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SQL date range

Posted on 2014-07-22
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I am very new to SQL.  I need a SQL statement for the proper way to do a date range.  Below is what is currently in place.  I need it to run for the previous weeks Sunday-Saturday

WHERE
B.CANCEL.DATE = CURRENT_DATE

My Best Guess

WHERE
B.CANCEL.DATE = >=CURRENT_DATE-8
AND B.CANCEL.DATE = <+CURRENT_DATE-2
0
Question by:garyrobbins
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LVL 8

Expert Comment

ID: 40211875
Here is some logic to find last last weeks Saturday and Sunday... Questions still exists about data scenarios that can occur. I choose to pick a Saturday and logic backs up to prior Saturday (just guessing)

declare @date1 datetime;
set @date1='2014-07-26'
--Saturday Prev
select DATEADD(WEEK, DATEDIFF(WEEK, 0, @date1), 0) + 5  - 7
generates 2014-07-19
--Sunday Prev
select DATEADD(WEEK, DATEDIFF(WEEK, 0, @date1), 0) + 6 - 7
generates 2014-07-20

Change date and watch logic and then use in where clause or select? as needed
0

LVL 48

Accepted Solution

PortletPaul earned 500 total points
ID: 40213929
I am assuming you use MySQL.
------
The simple answer is you were almost correct, a better method is:

WHERE    B.CANCEL.DATE >= CURRENT_DATE - 8
AND B.CANCEL.DATE <  CURRENT_DATE - 1 /* just less than, but note the minus one */

the safest way to define a date range is to use a combination of  >= with <

e.g. let's say you want everything for February 2014, this is the way to do it:

WHERE    B.CANCEL.DATE >= '2014-02-01' /* >= 1st of Feb */
AND B.CANCEL.DATE <  '2014-03-01' /* but less than the first of March */

for more on this topic please see: "Beware of Between"

------------------------
Now, automating "the previous weeks Sunday-Saturday"

This is a bit more complex because we must use date functions. But remember we want to use the method above, which will mean:

WHERE B.CANCEL.DATE >= "previous week's Sunday"
AND B.CANCEL.DATE <  "the Sunday after that"

i.e. because we use less than we also use Sunday, not Saturday, for the end date.
----

You already know a date function: CURRENT_DATE

we need 3 more date functions:
WEEKDAY() This function tells us which day of week using integers  (0 = Monday, 1 = Tuesday, â€¦ 6 = Sunday).
DATE_ADD() this allows us to add time intervals to a date/time value
DATE_SUB() which does the opposite (subtracts)

plue, we also need an IF() function so we can test if a date is Sunday then deduct 7 otherwise deduct a different value.

Putting these together we can get "previous week's Sunday" using this:
DATE_SUB( CURRENT_DATE , INTERVAL (if(WEEKDAY( CURRENT_DATE )=6,7,WEEKDAY( CURRENT_DATE )+8)) DAY)

What this combination does is if current_date weekday is 6 (sunday) deduct 7, otherwise deduct the weekday plus 8 from the current date
The next combination does the same, plus adds 7 days to that result.

and, for "the Sunday after that" using:
DATE_ADD(DATE_SUB( CURRENT_DATE , INTERVAL (if(WEEKDAY( CURRENT_DATE )=6,7,WEEKDAY( CURRENT_DATE )+8)) DAY), INTERVAL 7 DAY)

So, to automate your where clause you could use:
``````where B.CANCEL.DATE >= DATE_SUB( CURRENT_DATE , INTERVAL (if(WEEKDAY( CURRENT_DATE )=6,7,WEEKDAY( CURRENT_DATE )+8)) DAY)
and B.CANCEL.DATE <  DATE_ADD(DATE_SUB( CURRENT_DATE , INTERVAL (if(WEEKDAY( CURRENT_DATE )=6,7,WEEKDAY( CURRENT_DATE )+8)) DAY), INTERVAL 7 DAY)
``````
Here's some further information to see it working:
``````**MySQL 5.6.6 m9 Schema Setup**:

CREATE TABLE tbl
(`dt` datetime)
;

INSERT INTO tbl
(`dt`)
VALUES
('2014-07-13 00:00:00'),
('2014-07-14 00:00:00'),
('2014-07-15 00:00:00'),
('2014-07-16 00:00:00'),
('2014-07-17 00:00:00'),
('2014-07-18 00:00:00'),
('2014-07-19 00:00:00'),
('2014-07-20 00:00:00'),
('2014-07-21 00:00:00'),
('2014-07-22 00:00:00'),
('2014-07-23 00:00:00'),
('2014-07-24 00:00:00'),
('2014-07-25 00:00:00'),
('2014-07-26 00:00:00'),
('2014-07-27 00:00:00'),
('2014-07-28 00:00:00'),
('2014-07-29 00:00:00'),
('2014-07-30 00:00:00'),
('2014-07-31 00:00:00'),
('2014-08-01 00:00:00'),
('2014-08-02 00:00:00'),
('2014-08-03 00:00:00'),
('2014-08-04 00:00:00')
;

**Query 1**:

select
date_format(dt ,'%a, %d/%M/%Y') as dt
, if(WEEKDAY(dt)=6,7,WEEKDAY(dt)+8) as deduct_days
, date_format(DATE_SUB(dt, INTERVAL (if(WEEKDAY(dt)=6,7,WEEKDAY(dt)+8)) DAY) ,'%a, %d/%M/%Y')                           as prev_week_sun
, date_format(DATE_ADD(DATE_SUB(dt, INTERVAL (if(WEEKDAY(dt)=6,7,WEEKDAY(dt)+8)) DAY), INTERVAL 7 DAY) ,'%a, %d/%M/%Y') as sun_after_that
from tbl

**[Results][2]**:

|                  DT | DEDUCT_DAYS |     PREV_WEEK_SUN |      SUN_AFTER_THAT |
|---------------------|-------------|-------------------|---------------------|
|   Sun, 13/July/2014 |           7 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Mon, 14/July/2014 |           8 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Tue, 15/July/2014 |           9 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Wed, 16/July/2014 |          10 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Thu, 17/July/2014 |          11 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Fri, 18/July/2014 |          12 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Sat, 19/July/2014 |          13 | Sun, 06/July/2014 |   Sun, 13/July/2014 |
|   Sun, 20/July/2014 |           7 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Mon, 21/July/2014 |           8 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Tue, 22/July/2014 |           9 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Wed, 23/July/2014 |          10 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Thu, 24/July/2014 |          11 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Fri, 25/July/2014 |          12 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Sat, 26/July/2014 |          13 | Sun, 13/July/2014 |   Sun, 20/July/2014 |
|   Sun, 27/July/2014 |           7 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
|   Mon, 28/July/2014 |           8 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
|   Tue, 29/July/2014 |           9 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
|   Wed, 30/July/2014 |          10 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
|   Thu, 31/July/2014 |          11 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
| Fri, 01/August/2014 |          12 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
| Sat, 02/August/2014 |          13 | Sun, 20/July/2014 |   Sun, 27/July/2014 |
| Sun, 03/August/2014 |           7 | Sun, 27/July/2014 | Sun, 03/August/2014 |
| Mon, 04/August/2014 |           8 | Sun, 27/July/2014 | Sun, 03/August/2014 |

[1]: http://sqlfiddle.com/#!9/a3577/1
``````
0

LVL 48

Expert Comment

ID: 40214041
It is possible to make those calculations into your own functions, e.g.
``````    CREATE FUNCTION `PrevWkStartingSunday`() RETURNS DATETIME

/* determines Sunday that begins the previous week
if current_date weekday is 6 (sunday) deduct 7,
otherwise deduct the weekday plus 8 from the current date */

RETURN DATE_SUB(CURRENT_DATE, INTERVAL (IF(WEEKDAY(CURRENT_DATE)=6,7,WEEKDAY(CURRENT_DATE)+8)) DAY)
;

CREATE FUNCTION `PrevWkEndingSunday`() RETURNS DATETIME

/* determines Sunday that begins the previous week
if current_date weekday is 6 (sunday) deduct 7,
otherwise deduct the weekday plus 8 from the current date
then add 7 days to get the following Sunday */

RETURN DATE_ADD(DATE_SUB( CURRENT_DATE , INTERVAL (if(WEEKDAY( CURRENT_DATE )=6,7,WEEKDAY( CURRENT_DATE )+8)) DAY), INTERVAL 7 DAY)
;

**Query 1**:

select
date_format(dt ,'%a, %d/%M/%Y') as dt
, PrevWkStartingSunday()
, PrevWkEndingSunday()
from tbl
where dt >= PrevWkStartingSunday()
and dt <  PrevWkEndingSunday()

**[Results][2]**:

|                DT |      PREVWKSTARTINGSUNDAY() |        PREVWKENDINGSUNDAY() |
|-------------------|-----------------------------|-----------------------------|
| Sun, 13/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Mon, 14/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Tue, 15/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Wed, 16/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Thu, 17/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Fri, 18/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |
| Sat, 19/July/2014 | July, 13 2014 00:00:00+0000 | July, 20 2014 00:00:00+0000 |

[1]: http://sqlfiddle.com/#!9/a28ce/1
``````
0

Author Comment

ID: 40230225
I've requested that this question be deleted for the following reason:

0

LVL 48

Expert Comment

ID: 40230227
There was no suitable feedback either. Complete silence from your end.

Was it MySQL? I don't know to this day.

Take a look, there's 2 answers, but no indication from you why they are not suitable.
0

Author Closing Comment

ID: 40231830
Thank You very much for the Feedback,

I didn't realize the importance of selecting and grading a solution accurately.  Still learning the appropriate etiquette on this sight, I was just trying to clear the Question.  I will pay attention to that in the future.  Again Thank You for your time and feed back it was very thorough.
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