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Re: [xml-dev] XML Schema - How to define any element of type Foe

From: Hans-Juergen Rennau <>
To: "" <>, "" <>
Date: 2017-03-09T11:55:12Z
ID: <>
Thread: Re: [xml-dev] 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")

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.

Cheers,

Hans

"[email protected]" <[email protected]> schrieb am 10:08 Donnerstag, 9.März 2017:

Hello,

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">