BizTalk 2009 Community ODBC Adapter  Generating Schemas from MySQL Stored Procedure
By : omi
Date : March 29 2020, 07:55 AM

Merge 2 input XML schemas into one CSV file  biztalk 2009
By : Ed Y.
Date : March 29 2020, 07:55 AM

Generating a row within a cross product given an integer
By : Twisted Candle
Date : March 29 2020, 07:55 AM
With these it helps Induction is your friend when looking for solutions to this kind of problem. For the easy case, it looks like code :
easy( a, i ) ≡ easyHelper( a, a.length, i )
easyHelper( a, n, i ) ≡ easyInduction( easyHelper, 0 )( a, n, i )
easyInduction( f, b )( a, 0, 0 ) ≡ []
easyInduction( f, b )( a, 0, i + 1 ) ≡ undefined
easyInduction( f, b )( a, n + 1, i ) ≡
let t = a[n + 1].length  b
in f( a, n, ⌊i / t⌋ ) ++ [ a[n + 1][i mod t] ]
hard( a, i ) ≡ hardHelper( a, a.length ).second( i )
hardHelper( a, 0 ) ≡ ( 1, { case 0 => [] } )
hardHelper( a, n + 1 ) ≡
let t = a[n + 1].length
and ( k, f ) = hardHelper( a, n )
and k' = k * ( t  1 )
and f'( i ) =
if ( i < k' )
then easyInduction( f, 1 )( a, n + 1, i )
else if ( k' <= i < k' + k )
then easyInduction( f, 1 )( a[ 0..n ] ++ [ [ null ] ], n + 1, i  k' )
else easyInduction( ( b, m, j ) => f( b, m, j + k ), 0 )( a, n + 1, i  k'  k )
in ( k', f' )

generating cross product for list of lists
By : gan
Date : March 29 2020, 07:55 AM
I wish this helpful for you What you want is a carthesian product. itertools has a function product that does exactly that: code :
import itertools
a = [[1, 2, 3], [4, 5], [3], [1]]
p = itertools.product(*a)
print list(p)
[(1, 4, 3, 1),
(1, 5, 3, 1),
(2, 4, 3, 1),
(2, 5, 3, 1),
(3, 4, 3, 1),
(3, 5, 3, 1)]

Edifact D16 XML schemas for use in BizTalk 2016
By : Cyber Army
Date : March 29 2020, 07:55 AM
will be helpful for those in need To be clear, what you are planning is not uncommon. This has been the solution for odd versions in both EDIFACT and X12 for ever.

