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Re: XML Schema - How to define any element of type Foe

  • From: Hans-Juergen Rennau <hrennau@yahoo.de>
  • To: "cmarchand@oxiane.com" <cmarchand@oxiane.com>, "xml-dev@l..." <xml-dev@l...>
  • Date: Thu, 9 Mar 2017 11:55:12 +0000 (UTC)

Re:  XML Schema - How to define any element of type Foe
Christophe, if you have to stick to XSD 1.0, a weak and very limited workaround for your problem could be the use of a substitution group:

(a) Introduce a global element declaration with an arbitrary name (e.g. "FoeElem") and type "Foe"
(b) For each element which you would like to allow at the critical place in type "parent": create a global element declaration with type "Foe" (or a type derived from it) and make it a member of substitutionGroup "FoeElem" (substitutionGroup="FoeElem")
(c) In your type "parent" type replace <xs:any type="Foe"/> by <xs:element ref="FoeElem"/>

It's a far cry from a wildcard constrained to have type Foe, as you have to forsee all allowed element names at XSD design time; but you can at least create a set of elements allowed to appear at this place. Note the difference to a choice: a choice is closed, but using a substitution group, additional elements can be allowed later without changing type "parent": by just adding more element declarations declaring this substitution group. Later addition might also be achieved by including or importing further XSDs providing the additional element declarations.


"cmarchand@oxiane.com" <cmarchand@oxiane.com> schrieb am 10:08 Donnerstag, 9.März 2017:


I have a complexType definition, Foe.

In another complexType, I want to allow any child element, but require
that each child element is valid against Foe definition.

Something like :

<xs:complexType name="Foe">
  <xs:attribute name="start" type="xs:positiveInteger" use="required"/>
  <xs:attribute name="end" type="xs:positiveInteger" use="required"/>
  <xs:assert test="@end ge @start">

<xs:complexType name="parent">
    <xs:any type="Foe"/>

How can I do this ?



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