I need to select all data from a table 'followup' between followup_date and first of next month (including 1rst of next month). The format of my date in DB (obtained from an API) is d-m-Y.
The follow-up date for example is: 18-07-2020.
I have the following query:
SELECT * from followup
WHERE DATEDIFF(STR_TO_DATE(`followup_date`,'%d-%m-%Y'), DATE_FORMAT(CURDATE() + INTERVAL 1 MONTH,'%Y-%m-01'))- 1 < 0 ;
I am getting -15 days as difference and getting records correctly, including 1rst. Is the query correct and efficient and will it work correctly for all months.
Requesting suggestions from experts for improvements, if any.
You should convert your followup_date column to a DATE type. You can then make your query sargable by removing the function calls on followup_date and simply comparing it with the target date:
SELECT * from followup
WHERE `followup_date` <= DATE_FORMAT(CURDATE() + INTERVAL 1 MONTH,'%Y-%m-01')
I suspect adding one day to LAST_DAY(CURDATE()) might be more efficient:
SELECT * from followup
WHERE `followup_date` <= LAST_DAY(CURDATE()) + INTERVAL 1 DAY
Demo on SQLFiddle
I have a MySQL DB table with multiple date type fields. I need to do different SELECT queries on this table but I am not sure which way is the best to find records from the same month.
I know I can do the following:
SELECT *
FROM table
WHERE MONTH(somedate) = 5
AND YEAR(somedate) = 2015
But I keep reading that isn't efficient and that I should go with using actual dates, i.e.
SELECT *
FROM table
WHERE somedate BETWEEN '2015-05-01' AND '2015-05-31'
However, all I would have is the month and the year as variables coming in from PHP. How do I easily and quickly calculate the last day of the month if I go with second option?
Don't calculate the last day of the month. Calculate the first day of the next month instead.
Your query can be like this
WHERE t.mydatetimecol >= '2015-05-01'
AND t.mydatetimecol < '2015-05-01' + INTERVAL 1 MONTH
Note that we're doing a less than comparison, not a "less than or equal to"... this is very convenient for comparing TIMESTAMP and DATETIME columns, which can include a time portion.
Note that a BETWEEN comparison is a "less than or equal to". To get a comparison equivalent to the query above, we'd need to do
WHERE t.mydatetimecol
BETWEEN '2015-05-01' AND '2015-05-01' + INTERVAL 1 MONTH + INTERVAL -1 SECOND
(This assumes that the resolution of DATETIME and TIMESTAMP is down to a second. In other databases, such as SQL Server, the resolution is finer than a second, so there we'd have the potential of missing a row with value of '2015-05-31 23:59:59.997'. We don't have a problem like that with the less than the first day of the next month comparison... < '2015-06-01'
No need to do the month or date math yourself, let MySQL do it for you. If you muck with adding 1 to the month, you have to handle the rollover from December to January, and increment the year. MySQL has all that already builtin.
date('t', strtotime("$year-$month-01")) will give days in the month
The title might be a bit misleading, but what I want is:
SELECT * FROM table ORDER BY pid ASC
And in one of the columns I have a DATE(). I want to compare the current date (not time) and return how many days are left till that date. Let's say the date is 2013-04-20 and today's date is 2013-04-16 I don't want to get any data if it's < current date. If it is I want it returned in days.
I've been looking around here and I've found no way to do it, and I can't for the love of me figure it out.
If you're looking for the difference between two date you can use the GETDATE function in MS SQL
SELECT DATEDIFF(DD, DateOne, DateTwo) FROM TABLE
This will return the difference in number of days between the two dates.
If you only want rows where the date field is less than or equal to today's date you can use:
SELECT DATEDIFF(DD, DateField, GETDATE())
FROM TableName
WHERE DateField <= GETDATE()
If you're using MySQL you can use DATEDIFF()
SELECT
DATEDIFF(NOW(), date_column) AS days_diff
FROM
tablename
Get the difference between two dates (ANSI SQL)
select the_date_column - current_date as days_left
from the_table
where the_date_column - current_date <= 4;
SQLFiddle: http://sqlfiddle.com/#!12/3148d/1
I store a date in my database as a string like this:
03/08/2013 --> 8th of march
I'm trying to select only the rows that are the same day as the current day:
SELECT * FROM wp_aerezona_booking WHERE DATE_SUB(CURDATE(),INTERVAL 1
DAY) <= STR_TO_DATE(date, '%m/%d/%Y')
The above is what I tried, but it is returning a lot of results and should only return 1.
This should work already:
SELECT * FROM wp_aerezona_booking
WHERE STR_TO_DATE('03/08/2013', '%m/%d/%Y') = CURDATE();
By using the DATE_SUB you are subtracting1 day from the current day. You're not looking at today but yesterday. Also the <= makes you look at yesterday and all days before that.
Then you don't want <=, but you want =. The former will get all results if date is less than or equal to yesterday's date. I'm not sure that you even want the DATE_SUB either.
If you want the same date as today's date then you have to use "=" operator with.
SELECT *
FROM wp_aerezona_booking
WHERE STR_TO_DATE(date, '%m/%d/%Y')= CURDATE()
I want to get first day of every corresponding month of current year. For example, if user selects '2010-06-15', query demands to run from '2010-06-01' instead of '2010-06-15'.
Please help me how to calculate first day from selected date. Currently, I am trying to get desirable using following mysql select query:
Select
DAYOFMONTH(hrm_attendanceregister.Date) >=
DAYOFMONTH(
DATE_SUB('2010-07-17', INTERVAL - DAYOFMONTH('2010-07-17') + 1 DAY
)
FROM
hrm_attendanceregister;
Thanks
Is this what you are looking for:
select CAST(DATE_FORMAT(NOW() ,'%Y-%m-01') as DATE);
You can use the LAST_DAY function provided by MySQL to retrieve the last day of any month, that's easy:
SELECT LAST_DAY('2010-06-15');
Will return:
2010-06-30
Unfortunately, MySQL does not provide any FIRST_DAY function to retrieve the first day of a month (not sure why). But given the last day, you can add a day and subtract a month to get the first day. Thus you can define a custom function:
DELIMITER ;;
CREATE FUNCTION FIRST_DAY(day DATE)
RETURNS DATE DETERMINISTIC
BEGIN
RETURN ADDDATE(LAST_DAY(SUBDATE(day, INTERVAL 1 MONTH)), 1);
END;;
DELIMITER ;
That way:
SELECT FIRST_DAY('2010-06-15');
Will return:
2010-06-01
There is actually a straightforward solution since the first day of the month is simply today - (day_of_month_in_today - 1):
select now() - interval (day(now())-1) day
Contrast that with the other methods which are extremely roundabout and indirect.
Also, since we are not interested in the time component, curdate() is a better (and faster) function than now(). We can also take advantage of subdate()'s 2-arity overload since that is more performant than using interval. So a better solution is:
select subdate(curdate(), (day(curdate())-1))
This is old but this might be helpful for new human web crawlers XD
For the first day of the current month you can use:
SELECT LAST_DAY(NOW() - INTERVAL 1 MONTH) + INTERVAL 1 DAY;
You can use EXTRACT to get the date parts you want:
EXTRACT( YEAR_MONTH FROM DATE('2011-09-28') )
-- 201109
This works well for grouping.
You can use DATE_FORMAT() function in order to get the first day of any date field.
SELECT DATE_FORMAT(CURDATE(),'%Y-%m-01') as FIRST_DAY_CURRENT_MONTH
FROM dual;
Change Curdate() with any other Date field like:
SELECT DATE_FORMAT(purchase_date,'%Y-%m-01') AS FIRST_DAY_SALES_MONTH
FROM Company.Sales;
Then, using your own question:
SELECT *
FROM
hrm_attendanceregister
WHERE
hrm_attendanceregister.Date) >=
DATE_FORMAT(CURDATE(),'%Y-%m-01')
You can change CURDATE() with any other given date.
There are many ways to calculate the first day of a month, and the following are the performance in my computer (you may try this on your own computer)
And the winner is LAST_DAY(#D - interval 1 month) + interval 1 day
set #D=curdate();
select BENCHMARK(100000000, subdate(#D, (day(#D)-1))); -- 33 seconds
SELECT BENCHMARK(100000000, #D - INTERVAL (day(#D) - 1) DAY); -- 33 seconds
SELECT BENCHMARK(100000000, cast(DATE_FORMAT(#D, '%Y-%m-01') as date)); -- 29 seconds
SELECT BENCHMARK(100000000, LAST_DAY(#D - interval 1 month) + interval 1 day); -- 26 seconds
I'm surprised no one has proposed something akin to this (I do not know how performant it is):
CONCAT_WS('-', YEAR(CURDATE()), MONTH(CURDATE()), '1')
Additional date operations could be performed to remove formatting, if necessary
use date_format method and check just month & year
select * from table_name where date_format(date_column, "%Y-%m")="2010-06"
SELECT LAST_DAY(date) as last_date, DATE_FORMAT(date,'%Y-%m-01') AS fisrt_date FROM table_name
date=your column name
The solutions that use last_day() and then add/subtract a month and a day are not interchangeable.
Example:
date_sub(date_add(last_day(curdate()), interval 1 day), interval 3 month)
always works for any supplied number of months you want to go back
date_add(date_sub(last_day(now()), interval 3 month), interval 1 day)
will fail in some cases, for instance if your current month has 30 days and the month you're subtracting back to (and then adding a day) has 31.
date_add(subdate(curdate(), interval day(?) day), interval 1 day)
change the ? for the corresponding date
This works fine for me.
date(SUBDATE("Added Time", INTERVAL (day("Added Time") -1) day))
** replace "Added Time" with column name
Use Cases:
If you want to reset all date fields except Month and Year.
If you want to retain the column format as "date". (not as "text" or "number")
Slow (17s):
SELECT BENCHMARK(100000000, current_date - INTERVAL (day(current_date) - 1) DAY);
SELECT BENCHMARK(100000000, cast(DATE_FORMAT(current_date, '%Y-%m-01') as date));
If you don't need a date type this is faster:
Fast (6s):
SELECT BENCHMARK(100000000, DATE_FORMAT(CURDATE(), '%Y-%m-01'));
SELECT BENCHMARK(100000000, DATE_FORMAT(current_date, '%Y-%m-01'));
select big.* from
(select #date := '2010-06-15')var
straight_join
(select * from your_table where date_column >= concat(year(#date),'-',month(#date),'-01'))big;
This will not create a full table scan.