Ajama commented on my article Oracle Question and Answer Site asking how to write a SQL function to return the calendar week in a month.

Basically if your week starts on a Sunday, anything leading up to the first Sunday would be week 1, then the first complete week, Sunday through Saturday would be week 2, then week 3, etc. If the month happens to start on a Sunday week 1 would be a complete week.

If you are unfamiliar with manipulating dates in Oracle, you may want to check out my other articles on Oracle, SQL, Dates and Timestamps and Performing Math on Oracle Dates.

Starting simple I worked out how to find the first day of the month. The `TRUNC`

function will allow you to truncate a date to a certain precision, so I started by truncating the date to the month. Today is February 1, so since we’re using `sysdate`

for some testing we coincidentally get back February 1.

`SQL> select trunc(sysdate, 'MM') from dual;`

```
```

`TRUNC(SYS`

---------

01-FEB-06

Now we have a starting point. From here I used the `NEXT_DAY`

function to find the beginning of the first week of the month. This is found by looking at the first of the month and the 6 days leading up to it. Since `next_day`

excludes the day it is calculating from we subtract 7 for the start date.

`SQL > select next_day(trunc(sysdate, 'MM') - 7, 'Sunday') from dual;`

```
```

`NEXT_DAY(`

---------

29-JAN-06

Now we have established the beginning of the first week in the month. The first week starts at the end of the previous month as it is only partially in February. With this date in mind we can now calculate how many days we are from then:

`SQL> select sysdate - next_day(trunc(sysdate, 'MM') - 7, 'Sunday') from dual;`

```
```

`SYSDATE-NEXT_DAY(TRUNC(SYSDATE,'MM')-7,'SUNDAY')`

------------------------------------------------

3.55945602

This tells us we are 3.559 days from the beginning of the first week in the month. To express that in weeks we simply divide by 7.

`SQL> select (sysdate - next_day(trunc(sysdate, 'MM') - 7, 'Sunday'))/7 from dual;`

```
```

`(SYSDATE-NEXT_DAY(TRUNC(SYSDATE,'MM')-7,'SUNDAY'))/7`

----------------------------------------------------

.508627646

This tells us that we are about .5 weeks from that first Sunday. We don’t care about the fractional part so we can use the other form of the `TRUNC`

function to to truncate this to a whole number. That would put us in the 0th week of the month, but since we want to start counting at 1 we add 1 to the number.

`SQL> select trunc(((sysdate - next_day(trunc(sysdate, 'MM') - 7, 'Sunday'))/7),0) + 1`

from dual;

```
```

`TRUNC(((SYSDATE-NEXT_DAY(TRUNC(SYSDATE,'MM')-7,'SUNDAY'))/7),0)+1`

-----------------------------------------------------------------

1

This tells us accurately that right now we’re in the first week of the month. Now we want to try this logic out with some other dates. It’s easiest to do that if we just create a function for it.

`CREATE OR REPLACE FUNCTION week_in_month`

(check_date DATE DEFAULT sysdate, week_start CHAR DEFAULT 'Sunday')

RETURN number

IS

week_number NUMBER;

BEGIN

select trunc(((check_date - next_day(trunc(check_date, 'MM') - 7, week_start))/7),0) + 1 into week_number from dual;

RETURN week_number;

END;

/

Here I’ve created a function week_in_month to execute this SQL with two parameters. The first, `check_date`

can be any Oracle date. The second parameter, `week_start`

is the day that your week starts. This must be spelled out, like ‘Sunday’, ‘Monday’, etc. The function will return the week number from start of month.

If the check_date is not specified, the current date and time will be used. If the week_start is not specified we’ll default to Sunday.

Now let’s try our function without parameters.

`SQL> select week_in_month from dual;`

```
```

`WEEK_IN_MONTH`

-------------

1

The function correctly identifies that we’re in week 1 of the month by our original definition. Now we’ll look at a few other examples. Remember, where we have not specified a starting day our function will assume Sunday.

`SQL> select week_in_month(to_date('01/29/06','MM/DD/YY')) from dual;`

```
```

`WEEK_IN_MONTH(TO_DATE('01/29/06','MM/DD/YY'))`

---------------------------------------------

5

Here we see that 1/29/06 was in the 5th week of January.

`SQL> select week_in_month(to_date('02/06/06','MM/DD/YY')) from dual;`

```
```

`WEEK_IN_MONTH(TO_DATE('02/06/06','MM/DD/YY'))`

---------------------------------------------

2

This shows that February 6 will be in the second week in February.

`SQL> select week_in_month(to_date('02/06/06','MM/DD/YY'), 'Wednesday') from dual;`

```
```

`WEEK_IN_MONTH(TO_DATE('02/06/06','MM/DD/YY'),'WEDNESDAY')`

---------------------------------------------------------

1

Here we have specified that our week starts on Wednesday, and based on that February 6th will be in the first week in February.

`SQL> select week_in_month(to_date('04/30/06','MM/DD/YY')) from dual;`

```
```

`WEEK_IN_MONTH(TO_DATE('04/30/06','MM/DD/YY'))`

---------------------------------------------

6

This is a somewhat rare instance, and it may look like there’s a problem with the logic, but because of how April of 2006 falls, the 30th would actually be in the 6th week of the month based on our criteria.

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