I need to get number of days between 2 dates, a given one and current date.
But in pure SQL, I mean without usign functions, it is possible?
For exaple
SELECT days (t.givenDate) - days (current date) FROM table t
Have you any idea?
Thaks a lot.
The built-in function is datediff(). The equivalent for the above is:
SELECT datediff(t.givenDate, curdate()) FROM table t;
Normally, givenDate would be in the past and you would want the arguments in the other order.
Related
How do I extract the month and date from a mySQL date and compare it to another date?
I found this MONTH() but it only gets the month. I looking for month and year.
in Mysql Doku:
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_extract
SELECT EXTRACT( YEAR_MONTH FROM `date` )
FROM `Table` WHERE Condition = 'Condition';
While it was discussed in the comments, there isn't an answer containing it yet, so it can be easy to miss.
DATE_FORMAT works really well and is flexible to handle many different patterns.
DATE_FORMAT(date,'%Y%m')
To put it in a query:
SELECT DATE_FORMAT(test_date,'%Y%m') AS date FROM test_table;
If you are comparing between dates, extract the full date for comparison. If you are comparing the years and months only, use
SELECT YEAR(date) AS 'year', MONTH(date) AS 'month'
FROM Table Where Condition = 'Condition';
SELECT * FROM Table_name Where Month(date)='10' && YEAR(date)='2016';
You may want to check out the mySQL docs in regard to the date functions. http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
There is a YEAR() function just as there is a MONTH() function. If you're doing a comparison though is there a reason to chop up the date? Are you truly interested in ignoring day based differences and if so is this how you want to do it?
There should also be a YEAR().
As for comparing, you could compare dates that are the first days of those years and months, or you could convert the year/month pair into a number suitable for comparison (i.e. bigger = later). (Exercise left to the reader. For hints, read about the ISO date format.)
Or you could use multiple comparisons (i.e. years first, then months).
I want to get the previous month date for specific dates in SQL. For example: 6.21.19 has a previous month date of 5.21.19.
I am just trying to get comps from this.
MONTH( curdate() ) -1
I need to return the previous month date.
Welcome to the board Arie. Judging from your question and responses, you need a range of dates and their prior month relations. The easiest way would be for all of the dates you need to look up to be in a table, then the answers provided so far would work. Since that doesn't appear to be the case, I'm guessing you are creating date ranges on the fly.
So lets assume you need exactly the data shown in your example, there are two parts to this, first you need to get a list of days that you want to look up, then you need to get the day in the prior month. There are lots of ways to get a sequence of days, but for simplicity I'll use a recursive CTE. Once I have the date range, I'll just select the dates and their prior month date as well.
with Date_CTE as (select cast('6/1/2019' as datetime) as repDate
union all
select dateadd(day, 1, repdate) as repDate
from Date_CTE
where repDate < '06/07/2019')
select repDate, dateadd(month, -1, repDate) as PriorDate
from Date_CTE
CTEs are helpful functions and you can get more details on them here, but it's worth noting there are many ways to do this. Hope this gets you pointed in the right direction.
SELECT yourDateColumn, yourDateColumn-interval 1 month as prevMonthDate
I have this problem if anyone can help.
There is a field (date) in my table (table1) that is a date in the format 3/31/1988 (M/D/y), and my necessity is to define how many days have passed since that date.
I have tried to give this instruction
SELECT DATEDIFF(CURDATE(), date) AS days
FROM table1
But it gives back 'null' and I think this happens because the two date formats are different (CURDATE() is YMD.....
Is it correct? can anyone help me?
Thank you in advance
You can use STR_TO_DATE():
SELECT DATEDIFF(CURDATE(),STR_TO_DATE(date, '%m/%d/%Y')) AS days
FROM table1
SQLFiddle Demo
Your DATE field should have DATE or DATETIME format to be used as DATEDIFF argument correctly.
Also DATE is MySQL keyword and I am not sure that you can use it as valid field name.
You can use this for accurate result
SELECT DATEDIFF(CURDATE(), DATE_FORMAT(FROM_UNIXTIME(UNIX_TIMESTAMP(`date`)), '%Y-%m-%d')) AS days FROM `table1`
If you want to consider results without - signs that you have to follow parameters position as below :
SELECT DATEDIFF(Big_Date,Small_Date) AS days FROM table1.
positive results e.g 5 (with no sign), if you place a Small date as the first parameter then it will results minus sign e.g -5.
If I have MySQL query like this, summing word frequencies per week:
SELECT
SUM(`city`),
SUM(`officers`),
SUM(`uk`),
SUM(`wednesday`),
DATE_FORMAT(`dateTime`, '%d/%m/%Y')
FROM myTable
WHERE dateTime BETWEEN '2011-09-28 18:00:00' AND '2011-10-29 18:59:00'
GROUP BY WEEK(dateTime)
The results given by MySQL take the first value of column dateTime, in this case 28/09/2011 which happens to be a Saturday.
Is it possible to adjust the query in MySQL to show the date upon which the week commences, even if there is no data available, so that for the above, 2011-09-28 would be replaced with 2011/09/26 instead? That is, the date of the start of the week, being a Monday. Or would it be better to adjust the dates programmatically after the query has run?
The dateTime column is in format 2011/10/02 12:05:00
It is possible to do it in SQL but it would be better to do it in your program code as it would be more efficient and easier. Also, while MySQL accepts your query, it doesn't quite make sense - you have DATE_FORMAT(dateTime, '%d/%m/%Y') in select's field list while you group by WEEK(dateTime). This means that the DB engine has to select random date from current group (week) for each row. Ie consider you have records for 27.09.2011, 28.09.2011 and 29.09.2011 - they all fall onto same week, so in the final resultset only one row is generated for those three records. Now which date out of those three should be picked for the DATE_FORMAT() call? Answer would be somewhat simpler if there is ORDER BY in the query but it still doesn't quite make sense to use fields/expressions in the field list which aren't in GROUP BY or which aren't aggregates. You should really return the week number in the select list (instead of DATE_FORMAT call) and then in your code calculate the start and end dates from it.
I have several rows in a table, each containing a start date and an end date. The user has a checkbox for each month of the year. I need to determine which rows contain a date range that includes any of the user's chosen months.
It's easy to check the start & end months by, for example, MONTH(start_date) IN ($month_list), but this approach won't match any months between the two dates.
So I suppose what I'm asking is: is there a way of obtaining the inclusive months from a date range purely in SQL?
I assume you would want to include data rows where the date range spans or intersects with the selected periods - in which case, I'd shove the user selected periods into a table and do a fuzzy join, something like.....
SELECT DISTINCT at.*
FROM a_table at, user_periods up
WHERE at.start_date<=up.end_date
AND at.end_date>=up.start_date
AND up.trans_id=$SOME_VAR
(the trans_id just allows the table to be used for multiple operations)
To minimise the effort here, the user_periods table should have an index on start_date and end_date, and similar for a_table.
Can something like this help?
WHERE
MONTH(start_date) < MONTH_YOU_ARE_CHECKING and
MONTH() > MONTH_YOU_ARE_CHECKING
If you need to check all at once you can do a list of all the months and after delete from the list the month that the user choose, and after compare against the list. It will be better with a pseudocode example :)
MONTHS = 1,2,3,4,5,6,7,8,9,10,11,12
USER_SELECTED_MONTHS= 1,6,8,9,12
LIST_TO CHECK = 2,3,4,5,7,10,11
so, now you can do:
MONTH(start_date) NOT IN (2,3,4,5,7,10,11)
What do you think, could it help you?
regards