todayofwk(X;Y)
  
  Convert a string, date+time, or date to a value corresponding to the day of the week. (Available as of version 18.02)
Syntax
todayofwk(X;Y)Input
| Argument | Type | Description | 
|---|---|---|
| X | any | A scalar value or the name of a column containing date-related values Valid values can be of the form: 
 | 
| Y | text | An optional argument specifying which date mode to use if the date has an
              ambiguous format. For example, the string "1/2/3" can be interpreted as January 2,
              2003 (using month/day/year format), or it can be interpreted as February 3, 2001
              (using year/month/day format). Valid values are: 
 The default value is  You can add the
                argument  You can also add an
                optional argument to tell  Valid values are: 
 | 
todayofwk(X;Y) can take a variety of input values and uses internal logic
        to parse the dates. For example, the function assumes that a 4-digit number represents a
        year, and that a month name or abbreviation (such as "Jan") is obviously a month. Any
        ambiguities are resolved using the setting of the Y argument. 
Return Value
Returns an integer value corresponding to the day of the week. If X
        contains a time stamp, it is discarded.
If the input is invalid and the Y argument includes
          'invalidna', the function returns a null value. If the input is invalid
        and the Y argument includes 'invaliderr', the function
        throws an error. In this way, todayofwk(X;Y) provides date validation if
        the input value is already in a date or date+time
        format.
For more information about date formats, see Dates and time.
Example
The following example converts strings containing dates in various formats to an integer
        corresponding to a day of the week. The Y argument is
          'mdymode', so that any ambiguous dates are parsed as month/day/year.
<table cols="date_text"> "31 Oct 2017" "February 3, 2016 04:45:01" "1/2/5" "15/3/12" "2021Q4" </table> <willbe name="date_todayofwk" value="todayofwk(date_text;'mdymode' 'allowyqm')"/>

The first date, 31 Oct 2017, is parsed correctly because there
        is no ambiguity about "Oct" being the month and 31 being the day. The second date,
          February 3, 2016 04:45:01, is parsed after dropping the time
        stamp. The third date, 1/2/5, is ambiguous, so the
          Y argument tells todayofwk to parse the date as
        month/day/year, or 01/02/2005.
          15/3/12, cannot be parsed as month/day/year because it is not
        a valid date, and returns a null value. 2021Q4 is converted to 6 (October
        1, the beginning of Q4, occurs on a Friday), because we used the 'allowyqm'
        option.
