I need help calculating the following in MYSQL
how many days in quarter based on current date in MYSQL?
If today is 8/27/2014 - how many days is that in the quarter?
Also how do I calculate how many days in a quarter between two dates in MYSQL?
If user chose 3/1/2014 - 8/27/2014 how many total days for those 3 quarters?
to get the number of days in this quarter use this.
SELECT
DATEDIFF(
MAKEDATE(YEAR(CURDATE()), 1) + INTERVAL QUARTER(CURDATE()) QUARTER - INTERVAL 1 DAY,
MAKEDATE(YEAR(CURDATE()), 1) + INTERVAL QUARTER(CURDATE()) QUARTER - INTERVAL 1 QUARTER
)
to get the difference between two dates just use this
SELECT DATEDIFF(2014-08-27','2014-03-01')
im not sure why you would be doing this in mysql.. there are built in functions in other languages to do this same operation. so I wouldn't recommend doing it in MySQL.
This is what I was looking for:
select datediff(MAKEDATE(YEAR(CurDate()), 1)
+ INTERVAL QUARTER(CurDate()) QUARTER - INTERVAL 1 QUARTER,
case when Quarter(CurDate()) = quarter(curdate())
then CurDate() -- get the number of days in quarter up to today date
else
-- return number of days in quarter totally
DATE (DATE_SUB( DATE_ADD( CONCAT( YEAR( CurDate() ), '-01-01'),
INTERVAL QUARTER(CurDate()) QUARTER ), INTERVAL 1 DAY)) end) * -1 days_in_quarter
Related
Please help me to get the number of days left in the quarter from current date in SQL
Thanks
You can create quarter end date and then do datediff as follows:
SELECT DATEDIFF(MAKEDATE(YEAR(CURDATE()), 1)
+ INTERVAL QUARTER(CURDATE()) QUARTER
- INTERVAL 1 DAY,
CURDATE())
I am trying to get records from last one year and upto last date of provious month i.e. not including the current month. Here's my query:
SELECT `customer_id`, `customer_name`, `customer_date`
FROM `customers`
WHERE DATE(`customer_date`) <= (CURDATE() - INTERVAL 1 MONTH)
AND DATE(customer_date) >= (CURDATE() - INTERVAL 12 MONTH)
This query fetches records from 1 May, 2018 to 8 April 2019.
INTERVAL 1 MONTH fetches records 30 days ago. What I need to do something here?
I want to exclude current month records, so query should return records upto 30 April 2019. How do we do that?
You must correctly calculate the first and last days of range with help LAST_DAY() function. For example:
Calculate the first day of the range
SELECT LAST_DAY(CURDATE() - INTERVAL 13 MONTH) + INTERVAL 1 DAY
Output:
2018-05-01
Calculate last day of the range
SELECT LAST_DAY(CURDATE() - INTERVAL 1 MONTH)
Output:
2019-04-30
The full query might look like:
SELECT `customer_id`, `customer_name`, `customer_date`
FROM `customers`
WHERE `customer_date` >= LAST_DAY(CURDATE() - INTERVAL 13 MONTH) + INTERVAL 1 DAY
AND `customer_date` <= SELECT LAST_DAY(CURDATE() - INTERVAL 1 MONTH)
To get data up to the previous month:
where customer_date < curdate() + interval (1 - day(curdate()) day
Why? First note that there is no function call on the customer_date. So, this expression is index-compatible and can use an index.
Second, this structure works both for dates and date/times. That is very handy, because it may not always be obvious if a column has a time component (people are not very good about naming columns to capture this information).
You claim that the "12 months" ago portion works. That doesn't look correct to me. For the complete logic:
where customer_date < curdate() + interval (1 - day(curdate()) day and
customer_date >= (curdate() + interval (1 - day(curdate()) day) - interval 1 year)
SELECT `customer_id`, `customer_name`, `customer_date` FROM `customers` WHERE
MONTH(`customer_date`) <= (CURDATE() - INTERVAL 1 MONTH) AND
MONTH(customer_date) >= (CURDATE() - INTERVAL 12 MONTH)
hope this help
How do I SELECT the first week of a previous month I've tried
$myQuery = "SELECT repairId , startDate,catId,statusId FROM repair
WHERE supermarketId = '$supermarket'
AND startDate>=(CURDATE()- 1 WEEK - INTERVAL 2 week)";
This was used to try and select the third week but this didn't work
Does this work for you:
$myQuery = "SELECT repairId , startDate,catId,statusId FROM repair
WHERE supermarketId = '$supermarket'
AND startDate>= curdate() - interval 1 month
- interval weekday(curdate() - interval 1 month) day
- interval (day(curdate() - interval 1 month
- interval weekday(curdate() - interval 1 month) day) div 7) week
AND startDate < curdate() - interval 1 month
- interval weekday(curdate() - interval 1 month) day
- interval (day(curdate() - interval 1 month
- interval weekday(curdate() - interval 1 month) day) div 7) week _+ interval 1 week";
?
The idea here is that we first go back a month, then find the start of the week (assuming Monday, for Sunday we will need some extra tweaking), then figure out how many whole weeks it has been from the start of the month, and subtract that many weeks from the date so far. This takes us back to the start of the first week of the month. For the end of the range we just add one week to the start.
Ah, your question became more detailed.
Not really familiar with sql, there might be better but something like:
SELECT repairId , startDate,catId, statusId FROM repair
WHERE EXTRACT(YEAR_MONTH FROM start_date) = EXTRACT(YEAR_MONTH FROM NOW()) - 1 AND CAST(EXTRACT(DAY FROM start_date) / 7 + 1 as INT) = ?;
Basically, extract the year month components to compare year and month and then extract the day of month use the flooring caused by integer truncation to get the week to compare with whatever week you are looking for
mysql return rows matching year month
I have a table with a timestamp field. How do I get data from the last 3 months?
In particular, March is my current month let say, 03/2012. I need to return records from the months March, February, and January only.
3 months before today:
select * from table where timestamp >= now()-interval 3 month;
Start with first of month:
select * from table where timestamp >= last_day(now()) + interval 1 day - interval 3 month;
To get the first day of the current month, you could use this:
DATE_FORMAT(CURDATE(), '%Y-%m-01')
if current date is 2013-03-13, it will return 2013-03-01, and we can just substract 2 months from this date to obtain 2013-01-01. Your query could be like this:
SELECT *
FROM yourtable
WHERE data >= DATE_FORMAT(CURDATE(), '%Y-%m-01') - INTERVAL 2 MONTH
I know this is an old question, but to possibly save others time and to sum the above answers up for the case of needing (1) dates from current month and (2) dates from the prior 2 months (common when displaying data statistics):
WHERE ((timestamp >= NOW() - DATE_FORMAT(CURDATE(), '%Y-%m-01'))
OR (timestamp >= DATE_FORMAT(CURDATE(), '%Y-%m-01') - INTERVAL 2 MONTH))
Assuming you're using SQL Server (Oracle, MySQL and others have similar date functions), you can use the dateadd function to add or subtract an interval to the current date.
If you want a full three months, you can subtract 3 months from today : DATEADD(m,-3,getdate())
But, as you state, you only want data from January, February and March. You have to make some calculation based on today's date: dateadd(m,-2, CONVERT(datetime, CONVERT(VARCHAR(2), MONTH(getdate())) + '/01/' + CONVERT(VARCHAR(4), YEAR(getdate()))))
And in the end, get a query like
SELECT fields
FROM table
WHERE timestampfield > DATEADD(m,-2, CONVERT(datetime, CONVERT(VARCHAR(2), MONTH(getdate())) + '/01/' + CONVERT(VARCHAR(4), YEAR(getdate()))))
--- edit ---
erf, I just noticed the "mysql" tag... you can get more information on MySQL date functions here : https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
Another possibility would be:
SELECT * WHERE your_date_column > LAST_DAY(CURRENT_DATE - INTERVAL 3 MONTH);
Use this code here to get the previous 3 months from a certain date
SELECT * FROM table WHERE date_column>= DATE_FORMAT(current_date(), '%Y-%m-01') - INTERVAL 3 MONTH and date_column< DATE_FORMAT(current_date(), '%Y-%m-01')
WHERE ((timestamp >= NOW() - DATE_FORMAT(CURDATE(), '%Y-%m-01'))
OR (timestamp >= DATE_FORMAT(CURDATE(), '%Y-%m-01') - INTERVAL 2 MONTH))
I found the following code to help in creating a weekly report based on a start date of Friday. The instructions say to replace ".$startWeekDay." with a 4. When I put '".$startDay."' as '2013-01-30', I get errors.
Also I get a report by day rather than week as I desire.
SELECT SUM(cost) AS total,
CONCAT(IF(date - INTERVAL 6 day < '".$startDay."',
'".$startDay."',
IF(WEEKDAY(date - INTERVAL 6 DAY) = ".$startWeekDay.",
date - INTERVAL 6 DAY,
date - INTERVAL ((WEEKDAY(date) - ".$startWeekDay.")) DAY)),
' - ', date) AS week,
IF((WEEKDAY(date) - ".$startWeekDay.") >= 0,
TO_DAYS(date) - (WEEKDAY(date) - ".$startWeekDay."),
TO_DAYS(date) - (7 - (".$startWeekDay." - WEEKDAY(date)))) AS sortDay
FROM daily_expense
WHERE date BETWEEN '".$startDay."' AND '".$endDay."'
GROUP BY sortDay;
The following code is what I am using
SELECT count(DISTINCT (
UserID)
) AS total, CONCAT(IF(date(LastModified) - INTERVAL 6 day < date(LastModified),
date(LastModified),
IF(WEEKDAY(date(LastModified) - INTERVAL 6 DAY) = 4,
date(LastModified) - INTERVAL 6 DAY,
date(LastModified) - INTERVAL ((WEEKDAY(date(LastModified)) - 4)) DAY)),
' - ', date(LastModified)) AS week
FROM `Purchase`
WHERE `OfferingID` =87
AND `Status`
IN ( 1, 4 )
GROUP BY week
The output I get is
total week
3 2013-01-30 - 2013-01-30
1 2013-01-31 - 2013-01-31
I'm not sure exactly how you want to display your week, the sql above is attempting to display date ranges. If this isn't a requirement, your query could be very simple, you can just offset your time by two days (since friday is two days away from the natural star of the week) and use the week function to get the week number.
The query would look like this:
select count(distinct (UserID)) as total
, year( LastModified + interval 2 day ) as year
, week( LastModified + interval 2 day ) as week_number
FROM `Purchase`
WHERE `OfferingID` =87
AND `Status`
IN ( 1, 4 )
group by year, week_number;