I've got an mysql-db with charges in it, with a datetime. Every month i want to create an invoice with all the charges from last month or earlier. So if it´s may 2nd, 5th or 30th 2012, i still only want the invoices from april 2012 or earlier. I've tried with date_sub, but it just subtracts a month, so it only invoices up to the same day of the previous month. How should i do this?
get * from Ads WHERE AdEnd > ??
Ty!
I always found that if you subtract the day of the month in days from the current date, you'll get the last day of the previous month. For example, on the 15th of the month, subtract 15 days, and you'll end up with the last day of the previous month:
SELECT (DATE_SUB(CURDATE(),INTERVAL DAYOFMONTH(CURDATE()) DAY))
If CURDATE() is 2012-05-05 then the above returns 2012-04-30.
So, to get Ads up to the last day of last month, do something like this:
SELECT *
FROM Ads
WHERE AdEnd <= (DATE_SUB(CURDATE(),INTERVAL DAYOFMONTH(CURDATE()) DAY))
This works if AdEnd is a DATE, but if it's a DATETIME, you'll just want to do less than the first of the month, so you subtract one less day to get the first of the month like this:
SELECT *
FROM Ads
WHERE AdEnd < (DATE_SUB(CURDATE(),INTERVAL DAYOFMONTH(CURDATE()) - 1 DAY))
Try date_diff, for example:
SELECT DATEDIFF('2008-11-30','2008-11-29') AS DiffDate
yields:
1
Related
I want to get records from last month's same day by keeping in mind for days which will be skipped and should be included in next day.
e.g. on 31st of march, february 31 does not exist so it should skip the query and also if i want to get records on april 30 it will give the results as required, but then on 1st of may, 31st of march will be skipped.
Currently, I am using
SELECT * FROM registrations WHERE orderdate = DATE_SUB(CURDATE(), INTERVAL 1 month)
How can I tackle this in mysql query?
Sorry if i am not able to communicate my query.
Consider the following logic, which retains a record only if subtracting one month did not result in an earlier day value:
SELECT *
FROM registrations
WHERE
orderdate = DATE_SUB(CURDATE(), INTERVAL 1 month) AND
DAY(orderdate) = DAY(DATE_SUB(CURDATE(), INTERVAL 1 month));
I'm trying to get data going back three months for monthly reports. I got to the point where i can get all the data. The problem is that it goes back 3 months based on the current date.
Example: If today is 7th of November it will give me the data up until the 7th of August.
I need it to give me the data going back three months but starting from the first of the month.
Example: today is the 7th of November, I'll need the data starting from the 1st of August.
Here is the code I'm using to get the data from three months back:
SELECT * FROM 'closed_wo_journal' WHERE date_time_stamp > CURDATE() - INTERVAL 3 MONTH
The date_trunc function is just what the doctor ordered:
SELECT *
FROM closed_wo_journal
WHERE date_time_stamp > DATE_TRUNC('MONTH', CURDATE() - INTERVAL '3 MONTH')
Here is logic to get the first of the month, three months ago:
SELECT j.*
FROM closed_wo_journal j
WHERE j.date_time_stamp >= ( CURDATE() - INTERVAL (1 - DAY(CURDATE()) DAY) ) - INTERVAL 3 MONTH )
How to get last day of the month in MySQL by providing month and year as input.
Similar example, To get last day of the month by date as input
Eg: SELECT LAST_DAY(?) as lastDate
Format:
Input month and year: '07/2015'
Output Result: day /*Eg: 30,31*/
I have tried with date Format
SELECT LAST_DAY(DATE_FORMAT('09/2015','%m/%Y')) as lastDate but it din't work
You just need to convert your values to a date. Here is one way:
select last_day(date(concat_ws('-', year, month, 1)))
Get last day of current month
SELECT LAST_DAY('2020-06-18') AS 'Result';
Get last day of next month
SELECT LAST_DAY(CURDATE() + INTERVAL 1 MONTH);
Get Last day of previous month
SELECT LAST_DAY(CURDATE() - INTERVAL 1 MONTH);
set #year:=2010;
set #month:=10;
select last_day(concat(#year,'-',#month,'-01')) as last_day;
SELECT DATE_FORMAT(LAST_DAY('2015-9-1'),'%d')
just add an any number date (day) in the query, and then get last day only using date format %d
I want to select all dates within the last week.
But not by simply counting the current date - 7, as all posts Ive come across suggest.
This is how I actually have it for the SUM now....
SELECT SUM(total) FROM payforms WHERE user_id = 1 GROUP BY WEEK(date)
This gives me a nice total...
But I want to retrieve all individual records within the last week.
So I can use a BETWEEN query....but how do I get it to look in the current week.
Example...
Tuesday I want it to only find values from Sun, Mon, Tue.
On wednesday, I want it to find Sun, Mon, Tue, Wed.
On Saturday it finds the whole previous week. etc
So to be more clear....
I dont want it to find last Monday, on a monday.
On Mondays it should only display Mondays, if you get what I mean.
Can I do this??
Thanks
select * from payforms where yearweek(date) = yearweek(now());
Although I think MySQL weeks start on Sunday
You can make use of the WEEKDAY date function, which returns the day index of the week.
If I understand your question correctly, you could use this query:
SELECT date, SUM(total)
FROM payforms
WHERE date >= CURDATE() - INTERVAL DAYOFWEEK(CURDATE())-1 DAY
GROUP BY date
Please see fiddle here.
This will SUM all totals for every day of the current week, starting on last Sunday. If you want it to start on last Monday you can use this:
WHERE date >= CURDATE() - INTERVAL WEEKDAY(CURDATE()) DAY
I did not find any of examples for MySQL - all of them were quite complicated.
How can I SELECT the first Sunday of the month?
So choose the first day of the month: 2012-01-01 (or whatever month and year you want).
Get the weekday index of the date. Indexes here are from 0 to 6.
Subtract that index from 6 and you will get how many days you need to add until the date is Sunday.
Add that amount of days to the chosen day.
SELECT DATE_ADD("2012-01-01 10:00:00", INTERVAL (6 - WEEKDAY("2012-01-01 10:00:00")) DAY);
Or:
SELECT DATE_ADD("2012-01-01", INTERVAL (6 - WEEKDAY("2012-01-01")) DAY);
So choose the first day of the month: 2021-08-01 (or whatever month and year you want).
SELECT ADDDATE( '2021-08-01' , MOD((8-DAYOFWEEK('2021-08-01')),7))