SQL query to group by day

Question

I want to list all sales, and group the sum by day.

Sales (saleID INT, amount INT, created DATETIME)

NOTE: I am using SQL Server 2005.

Answer 1

use linq

from c in Customers
group c by DbFunctions.TruncateTime(c.CreateTime) into date
orderby date.Key descending
select new  
{
    Value = date.Count().ToString(),
    Name = date.Key.ToString().Substring(0, 10)
}

Answer 2

actually this depends on what DBMS you are using but in regular SQL convert(varchar,DateColumn,101) will change the DATETIME format to date (one day)

so:

SELECT 
    sum(amount) 
FROM 
    sales 
GROUP BY 
    convert(varchar,created,101)

the magix number 101 is what date format it is converted to

Answer 3

if you're using SQL Server,

dateadd(DAY,0, datediff(day,0, created)) will return the day created

for example, if the sale created on '2009-11-02 06:12:55.000', dateadd(DAY,0, datediff(day,0, created)) return '2009-11-02 00:00:00.000'

select sum(amount) as total, dateadd(DAY,0, datediff(day,0, created)) as created
from sales
group by dateadd(DAY,0, datediff(day,0, created))

Answer 4

For PostgreSQL:

GROUP BY to_char(timestampfield, 'yyyy-mm-dd')

or using cast:

GROUP BY timestampfield::date

if you want speed, use the second option and add an index:

CREATE INDEX tablename_timestampfield_date_idx ON  tablename(date(timestampfield));

Answer 5

If you're using MySQL:

SELECT
    DATE(created) AS saledate,
    SUM(amount)
FROM
    Sales
GROUP BY
    saledate

If you're using MS SQL 2008:

SELECT
    CAST(created AS date) AS saledate,
    SUM(amount)
FROM
    Sales
GROUP BY
    CAST(created AS date)

Answer 6

For oracle you can

group by trunc(created);

as this truncates the created datetime to the previous midnight.

Another option is to

group by to_char(created, 'DD.MM.YYYY');

which achieves the same result, but may be slower as it requires a type conversion.