Getting days of the week from dateTime in SPARQL

Okay, SPARQL 1.1... (Note SPARQL 1.1 has floor. You can calculate a % b (modulo) using a - (floor(a/b) * b).)

Using the Gaussian function in @harschware's link, this should give an ID for the current day of the week:

(EDIT: formula now (i) adjusts for January, February, (ii) ensures positive result, (iii) doesn't assume a four digit year.)

PREFIX  xsd:  <http://www.w3.org/2001/XMLSchema#>
SELECT ?date ?dayID 
WHERE { 
    BIND (now() AS ?date)
    BIND (day(?date) AS ?day)
    BIND (month(?date) AS ?month)
    BIND (year(?date) AS ?year)

    ############################################
    ## Gaussian algorithm for day of the week ##

    # Subtract 1 to year if January or February
    BIND (IF(?month<=2, 1, 0) AS ?jf)
    BIND (?year - ?jf AS ?adjyear)

    #century and year
    BIND (floor(?adjyear/100) as ?c) #?c is the century
    BIND (?adjyear - (?c * 100) as ?y) #?y is the year

    # x2 = (month + 9) % 12 + 1
    BIND (?month+9 AS ?x1)
    BIND (xsd:integer(?x1 - floor(?x1/12)*12 + 1) AS ?x2)

    # body of formula
    BIND ((?day + floor((2.6 * ?x2) - 0.2) + ?y + floor(?y/4) + floor(?c/4) - (2 * ?c)) AS ?i)
    BIND (?i - (floor(?i/7) * 7) AS ?dayIDx) # ?i % 7

    # ensure result is positive
    BIND (xsd:integer(IF(?dayIDx < 0, ?dayIDx + 7 , ?dayIDx)) AS ?dayID)

    ############################################
}

?dayID gives you 0–6 where 0 = Sunday, 1 = Monday, etc.

You can test this here, with todays "day" here.

If you wanted to return names of the days, one quick trick would be to add triples of the following form to the store (assuming you have write access):

:Sunday :dayID 0 ; rdfs:label "Sunday"@en .
:Monday :dayID 1 ; rdfs:label "Monday"@en .
...

and then add:

?day :dayID ?dayID ; rdfs:label ?dayName .

to your query to retrieve the day name.

If you don't have write access, you could encode the day names into the query:

PREFIX  xsd:  <http://www.w3.org/2001/XMLSchema#>
SELECT ?date ?dayName
WHERE { 
    #calculate ?dayID as above

    BIND (
      IF(?dayID = 0, "Sunday", 
       IF(?dayID = 1, "Monday",
        IF(?dayID = 2, "Tuesday",
         IF(?dayID = 3, "Wednesday",
          IF(?dayID = 4, "Thursday",
           IF(?dayID = 5, "Friday",
            IF(?dayID = 6, "Saturday", "Unknown")
      )))))) AS ?dayName
    )
}

Which gives the answer here. (Thanks to @database_animal for pointing me to the if–then–else ladder pattern.)