.

GROUP BY CAST(myDateTime AS DATE)

Group by date only on a Datetime column

Cast the datetime to a date, then GROUP BY using this syntax:

SELECT SUM(foo), DATE(mydate) FROM a_table GROUP BY DATE(a_table.mydate);

Or you can GROUP BY the alias as @orlandu63 suggested:

SELECT SUM(foo), DATE(mydate) DateOnly FROM a_table GROUP BY DateOnly;

Though I don't think it'll make any difference to performance, it is a little clearer.

How to GROUP BY date with a timestamp field in Postgres?

You need to truncate in the GROUP BY as well:

SELECT created_at::date, COUNT(*)
FROM table_name
WHERE product_uuid = '586d8e21b9529d14801b91bd' AND
      created_at > now() - interval '30 days'
GROUP BY created_at::date
ORDER BY created_at::date ASC;

Your version is aggregating by each date/time value but only showing the date component.

Also, I would recommend that you use current_date rather than now() so the first date is not truncated.

How do I group by day instead of date?

Time is a quite complex object to group by. Assuming you want to group by the creation date, instead of the full Time, start creating a custom method in your model to return the group criteria.

The method should return the creation date, possibly as string.

def group_by_criteria
  created_at.to_date.to_s(:db)
end

Then, group by that method.

list.group_by(&:group_by_criteria).map {|k,v| [k, v.length]}.sort

Group by day and hour

You can use lubridate::ymd_hms to convert the date variable to date-time, group by day and hour from it and take mean value of price for each hour.

library(dplyr)

prices_2019 %>%
  mutate(date = lubridate::ymd_hms(date),
         date_hour = format(date, "%Y-%m-%d %H")) %>%
  group_by(date_hour) %>%
  summarize(mean_price = mean(price))

How to group data based on continuous date range?

the date range for price is 23.9 is not right because price not same for all the days in that range.

Because there are two same price in different overlapping date ranges, so you might get only one row when you used aggregate function.

This is a gap-and-island problem, we can try to use ROW_NUMBER window function to get the gap of overlapping date and then group by that.

SELECT  Product_Code,
        min(Pricing_Date) AS Min_Date ,
        max(Pricing_Date) AS Max_Date, 
        price
FROM (
    SELECT *,
           ROW_NUMBER() OVER(ORDER BY PRICING_DATE) - ROW_NUMBER() OVER(PARTITION BY  PRODUCT_CODE,PRICE ORDER  BY PRICING_DATE) grp
    FROM PRICE_DATA 
) t1
GROUP BY grp,Product_Code,price
ORDER BY  min(Pricing_Date)

sqlfiddle

Explain

The gap-and-island problem is a feature

continuous(overlapping) data is that a set (continuous range of sequence) - (values ​​based on a certain order of conditions sequence) yields the same grouping.

so that We can use

• ROW_NUMBER() OVER(ORDER BY PRICING_DATE) making a continuous range of values.
• ROW_NUMBER() OVER(PARTITION BY PRODUCT_CODE,PRICE ORDER BY PRICING_DATE) making values ​​based on a certain order of conditions.

Then we will get a grouping column with overlapping data as sqlfiddle

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