Get members in multiple groups

You're talking about union() here, but I assume you really want the function intersection(), which takes two lists and returns only those elements that appear in both.  This of course could be daisychained like intersection(local!list1, intersection(local!list2, local!list3)), etc.

So how would I implement that across an unknown number of groups? Some kind of recursive expression? 

I was able to cook up this funcitonal workaround that sidesteps Appian's nearly-impossibly-sloppy handling of arrays-of-arrays that seem to want to actively prevent using intersection() inside reduce().  It does require creating a sub-rule, but allows us to use the inherent recursive functionality of reduce(), and not need to use our own recursion which is always riskier.

parent rule:

a!localVariables(
  
  local!allUsers: a!forEach(
    ri!groups,
    joinarray(
      a!groupMembers(
        group: fv!item,
        direct: true(),
        memberType: "USER"
      ).data,
      ";"
    )
  ),

  reduce(
    rule!TEST_manualIntersection,
    local!allUsers[1],
    ldrop(local!allUsers, 1)
  )
)

Sub rule (test_manualIntersection):

intersection(
  split(ri!textList1, ";"),
  split(ri!textList2, ";")
)

/* both RIs are non-array text variables */

hmm, almost works. I ran it on a set of groups with 2 members common among them and only got 1 of those two members back. All the other members were filtered out. I've been trying to get reduce to work, and having no luck there. 

seems to return only 1 no matter how many common users there are as far as I can tell. 

It seems to just grab the first user it encounters that exists in all the groups

urgh.  that happens for me too now that i test it more broadly.

the only other thing i can think of is to abandon intersection() and reduce() entirely and implement your own stuff.  the steps i'd thought through earlier would be along the lines of this:

1) iterate over all groups grabbing all members

2) make a single array comprising all unique usernames among all groups

3) iterate over the list of all usernames and for each one, loop over all original group lists checking that the current user is in all of them (if not, discard)

4) you should now end up with a list of all usernames in all groups provided.

This would likely get a *little* processing-intensive if you feed in an obscenely large number of groups, but then again the previous solution probably would too.

try this out:

a!localVariables(
  
  local!allUsers: a!forEach(
    ri!groups,
    a!groupMembers(
      group: fv!item,
      direct: true(),
      memberType: "USER"
    ).data
  ),
    
  local!uniqueUsers: rule!GCO_RULE_General_distinct(  /* helper rule that unions the input array against itself */
    touniformstring(a!flatten(local!allUsers))
  ),
  
  
  a!flatten(
    a!forEach(
      local!uniqueUsers,
      
      a!localVariables(
        local!currentUser: fv!item,
        if(
          and(
            a!forEach(
              local!allUsers,
              contains(touniformstring(fv!item), local!currentUser)
            )
          ),
          local!currentUser,
          {}
        )
      )
    )
  )
)

I was thinking of something like this, but I am not entirely sure how to use all() with contains(). It's so hard getting appian to do what you want it to do when you're used to programming in other languages. Let me try the second one 

a!localVariables(
  local!allUsers: a!forEach(
    items: ri!groupList,
    expression: a!groupMembers(
      group: fv!item,
      direct: true(),
      memberType: "USER"
    ).data
  ),
  local!uniqueUsers: a!forEach(
    items: local!allUsers[1],
    expression: if(
      all(fn!contains, local!allUsers, fv!item),
      fv!item,
      null()
    )
  ),
  local!uniqueUsers
)

The example I posted above should be working, FWIW.  I don't think "all()" is needed.

Almost works, but if we don't have any valid users across those groups, I'm getting a big list of empty lists. 

So if I have the following groups:

{a,b,c}

{b,c,d}

{z,y,x}

{a,b,c,d}

the result I'm getting is something like {{},{},{},{},{},{},{}} (I'm assuming it's one for each unique value, but can't tell for sure)

that's a known side-effect of a!forEach when it returns all empty lists.  that can be fixed by wrapping the whole enchilada in a!flatten.  i'll update my example above to reflect this.

ooh, that works! Thanks!

ooh, that works! Thanks!