MS Access is a database management system that allows you to store, manipulate, and query data. One of the common tasks you may encounter when working with data is to filter or group records by date or time intervals, such as weeks, months, or years.
In this article, we will show you how to identify if a week is within the last 5 weeks in MS Access, using the DatePart and DateDiff functions. We will also provide some examples and a table to illustrate the results.
The DatePart Function
The DatePart function returns a numeric value that represents a specific part of a given date or time value. For example, you can use the DatePart function to extract the year, month, day, hour, minute, second, or weekday from a date or time value.
The syntax of the DatePart function is:
DatePart(interval, date, [firstdayofweek], [firstweekofyear])
intervalis a string expression that specifies the part of the date or time value to return. It can be one of the following values:
| Interval | Description |
|---|---|
| yyyy | Year |
| q | Quarter |
| m | Month |
| y | Day of year |
| d | Day |
| w | Weekday |
| ww | Week |
| h | Hour |
| n | Minute |
| s | Second |
dateis a date or time expression that you want to evaluate.firstdayofweekis an optional numeric expression that specifies the first day of the week. It can be one of the following values:
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | Use the NLS API setting. |
| vbSunday | 1 | Sunday (default) |
| vbMonday | 2 | Monday |
| vbTuesday | 3 | Tuesday |
| vbWednesday | 4 | Wednesday |
| vbThursday | 5 | Thursday |
| vbFriday | 6 | Friday |
| vbSaturday | 7 | Saturday |
firstweekofyearis an optional numeric expression that specifies the first week of the year. It can be one of the following values:
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | Use the NLS API setting. |
| vbFirstJan1 | 1 | Start with the week in which January 1 occurs (default). |
| vbFirstFourDays | 2 | Start with the first week that has at least four days in the new year. |
| vbFirstFullWeek | 3 | Start with the first full week of the year. |
For example, the expression DatePart("ww", #2024/01/30#, 2, 2) returns the value 5, which means that January 30, 2024 is the 5th week of the year, assuming that the first day of the week is Monday and the first week of the year is the one that has at least four days in the new year.
The DateDiff Function
The DateDiff function returns a numeric value that represents the difference between two date or time values, measured in a specified interval. For example, you can use the DateDiff function to calculate the number of days, weeks, months, or years between two dates.
The syntax of the DateDiff function is:
DateDiff(interval, date1, date2, [firstdayofweek], [firstweekofyear])
intervalis a string expression that specifies the interval to use to calculate the difference betweendate1anddate2. It can be the same values as theintervalargument of the DatePart function.date1anddate2are date or time expressions that you want to compare.firstdayofweekandfirstweekofyearare optional numeric expressions that specify the first day of the week and the first week of the year. They can be the same values as thefirstdayofweekandfirstweekofyeararguments of the DatePart function.
For example, the expression DateDiff("w", #2024/01/01#, #2024/01/30#, 2) returns the value 21, which means that there are 21 weekdays between January 1, 2024 and January 30, 2024, assuming that the first day of the week is Monday.
How to Identify if a Week is Within the Last 5 Weeks
To identify if a week is within the last 5 weeks in MS Access, you can use a combination of the DatePart and DateDiff functions. The basic logic is to compare the week number and the year of the date value with the current week number and the current year, and check if the difference is less than or equal to 5.
The general formula is:
DateDiff("ww", DatePart("ww", date, firstdayofweek, firstweekofyear), DatePart("ww", Date(), firstdayofweek, firstweekofyear)) <= 5 And DatePart("yyyy", date) = DatePart("yyyy", Date())
dateis the date value that you want to check.firstdayofweekandfirstweekofyearare the numeric expressions that specify the first day of the week and the first week of the year. You can adjust them according to your preference.
For example, if you have a table named Transactions that has a field named TransactionDate, and you want to filter the records that have a transaction date within the last 5 weeks, assuming that the first day of the week is Monday and the first week of the year is the one that has at least four days in the new year, you can use the following query:
SELECT * FROM Transactions WHERE DateDiff("ww", DatePart("ww", TransactionDate, 2, 2), DatePart("ww", Date(), 2, 2)) <= 5 And DatePart("yyyy", TransactionDate) = DatePart("yyyy", Date())
This query will return the records that have a transaction date between December 25, 2023 and January 30, 2024, if the current date is January 30, 2024.
Examples and Table
To illustrate the results of the formula, we have created a table that shows some sample date values and whether they are within the last 5 weeks or not, based on the current date of January 30, 2024.
| Date | Week Number | Year | Within Last 5 Weeks |
|---|---|---|---|
| 2024/01/30 | 5 | 2024 | Yes |
| 2024/01/23 | 4 | 2024 | Yes |
| 2024/01/16 | 3 | 2024 | Yes |
| 2024/01/09 | 2 | 2024 | Yes |
| 2024/01/02 | 1 | 2024 | Yes |
| 2023/12/31 | 53 | 2023 | No |
| 2023/12/26 | 52 | 2023 | Yes |
| 2023/12/19 | 51 | 2023 | No |
| 2023/12/12 | 50 | 2023 | No |
| 2023/12/05 | 49 | 2023 | No |
As you can see, the formula correctly identifies the date values that are within the last 5 weeks, regardless of the month or year boundaries.
