如何创建通用 XSD 类型

问题描述

我有一个 java 应用程序，我可以将一个 XSD 类型映射到另一个具有相同类型的类型.现在我需要有一个 anyType xsd，我可以将任何类型映射到该 xsd.就像我们在 java 中有 Object 类型一样，是否可以像在 XSD 中一样创建.

I have a java application where i can map a XSD type to another with same type. Now i have requirement to have one anyType xsd to which i can map any type. Like as we have Object type in java, is it possible to create like in XSD.

在复杂类型级别是可能的.

At complex type level is it possible.

推荐答案

是的，有可能.类型是 xsd:anyType.这是一个例子:

Yes, it's possible. The type is xsd:anyType. Here's an example:

<xsd:element name="anything" type="xsd:anyType"/>

(取自primer)

这里有一个更复杂的例子:

Here's a more complex example:

<xsd:complexType>
  <xsd:complexContent>
    <xsd:restriction base="xsd:anyType">
      <xsd:attribute name="currency" type="xsd:string"/>
      <xsd:attribute name="value"    type="xsd:decimal"/>
    </xsd:restriction>
  </xsd:complexContent>
</xsd:complexType>

(同样来自入门书——值得一看)

(From the primer as well - it's worth looking at at it)

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