Re: How to assess the correctness of a Format1 --> Format2 mapping?

Two tools for this process:

1. Invisible XML: https://www.w3.org/community/ixml/
2. Delta.XML: https://www.deltaxml.com/

Thanks,
Dimitre

On Mon, Jan 23, 2023 at 11:24 AM Roger L Costello <[email protected]>
wrote:

> Hi Folks,
>
> There exists a proprietary data format called Format1.
>
> A smart group of people created a non-proprietary data format called
> Format2. It's used to store the same kinds of data as Format1. Of course,
> Format2 uses different fields and structures.
>
> I defined a mapping from Format1 to Format2.
>
> I want to know if my Format1 --> Format2 mapping is correct.
>
> Here's my idea on how to ascertain the correctness of the Format1 -->
> Format2 mapping:
>
> Use a technology to parse Format1 instances to XML and to parse Format2
> instances to XML. Compare the resulting XML's. If they are identical, then
> the mapping is correct.
>
> Data Format Description Language (DFDL) is a standard language for parsing
> data formats. Here's the idea using DFDL:
>
> -  Create a DFDL schema for Format1.
> -  Create a DFDL schema for Format2.
> -  Design the DFDL schemas to produce the same XML format.
> -  To ascertain the correctness of the Format1 --> Format2 mapping do
> this:
>         -  Use the Format1 DFDL schema to parse a Format1 document to
> XML.
>         -  Use the Format2 DFDL schema to parse a Format2 document to XML.
>         -  Compare the XML's:
>             -  If they are identical, then the mapping is correct.
>             -  If they are within a certain tolerance, then the mapping is
> correct.
>                E.g. of an acceptable tolerance: the Format2 XML omits an
>                optional element.
>             -  Otherwise the mapping is not correct.
>
> Am I thinking clearly about this problem (ascertain the correctness of the
> Format1 --> Format2 mapping)?
>
> Is mapping instances of the Format1 and Format2 data formats to a common
> XML format logical?
>
> What are the flaws in my thinking?
>
> /Roger
>
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