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About simonnito

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  1. Actually I created another xsd files containing only the common global variables. Thus file3.xsd looks like below:<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified"> <!-- D├ęclarations des variables globales --> <xs:attribute name="tableDeCode" type="xs:string"/> <xs:element name="ctpays"> <xs:complexType> <xs:simpleContent> <xs:extension base="xs:string"> <xs:attribute ref="tableDeCode" use="required"/> </xs:extension> </xs:simpleContent> </xs:com
  2. Hi,I am building a xsd schema splitted in severals xsd schema. My problem is the following (with xmlspy):I must use the same element and the same attribute in different xsd schema, but I do not know how to declare this element and this attribute to make them as global variables for any includes xsd files.For instance :the file main.xsd:<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" id="NewSchema"> <xs:include schemaLocation="file1.xsd"/> <xs:include schemaLocation="file2.xsd"/> <xs:attribute name="tableDeCode" type="xs:string"/> <xs:element name="ctpays">
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