Get common values from more than two arrays

Question

Hi , 
 I have list of arrays coming from foreach, now i want unique elements for the list of arrays. 
 for example 
 localvariables( 
 local!arrays:foreach( 
 items:&#123;5,6,7&#125;, 
 expresssion:enumerate(fv!item)+1 
 ) 
 local!arrays 
 ). 
 form this output,  i want unique elements from the list of arrays. 
 can anyone help me in this? 
 intersection is working only for separate arrays stored in each local variable but not for arrays executed in foreach.

shoaib.akhtar · Answer

if you are looking for distinct values this could be the possible solution. 
 a!localVariables(

 local!arrays:a!forEach(

 items:{5,6,7},

 expression:enumerate(fv!item)+1

 ),

 rule!APN_distinct(
 a!flatten(local!arrays)
 )

)

Mike Schmitt · Answer

You don't want "unique values" then, you want "values appearing in all". "Unique Values" would be returned by a union() operation and in this case would return all values from 1 - 7. 
 AFAIK intersection() accepts any number of arrays. 
 Additionally it should be noted that when you create an Array of Arrays like the initial Elements list, Appian doesn't handle this very intuitively in some cases, often times acting more as if you're referring to a flattened list instead of calling individual elements. For example if we try to union() just the 1st and 3rd arrays, we get something somewhat nonsensical like this: 
 I notice that if you wrap each element in a!flatten(), it behaves better - maybe because it strips out the "list of any type" typecasting that the internal arrays tend to get (which tends to confuse things in cases like this) - receiving this output instead: 
 Extending this to your need for the intersection() function, it seems to correct the behavior in a similar fashion: a!localVariables(
 local!elements: a!forEach(
 {6, 5, 7},
 enumerate(fv!item) + 1
 ),
 
 intersection(
 a!flatten(local!elements[1]),
 a!flatten(local!elements[2]),
 a!flatten(local!elements[3])
 )
)

Mathieu Drouin · Answer

Pretty sure it's just a wrapper function that uses union()

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Mike Schmitt · Answer

Druva said: This works for me but the list of arrays are not fixed 
 The issue here is, for intersection() and similar functions, the different arrays need to be passed in as separate parameters for the function to work (and not, like, a list of lists). Thus I'm not sure how possible it is to call intersection on a list of lists of indefinite length, unless via some very advanced expression function trickery i'm not thinking of at the moment.